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Physics

Rotational Mechanics

Kinematics of Rotational Motion

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⚡ Quick Summary
Describes the motion of a rigid body rotating about a fixed axis, focusing on angular position, angular velocity, and angular acceleration, and their relationships, analogous to linear motion.
<ul><li>ω = dθ/dt</li><li>α = dω/dt = d²θ/dt²</li><li>If α is constant: <ul><li>ω = ω₀ + αt</li><li>θ = ω₀t + (1/2)αt²</li><li>ω² = ω₀² + 2αθ</li></ul></li><li>1 rev = 2π radian</li></ul>
  • Axis of Rotation: A straight line about which a rigid body rotates, with planes of circles perpendicular to this line.
  • Angular Position (θ): The angle through which a particle or the entire rigid body has rotated from its initial position. Measured in radians.
  • Average Angular Velocity (ωavg): The change in angular position (Δθ) divided by the time interval (Δt): ωavg = Δθ / Δt
  • Instantaneous Angular Velocity (ω): The rate of change of angular position with respect to time: ω = dθ/dt. It's a vector quantity with direction along the axis of rotation.
  • Angular Speed: The magnitude of the angular velocity.
  • Uniform Angular Velocity: When a body rotates through equal angles in equal time intervals, ω = constant, and θ = ωt.
  • Angular Acceleration (α): The rate of change of angular velocity with respect to time: α = dω/dt = d2θ/dt2
  • Positive/Negative Rotation: Direction must be defined as either clockwise or counter-clockwise. Angular displacement, velocity and acceleration can be either positive or negative.
  • Analogy to Linear Motion: There's a direct analogy between linear motion (position, velocity, acceleration) and rotational motion (angular position, angular velocity, angular acceleration).

Relation between Linear and Angular Motion

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When a rigid body rotates, points on the body move in circles. The relationships connect the linear motion (speed, acceleration) of a point on the body to the angular motion (angular speed, angular acceleration) of the body.
['s = rθ', 'v = rω', 'a = rα']
For a rigid body rotating about a fixed axis: * Consider a point P on the rigid body at a distance *r* from the axis of rotation. As the body rotates through an angle *θ*, the point moves a distance *s* along a circle, where *s = rθ*. * The linear speed *v* of the point is related to the angular speed *ω* by *v = rω*. * The tangential linear acceleration *a* of the point is related to the angular acceleration *α* by *a = rα*. * The angular velocity (*ω*) and angular acceleration (*α*) are the same for all particles in the rigid body. However, the linear speed (*v*) and tangential acceleration (*a*) are different for different particles, depending on their distance (*r*) from the axis of rotation. * Particles on the axis of rotation (*r* = 0) have zero linear speed and zero tangential acceleration.

Torque

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Torque is the rotational equivalent of force. It depends on the force applied and the distance from the axis of rotation. The direction of torque is along the axis of rotation. The total torque determines the angular acceleration of a body.
Γ = rFsinθ, Γ = F.(OS), Γ<sub>total</sub> = r<sub>1</sub> × F<sub>1</sub> + r<sub>2</sub> × F<sub>2</sub> + ..., Γ = Iα
  • Definition of Torque (Γ): Γ = rFsinθ, where r is the distance from the axis of rotation to the point where the force is applied, F is the magnitude of the force, and θ is the angle between the force vector and the vector pointing from the axis of rotation to the point of application of the force. Alternatively, Γ = F.(OS), where OS is the lever arm or moment arm, representing the length of the common perpendicular to the force and the axis of rotation.
  • Direction of Torque: The direction of the torque is along the axis of rotation. Conventionally, torque is taken as positive if it tries to rotate the body anticlockwise when viewed through the axis.
  • Calculating Total Torque: If multiple forces are acting on a body, the total torque is the sum of the individual torques: Γtotal = r1 × F1 + r2 × F2 + ...
  • Special Cases for Torque Calculation:
    • Case I: If the force is parallel to the axis of rotation, the torque is zero.
    • Case II: If the force intersects the axis of rotation, the torque is zero.
    • Case III: If the force and the axis of rotation are perpendicular but do not intersect, the torque is equal to the magnitude of the force multiplied by the length of the common perpendicular to the force and the axis.
    • Case IV: If the force and the axis are skew but not perpendicular, take the component of force perpendicular to the axis and calculate the torque as in Case III. The torque of the parallel part is zero.
  • Relationship between Torque and Angular Acceleration: The total torque acting on a body determines its angular acceleration (α). Analogous to linear motion where force is proportional to linear acceleration, torque is proportional to angular acceleration: Γ = Iα, where I is the moment of inertia.

Torque and Angular Acceleration

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The net external torque on a rigid body about a fixed axis is equal to the moment of inertia of the body about that axis times the angular acceleration.
['G_ext = I * alpha', 'I = Σ m_i * r_i^2']
  • A rigid body rotating about a fixed axis AB.
  • Consider a particle P of mass m rotating in a circle of radius r.
  • Radial acceleration of the particle = v^2/r = w^2r.
  • Radial force on it = mw^2r.
  • Tangential acceleration of the particle = dv/dt = r*dw/dt = r*alpha.
  • Tangential force on it = m*r*alpha.
  • Torque of mw^2r about AB is zero.
  • Torque of mra is mr^2a.
  • Total torque of all the forces acting on all the particles of the body is G_total = Σ m_i r_i^2 alpha = I*alpha
  • I = Σ m_i r_i^2
  • I is the moment of inertia of the body about the axis of rotation.
  • G_total = Σ (r_i x F_i) where F_i is the resultant force on the ith particle.
  • F_i = Σ F_ij + F_i_ext
  • G_ext = I*alpha
  • Moment of inertia I = Σ m_i r_i^2 depends on the choice of the axis.

Equilibrium of a Cyclist

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A cyclist bends at an angle to maintain balance during a turn. This angle depends on the speed of the cyclist and the radius of the turn.
tan(θ) = v^2 / (rg)
The cyclist bends at an angle θ with the vertical such that tan(θ) = v^2 / (rg), where v is the speed, r is the radius of the turn, and g is the acceleration due to gravity.

Angular Momentum of a Particle about a Point

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Angular momentum is a measure of the amount of rotational motion a particle has about a point. It depends on the particle's momentum and its position relative to that point.
['l = r × p', 'l = mvr']
Angular momentum (l) of a particle about a point O is defined as l = r × p, where p is the linear momentum (mv) and r is the position vector of the particle from the point O. The magnitude is given by l = mvr, where r is the perpendicular distance of the line of motion from O.

Angular Momentum of a System of Particles

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The total angular momentum of a system of particles is the sum of the angular momenta of each individual particle.
L = Σ (ri × pi)
L = Σ li = Σ (ri × pi), where the sum is taken over all particles in the system.

Angular Momentum of a Particle in Circular Motion

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For a particle moving in a circle, the angular momentum about the center of the circle is simply the product of its mass, speed, and the radius of the circle.
l = mvr
If a particle is going in a circle of radius r with speed v, its angular momentum about the axis of the circle is l = mvr.

Angular Momentum of a Rigid Body Rotating About an Axis

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The angular momentum of a rigid body rotating about an axis is proportional to its moment of inertia and angular velocity.
L = Iω
For a rigid body rotating about an axis AB with angular velocity ω, the angular momentum about AB is L = Iω, where I is the moment of inertia of the body about AB.

Conservation of Angular Momentum

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The rate of change of angular momentum is equal to the total external torque acting on the system.
dL/dt = τ_total
dL/dt = Σ (ri × Fi) = τ_total, where τ_total is the total external torque.

Relationship between Torque and Angular Momentum

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The rate of change of angular momentum of a system is equal to the external torque acting on it. If the net external torque is zero, the angular momentum remains constant (conservation of angular momentum).
dL/dt = G_ext; If G_ext = 0, L = constant

Torque and Angular Momentum: The change in angular momentum of a system is equal to the integral of the external torque over time. Mathematically, this is represented as:
\[ \int_{t_1}^{t_2} \vec{G}_{ext} dt = \vec{L}_2 - \vec{L}_1 \]
where G_ext is the total external torque acting on the system, and L_1 and L_2 are the initial and final angular momenta, respectively.

Conservation of Angular Momentum: If the total external torque on a system is zero, its angular momentum remains constant. This is the principle of conservation of angular momentum.

Rigid Body Rotating About a Fixed Axis: For a rigid body rotating about a fixed axis, the relationship simplifies to:
\[ \vec{G}_{ext} = I \vec{\alpha} = I \frac{d\vec{\omega}}{dt} = \frac{d\vec{L}}{dt} \]
where *I* is the moment of inertia, *α* is the angular acceleration, and *ω* is the angular velocity.

Angular Impulse

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Angular impulse is the integral of torque over a time interval, and it equals the change in angular momentum.
J = integral(G dt) = L2 - L1

Angular Impulse: The angular impulse of a torque in a given time interval is defined as the integral of the torque over that time interval:
\[ \vec{J} = \int_{t_1}^{t_2} \vec{G} dt \]
This angular impulse is equal to the change in angular momentum:
\[ \vec{J} = \vec{L}_2 - \vec{L}_1 \]

Kinetic Energy of a Rigid Body Rotating About a Given Axis

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The kinetic energy of a rotating rigid body is the sum of the kinetic energies of all its particles, which can be expressed as 1/2 * I * ω^2, where I is the moment of inertia and ω is the angular speed.
K = 1/2 * I * omega^2

Kinetic Energy of a Rotating Rigid Body: The kinetic energy (K) of a rigid body rotating about an axis is given by:
\[ K = \frac{1}{2} I \omega^2 \]
where *I* is the moment of inertia of the body about the axis of rotation and *ω* is the angular speed.

The kinetic energy is derived from summing the kinetic energies of all the particles in the body:
\[ K = \sum \frac{1}{2} m_i v_i^2 = \frac{1}{2} \sum m_i (\omega r_i)^2 = \frac{1}{2} I \omega^2 \]

Power Delivered and Work Done by a Torque

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The power delivered by a torque is the rate at which it does work, which equals the rate of increase of kinetic energy. Work done by a torque is the integral of the torque over the angular displacement.
['P = dW/dt = dK/dt', 'P = Iωα = ωτ', 'dW = τ dθ', 'W = ∫ τ dθ (from θ1 to θ2)']
The torque produces angular acceleration and the kinetic energy increases. The rate of increase of the kinetic energy equals the rate of doing work on it, i.e., the power delivered by the torque.

Calculation of Moment of Inertia

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Moment of inertia is calculated by summing up the product of each particle's mass and the square of its distance from the axis of rotation. For continuous bodies, this becomes an integral.
['I = Σ mᵢrᵢ²', 'I = ∫ r² dm']
The moment of inertia of a system about a given line is defined as I = Σ mᵢrᵢ², where mᵢ is the mass of the ith particle and rᵢ is its perpendicular distance from the given line. For continuous bodies, the moment of inertia is calculated by integration: I = ∫ r² dm, where dm is a small element of mass and r is its perpendicular distance from the axis of rotation.

Moment of Inertia

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The moment of inertia of a body about an axis is a measure of its resistance to angular acceleration about that axis. It depends on the mass of the body and the distribution of mass relative to the axis of rotation.
['Moment of inertia of a uniform rod about a perpendicular bisector: I = Ml^2 / 12', 'Moment of inertia of a rectangular plate about a line parallel to an edge and passing through the centre: I = Ml^2 / 12 or I = Mb^2 / 12', 'Moment of inertia of a circular ring about its axis: I = MR^2', 'Mass of an element of rod: dm = (M/l)dx', 'Mass per unit length of rod = M/l', 'Mass per unit area of plate = M / (pi * R^2)']
(A) Uniform rod about a perpendicular bisector: Consider a uniform rod of mass M and length l. To calculate the moment of inertia about the bisector AB, take the origin at the middle point O of the rod. Consider an element of the rod between a distance x and x + dx from the origin. The mass per unit length is M/l, so the mass of the element is (M/l)dx. The perpendicular distance of the element from AB is x. The moment of inertia of this element about AB is dI = (M/l)dx * x^2. Integrating from -l/2 to l/2 gives the moment of inertia of the entire rod: I = integral(x^2 dm) = integral((M/l)x^2 dx) from -l/2 to l/2 = Ml^2 / 12. (B) Moment of inertia of a rectangular plate about a line parallel to an edge and passing through the centre: Consider a rectangular plate with sides l and b. Draw a line parallel to AB at a distance x from it and another at a distance x + dx. The strip enclosed between the two lines is a small element. The mass per unit area is M/(bl). The mass of the strip is (M/l)dx. The perpendicular distance of the strip from AB is x. The moment of inertia of the strip about AB is dI = (M/l)dx * x^2. The moment of inertia of the plate is I = integral(x^2 dm) = integral((M/l)x^2 dx) from -l/2 to l/2 = Ml^2 / 12. To find the moment of inertia about a line parallel to the other edge and passing through the centre, replace l by b, so I = Mb^2 / 12. (C) Moment of inertia of a circular ring about its axis: Suppose the radius of the ring is R and its mass is M. All elements of the ring are at the same perpendicular distance R from the axis. Therefore, the moment of inertia is I = integral(r^2 dm) = integral(R^2 dm) = R^2 * integral(dm) = MR^2. (D) Moment of inertia of a uniform circular plate about its axis: Let the mass of the plate be M and its radius R. Draw two concentric circles of radii x and x + dx, both centered at O and consider the area of the plate in between the two circles. This part of the plate is a circular ring of radius x. As the periphery of the ring is 2πx and its width is dx, the area is 2πxdx. The area of the plate is πR^2. The mass per unit area is M/(πR^2). The mass of the ring is dm = (M/(πR^2)) * 2πxdx = (2M/R^2) * xdx.

Moment of Inertia of a Circular Ring Element

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The moment of inertia of a small circular ring element within a solid cylinder about the axis of the cylinder.
dI = (2M/R^2) * x^3 dx
For a circular ring element of radius x and width dx within a solid cylinder of radius R and mass M, the moment of inertia dI about the axis is given by: dI = (2M/R^2) * x^3 dx

Moment of Inertia of a Solid Cylinder

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The moment of inertia of a solid cylinder about its axis.
I = (MR^2)/2
The moment of inertia of a solid cylinder of mass M and radius R about its axis is obtained by integrating the moment of inertia of elementary rings from 0 to R: I = ∫dI = (MR^2)/2

Moment of Inertia of a Hollow Cylinder

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The moment of inertia of a hollow cylinder about its axis.
I = MR^2
For a hollow cylinder of mass M and radius R, all elements are at the same distance R from the axis. Thus, the moment of inertia I is: I = MR^2

Moment of Inertia of a Uniform Hollow Sphere

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Moment of inertia calculation for a hollow sphere about its diameter, by integrating the moment of inertia of elemental rings.
dI = (M/2) * R^2 * sin³(θ) dθ
Consider a hollow sphere of mass M and radius R. An elemental ring at angle θ with width Rdθ has a mass dM = (M/2)sinθ dθ. The moment of inertia of this ring about the axis OX is dI = (M/2)R^2 sin³θ dθ. Integrating from 0 to π gives the total moment of inertia. Note that the radius of the ring is Rsinθ.

Moment of Inertia of a Uniform Solid Sphere about a Diameter

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The moment of inertia of a uniform solid sphere of mass M and radius R about a diameter is calculated using integration by considering the sphere as a collection of thin hollow spheres.
I = (2/5)MR²
To calculate the moment of inertia (I) of a uniform solid sphere of mass M and radius R about a diameter: 1. **Consider a thin hollow sphere:** Imagine the solid sphere as being made up of many thin, concentric hollow spheres. Let's take one such hollow sphere with radius x and thickness dx. 2. **Mass per unit volume:** The mass per unit volume (density) of the solid sphere is given by (3M) / (4πR³). 3. **Mass of the hollow sphere:** The volume of the thin hollow sphere is its surface area (4πx²) times its thickness (dx), i.e., 4πx²dx. Therefore, the mass (dm) of this hollow sphere is its volume multiplied by the density: dm = (3M / R³) * x² dx. 4. **Moment of inertia of the hollow sphere:** The moment of inertia (dI) of this hollow sphere about the diameter is given by (2/3) * dm * x² which simplifies to dI = (2M / R³) * x⁴ dx. 5. **Integrate to find the total moment of inertia:** Integrate dI with respect to x from 0 to R to cover the entire solid sphere: I = ∫dI = ∫(2M / R³) * x⁴ dx from 0 to R. This results in I = (2/5)MR².

Moment of Inertia Calculation Using Coordinates

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The moment of inertia of a body can be calculated by summing the product of each particle's mass and the square of its distance from the axis of rotation. For a sphere, the moment of inertia about the X, Y, and Z axes can be related, and since all diameters are equivalent, they are equal.
1. **Coordinate System:** Consider a particle P within the object, with coordinates (xᵢ, yᵢ, zᵢ) with respect to the center O. The distance from the particle to the X-axis is given by √(yᵢ² + zᵢ²). 2. **Moment of Inertia for a particle:** The moment of inertia of particle P about the X-axis (Iₓ) is mᵢ(yᵢ² + zᵢ²), where mᵢ is the mass of the particle. 3. **Total Moment of Inertia:** The total moment of inertia about the X-axis is found by summing the moment of inertia of all particles: Iₓ = Σ mᵢ(yᵢ² + zᵢ²). Similarly, Iᵧ = Σ mᵢ(zᵢ² + xᵢ²) and Iz = Σ mᵢ(xᵢ² + yᵢ²). 4. **Relationship Between Iₓ, Iᵧ and Iz:** Adding the three equations gives Iₓ + Iᵧ + Iz = 2Σ mᵢ(xᵢ² + yᵢ² + zᵢ²) = 2Σ mᵢR² = 2MR², where R is the radius (assuming all particles are at the same distance from the origin which is true for surface particles). 5. **Uniform Sphere:** For a uniform sphere, all diameters are equivalent. Therefore, Iₓ = Iᵧ = Iz. Thus, 3Iₓ = 2MR², which gives Iₓ = (2/3)MR² for a hollow sphere.

Parallel Axes Theorem

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The moment of inertia of a body about any axis is equal to the moment of inertia about a parallel axis through the center of mass plus the product of the mass of the body and the square of the distance between the two axes.
I = I₀ + Md²
Let *I* be the moment of inertia of the body about an axis AB, and *I₀* be the moment of inertia about a parallel axis CZ through the center of mass C. Let M be the mass of the body and d be the perpendicular distance between AB and CZ. Then, *I = I₀ + Md²*

Theorem of Perpendicular Axes

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For a plane body, the moment of inertia about an axis perpendicular to the plane is equal to the sum of the moments of inertia about two mutually perpendicular axes in the plane.
I<sub>z</sub> = I<sub>x</sub> + I<sub>y</sub>
Let X and Y-axes be chosen in the plane of the body and Z-axis perpendicular to this plane, three axes being mutually perpendicular. Then the theorem states that Iz = Ix + Iy.

Perpendicular Axes Theorem

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Relates moment of inertia about three mutually perpendicular axes. If two axes are in the plane of the lamina and one is perpendicular to it, then I_z = I_x + I_y
I_z = I_x + I_y
The perpendicular axes theorem states that I_z = I_x + I_y, where the z-axis is perpendicular to the plane containing the x and y axes.

Parallel Axes Theorem

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Used to find the moment of inertia about an axis parallel to an axis passing through the center of mass. I = I_0 + MR^2, where I_0 is moment of inertia about axis through center of mass, M is mass, R is the distance between the two axes.
I = I_0 + MR^2
Parallel axes theorem allows calculation of moment of inertia about any axis, given the moment of inertia about a parallel axis through the center of mass: I = I_0 + MR^2

Radius of Gyration

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The distance from the axis of rotation at which the entire mass of the body can be assumed to be concentrated to give the same moment of inertia.
I = Mk^2
Radius of gyration (k) is defined by I = Mk^2, where I is moment of inertia and M is mass.

Combined Rotation and Translation

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Motion where a rigid body rotates and translates simultaneously. Analyzing such motion often involves considering a non-inertial frame A, where the motion is pure rotation. If the axis of rotation passes through the center of mass, then G_cm = I_cm * alpha.
['a_cm = F_ext / M', 'alpha = G_ext / I_cm']
For combined rotation and translation, especially from a non-inertial frame A, if the axis of rotation passes through the center of mass, G_cm = I_cm * alpha. The general approach involves finding the acceleration of the center of mass (a_cm = F_ext/M) and the angular acceleration (alpha = G_ext/I_cm).

Rolling

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Combined rotation and translation, as in a wheel rolling on a surface.
[]
Rolling involves both rotation and translation.

Rolling Motion

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Rolling motion is a combination of rotation and translation. Pure rolling occurs when the point of contact with the surface is instantaneously at rest. Slipping occurs when the relationship between linear velocity and angular velocity deviates from the pure rolling condition (v = Rω).
['v_cm = Rω (Pure Rolling)', 'v_top = 2Rω = 2v_cm (Pure Rolling)']
  • Pure Rolling: The wheel rotates and moves forward such that the displacement of the center of the wheel (Δx) is related to the angle rotated by a spoke (Δθ) by the equation Δx = RΔθ. This leads to the relation vcm = Rω, where vcm is the linear speed of the center of mass and ω is the angular velocity of the wheel. In pure rolling, the velocity of the contact point is zero. The velocity of the topmost point is vtop = 2Rω = 2vcm.
  • Rolling with Forward Slipping: Occurs when the wheel moves through a distance greater than 2πR in one full rotation. In this case, vcm > Rω. The particles in contact have a velocity in the forward direction. An extreme case is when the wheel only translates with linear velocity v and doesn't rotate at all (ω = 0).
  • Rolling with Backward Slipping: Occurs when the wheel moves a distance shorter than 2πR while making one rotation. In this case, vcm < Rω. The particles in contact rub the road in the backward direction.

Kinetic Energy of a Body in Combined Rotation and Translation

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When a body is both rotating and translating, its total kinetic energy is the sum of the kinetic energy due to rotation about its center of mass and the kinetic energy due to the translation of its center of mass.
K = 1/2 * I * ω<sup>2</sup> + 1/2 * M * v<sub>0</sub><sup>2</sup>
  • Consider a body undergoing combined translational and rotational motion. In the center of mass frame, the body is in pure rotation with angular velocity ω. The center of mass is moving with velocity v0 in the lab frame.
  • The velocity of a particle of mass mi is given by vi = vi,cm + v0, where vi,cm is the velocity with respect to the center of mass frame and v0 is the velocity of the center of mass with respect to the lab frame.
  • The kinetic energy of the particle is 1/2 * mi * vi2 = 1/2 * mi * (vi,cm + v0) ⋅ (vi,cm + v0) = 1/2 * mi * vi,cm2 + 1/2 * mi * v02 + mi * (vi,cm ⋅ v0).
  • Summing over all particles, the total kinetic energy is K = Σ (1/2 * mi * vi2) = Σ (1/2 * mi * vi,cm2) + 1/2 * Σ (mi * v02) + (Σ mi * vi,cm) ⋅ v0.
  • Since Σ (1/2 * mi * vi,cm2) represents the kinetic energy in the center of mass frame (pure rotation), it equals 1/2 * I * ω2, where I is the moment of inertia about the center of mass. Also, Σ mi = M (total mass), and Σ mi * vi,cm = 0 (velocity of center of mass in center of mass frame).
  • Therefore, the total kinetic energy is K = 1/2 * I * ω2 + 1/2 * M * v02.

Pure Rolling

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In pure rolling, the point of contact between the rolling body and the surface is instantaneously at rest. This implies a specific relationship between the linear velocity of the center of mass and the angular velocity.
v = Rω; K = 1/2 * (I + MR<sup>2</sup>)ω<sup>2</sup> = 1/2 * I<sub>contact</sub> * ω<sup>2</sup>
  • Pure Rolling Condition: v0 = Rω, where v0 is the velocity of the center of mass, R is the radius of the rolling object, and ω is the angular velocity.
  • Kinetic Energy in Pure Rolling: K = 1/2 * I * ω2 + 1/2 * M * v02. Substituting v0 = Rω, we get K = 1/2 * I * ω2 + 1/2 * M * (Rω)2 = 1/2 * (I + MR2)ω2. Using the parallel axis theorem, I + MR2 = Icontact, where Icontact is the moment of inertia about the point of contact. Thus, K = 1/2 * Icontact * ω2.
  • Interpretation of Rolling: At any instant, a rolling body can be considered to be in pure rotation about an axis through the point of contact.

Rolling with Slipping

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Rolling with slipping occurs when the condition v = Rω is not met. Forward slipping happens when v > Rω, and backward slipping occurs when v < Rω.
v > ωr (forward slipping); v < ωr (backward slipping)
  • Forward Slipping: v > ωr. The velocity of the point of contact with respect to the road is vcontact, road = vcar, road - ωr > 0.
  • Backward Slipping: v < ωr. The velocity of the point of contact with respect to the road is vcontact, road = vcar, road - ωr < 0.

Angular Momentum of a Body in Combined Rotation and Translation

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The angular momentum of a body undergoing both rotation and translation is calculated with respect to a fixed point in the lab frame, considering both the motion of the center of mass and the rotation about the center of mass.
L = Σ m<sub>i</sub> (r<sub>i</sub> x v<sub>i</sub>)
  • Let O be a fixed point in the lab frame, taken as the origin.
  • Angular momentum of the body about O is given by L = Σ mi (ri x vi) where the summation is over all particles in the body.
  • Using ri = ri,cm + r0 and vi = vi,cm + v0, where r0 is the position vector of the center of mass and v0 is the velocity of the center of mass: L = Σ mi (ri,cm + r0) x (vi,cm + v0).

Angular Momentum

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Angular momentum is calculated considering the motion of the center of mass and the rotation around it.
['L = ∑ m<sub>i</sub> (r<sub>i</sub> × v<sub>i</sub>)', 'L = L<sub>cm</sub> + M (r<sub>0</sub> × v<sub>0</sub>)']
The angular momentum of a system of particles about a point O in an inertial frame is given by L = ∑ mi (ri × vi) where ri is the position vector of the ith particle with respect to O and vi is its velocity. This can be split into angular momentum about the center of mass (Lcm) and the angular momentum of the center of mass itself. L = Lcm + M (r0 × v0) where r0 is the position vector of the center of mass and v0 is its velocity.

Rolling Sphere Slowing Down

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A rolling sphere slows down due to deformation at the contact point, which shifts the normal force and creates a torque.
[]
When a sphere rolls on a horizontal surface, it slows down due to friction. The sphere and the surface deform near the point of contact, creating an area of contact rather than a single point. The normal force is shifted slightly forward of the center, resulting in a torque that opposes the rotation, causing angular deceleration.

Constant Angular Acceleration Equations

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Relates angular displacement with initial angular velocity, angular acceleration and time.
['q = w<sub>0</sub>t + 1/2 a t<sup>2</sup>', 'w = w<sub>0</sub> + a t']
For constant angular acceleration the following equations may be used: q = w0t + 1/2 a t2 and w = w0 + a t

Torque

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Torque is product of Moment of Inertia and Angular Acceleration.
['G = I * a']
G = I * a, relates Torque to Moment of Inertia and angular acceleration.

Torque and Angular Acceleration

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Relates applied torque to angular acceleration.
G = Iα
Torque (G) is related to angular acceleration (α) by the equation G = Iα, where I is the moment of inertia.

Kinematics of Rotational Motion

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Describes the motion of rotation with constant angular acceleration.
ω = ω₀ + αt
Relates final angular velocity (ω) to initial angular velocity (ω₀), angular acceleration (α), and time (t) by the equation ω = ω₀ + αt.

Linear and Angular Acceleration Relation

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Connects linear and angular acceleration when there is no slipping.
a = αR
The tangential acceleration (a) of a point on a rotating object is related to the angular acceleration (α) by the equation a = αR, where R is the radius.

Seesaw Balance

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Balancing torques for equilibrium on a seesaw.
Στ = 0
For rotational equilibrium, the sum of torques must be zero.

Equilibrium of a Rigid Body

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For a rigid body to be in equilibrium, the net force and the net torque acting on it must be zero. This ensures both translational and rotational equilibrium.
ΣF = 0, Στ = 0
Conditions for equilibrium: 1. **Translational Equilibrium:** The vector sum of all the forces acting on the body must be zero. Mathematically, this is expressed as ΣF = 0. 2. **Rotational Equilibrium:** The vector sum of all the torques acting on the body about any point must be zero. Mathematically, this is expressed as Στ = 0.

Angular Momentum of a System of Particles

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The angular momentum of a system is the sum of the angular momenta of its individual particles.
L = mvr
For a particle of mass *m* moving with velocity *v* at a distance *r* from the axis of rotation, the angular momentum *L* is given by L = mvr. For a system, the total angular momentum L is the sum of the angular momenta of individual particles.

Angular Momentum of a Particle

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Angular momentum of a particle is the product of its moment of inertia and angular velocity. It is a vector quantity and its direction is perpendicular to the plane of rotation.
L = r x p = m (r x v)
The angular momentum of a particle about a point A is given by L = r x p = m (r x v), where r is the position vector of the particle with respect to point A, p is the linear momentum of the particle, m is the mass, and v is the velocity of the particle. The direction of angular momentum is perpendicular to both r and v, following the right-hand rule.

Kinetic Energy of a Rotating Disc

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A rotating disc possesses kinetic energy due to its rotation, which depends on its moment of inertia and angular speed.
K = (1/2) * I * w^2, I = (1/4) * M * r^2 (for disc rotating about diameter)
The kinetic energy of a rotating disc is given by K = (1/2) * I * w^2, where I is the moment of inertia of the disc about the axis of rotation and w is the angular speed. For a uniform circular disc rotating about one of its diameters, I = (1/4) * M * r^2, where M is the mass of the disc and r is its radius.

Angular Momentum of a Rotating Disc

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A rotating disc possesses angular momentum which depends on its moment of inertia and angular speed.
L = I * w
The angular momentum of a rotating disc is given by L = I * w, where I is the moment of inertia of the disc about the axis of rotation and w is the angular speed.

Rotational Motion with Constant Torque

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When a constant torque is applied to a rotating object, it experiences angular acceleration/deceleration. The work done by the torque is related to the angular displacement.
w = w0 - αt, τ = I * α, θ = w0*t - (1/2) * α * t^2, W = θ * τ
The angular acceleration (α) is related to the initial angular velocity (w0), final angular velocity (w), and time (t) by w = w0 - αt. The torque (τ) is related to the moment of inertia (I) and angular acceleration by τ = I * α. The angular displacement (θ) is given by θ = w0*t - (1/2) * α * t^2. The work done by the torque is W = θ * τ.

Angular Momentum and Kinetic Energy in Rotational Motion

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Deals with the conservation of angular momentum and kinetic energy when a system involves both linear and rotational motion. Also it includes formula for moment of inertia.
['L = Iω (Angular Momentum)', 'K.E. = (1/2)Iω^2 (Kinetic Energy of Rotation)', 'T = mv^2/r (Tension in a string for circular motion)', 'I = ml^2/3 (Moment of inertia of a rod rotating about one end)']
  • Angular Momentum Conservation: In the absence of external torque, the angular momentum of a system remains constant. This principle is applied when a mass moves in a circular path with a changing radius.
  • Kinetic Energy Calculation: The kinetic energy of a rotating object is given by (1/2) * I * ω^2, where I is the moment of inertia and ω is the angular velocity. For a combination of linear and rotational motion, the total kinetic energy is the sum of the translational and rotational kinetic energies.
  • Moment of Inertia: The moment of inertia (I) depends on the mass distribution and the axis of rotation. For a uniform rod rotating about one end, I = (1/3) * m * l^2, where m is the mass and l is the length of the rod.

Moment of Inertia

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Moment of inertia is a measure of an object's resistance to changes in its rotation. It depends on the mass and the distribution of mass relative to the axis of rotation.
['I = Σ m_i r_i^2 (for discrete particles)', 'I = ∫ r^2 dm (for continuous body)', 'I = I_cm + Md^2 (Parallel Axes Theorem)', 'I_z = I_x + I_y (Perpendicular Axes Theorem)']
The moment of inertia (I) of a system of particles about a given axis is the sum of the products of the masses of the particles and the squares of their distances from the axis. For a continuous body, the moment of inertia is calculated by integrating over the entire body. Parallel Axes Theorem: The moment of inertia of a body about any axis is equal to the moment of inertia about a parallel axis through the center of mass plus the product of the mass of the body and the square of the distance between the two axes. I = Icm + md^2 Perpendicular Axes Theorem: For a planar lamina, the sum of the moments of inertia about two perpendicular axes in the plane of the lamina is equal to the moment of inertia about an axis perpendicular to the plane and passing through the intersection of the first two axes. I_z = I_x + I_y

Conservation of Momentum in Collisions

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In a collision, if there is no external force acting on the system, the total linear momentum of the system remains constant. Similarly, if there is no external torque acting on the system, the total angular momentum of the system about any axis remains constant.
['m1v1 + m2v2 = constant (Linear Momentum Conservation)', 'L_i = L_f (Angular Momentum Conservation)']
Linear Momentum Conservation: In the absence of external forces, the total linear momentum of a system remains constant. If m1v1 + m2v2 = M1V1 + m2V2 Angular Momentum Conservation: In the absence of external torques, the total angular momentum of a system remains constant. Li = Lf

Angular Momentum of a System

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The angular momentum of a system of particles about a point is the sum of the angular momenta of individual particles about that point.
['L = Iω', 'L = mv(a/4)']
If A is the centre of the rod, and AB is the line perpendicular to the plane, and the system consists of 'the rod plus the particle', and the rod is initially at rest, then:
  • The angular momentum of the particle about AB is L = mv(a/4).
  • After the collision, the particle comes to rest.
  • The angular momentum of the rod about A is L = Iω.
  • L = Lcm + r x V = Lcm, where L is the total angular momentum and Lcm is the angular momentum about the center of mass.

Rolling Motion

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When a body rolls without slipping, the point of contact with the surface is instantaneously at rest. The linear speed of the center of mass is related to the angular speed by v = ωr.
['v = ωr']
  • When a wheel rolls on the road, its angular speed ω about the center and the linear speed v of the center are related as v = ωr.

Rolling Without Slipping on Inclined Plane

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When a cylinder or sphere rolls down an inclined plane without slipping, its motion is governed by both linear and rotational equations. Friction plays a crucial role in enabling pure rolling.
['K = (1/2)Iω^2 + (1/2)mv^2', 'v = sqrt((4/3)glsinθ)', 'mg sinθ - f = ma', 'fr = Iα', 'a = (5/7)g sinθ', 'f = (2/7)mg sinθ', 'μ > (2/7)tanθ']
  • For a cylinder rolling without slipping:
    • Kinetic energy at the bottom: K = (3/4)mv^2
    • v = sqrt((4/3)glsinθ)
  • For a sphere rolling without slipping on an inclined plane:
    • Linear motion equation: mg sinθ - f = ma
    • Rotational motion equation: fr = (2/5)mr^2 (a/r) => f = (2/5)ma
    • Linear acceleration: a = (5/7)g sinθ
    • Friction force: f = (2/7)mg sinθ
    • Condition for pure rolling: μ > (2/7)tanθ

Rolling Without Slipping

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When an object rolls without slipping, the point of contact with the surface is instantaneously at rest. This implies a relationship between the linear and angular speeds.
v = ωr (where v is the linear speed, ω is the angular speed, and r is the radius)
When slipping ceases, the linear speeds of the points of contact of the two cylinders will be equal. If '1 and '2 be the respective angular speeds, we have '1 r1 = '2 r2. The change in the angular speed is brought about by the frictional force which acts as long as the slipping exists.

Torque and Angular Acceleration

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Torque causes angular acceleration. The relationship is analogous to Newton's second law for linear motion (F=ma).
τ = Iα (where τ is torque, I is the moment of inertia, and α is the angular acceleration)
The torque on the first cylinder is fr1 and that on the second is fr2. Assuming ω1r1 > ω2r2, the corresponding angular impulses are -fr1t and fr2t.

Relationship between Linear and Angular Acceleration during Rolling

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In pure rolling, the linear acceleration of the center of mass is related to the angular acceleration about the center.
a = rα (where a is linear acceleration, r is the radius, and α is angular acceleration)
The angular acceleration about the centre of the sphere is α = a/r, as there is no slipping.

Rolling Motion and Conservation Laws

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When a sphere on a rough surface is subjected to a force (like from a collision), friction acts to oppose its motion. This friction causes both linear deceleration and angular acceleration. Pure rolling starts when the linear velocity (v) equals the product of the radius (r) and angular velocity (ω), i.e., v = rω. Conservation of linear and angular momentum can be applied to analyze the motion of a system undergoing collisions and rolling.
['v(t) = v₀ - (f/M)t', 'Γ = fr', 'I = (2/5)MR² (for a sphere)', 'α = 5f / (2MR)', 'ω(t) = ω₀ + αt', 'v(t) = rω(t) (condition for pure rolling)', 'L = Iω + Mrv (Angular Momentum)']
  • Friction's Role: On a rough surface, friction (f) opposes motion, acting backward on the sphere. If the friction is kinetic, its magnitude is μN = μMg, where μ is the coefficient of kinetic friction, M is the mass of the sphere, and g is the acceleration due to gravity. This friction causes deceleration a = f/M.
  • Velocity Changes: The linear velocity as a function of time is v(t) = v₀ - (f/M)t, where v₀ is the initial velocity.
  • Torque and Angular Acceleration: Friction creates a torque Γ = fr about the center of the sphere. This torque causes angular acceleration α = Γ/I, where I is the moment of inertia. For a sphere, I = (2/5)MR². Therefore, α = f / ((2/5)MR) = 5f / (2MR).
  • Angular Velocity Changes: The angular velocity as a function of time is ω(t) = ω₀ + αt. If the sphere starts from rest (ω₀ = 0), then ω(t) = (5f / 2Mr) * t.
  • Pure Rolling Condition: Pure rolling occurs when v(t) = rω(t).
  • Alternative Approach - Conservation of Angular Momentum: Choosing the initial point of contact (A) as the reference, the torque due to friction is zero. If normal force and weight balance each other, the net torque about A is zero, implying angular momentum is conserved about A.

Torque and Equilibrium

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This section discusses concepts of translational and rotational motion, torque, angular momentum, and equilibrium. It explores scenarios involving rotating bodies, pendulums, and the effects of force and mass distribution on rotational behavior.
τ = r x F (Torque), a = αr (Linear and Angular Acceleration), v = ωr (Linear and Angular Velocity)
  • Pure Translation vs. Pure Rotation: An object can undergo pure translation (movement without rotation) or pure rotation (rotation about a fixed axis).
  • Simple Pendulum: A simple pendulum exhibits rotational motion about the fixed point of suspension.
  • Relationship between Linear and Angular Quantities: The relationships a = αr and v = ωr connect linear and angular acceleration and velocity, respectively. However, algebraic manipulation like a/α = v/ω might not always be valid.
  • Angular Velocity: The angular velocity of a rotating object can be considered about its center or another fixed point, although the values may differ.
  • Torque: The torque of a force F about a point is defined as τ = r x F, where r is the position vector from the point to the point of application of the force.
  • Equilibrium: For a body to be in equilibrium, the sum of all forces and the sum of all torques acting on the body must be zero. Translational equilibrium implies the net force is zero; rotational equilibrium implies the net torque is zero.
  • Center of Mass and Stability: The position of the center of mass influences stability. An object will topple if the vertical line through its center of mass falls outside its support base.
  • Moment of Inertia and Rotational Motion: The distribution of mass affects the moment of inertia, which determines the ease with which an object can be rotated.
  • Conservation of Angular Momentum: Changes in the distribution of mass in a rotating system can affect the angular speed.
  • Rolling Motion: The moment of inertia influences the speed of rolling objects down an inclined plane. Objects with lower moment of inertia will accelerate faster.

Relationship between linear and angular velocity

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The linear speed (v) of a particle in a rotating body is related to its distance (r) from the axis of rotation and the angular velocity (ω) of the body.
w = v × r
The relationship is expressed as `w = v (cross product) r`. This implies a dependency between angular velocity, linear velocity and the distance from the axis of rotation.

Torque

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Torque is a force acting on a particle that causes rotation.
G = r × F, and r . G = 0, F . G = 0
If F is a force acting on a particle with position vector r, then the torque (G) of this force about the origin is related to F and r.

Moment of Inertia

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Moment of inertia is a measure of an object's resistance to changes in its rotation.
Example: Moment of inertia of a semicircular wire = (2/5)Mr^2
The moment of inertia depends on the mass distribution of the object and the axis of rotation. For example, the moment of inertia of a uniform semicircular wire of mass M and radius r about a line perpendicular to the plane of the wire through the center is given.

Torque of Normal Force

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The torque of the normal force on a block depends on the position where the normal force acts and the angle. Without knowing the exact point of application, a general formula can be tricky. The torque is related to force and perpendicular distance from the axis of rotation.
τ = r x F, |τ| = r F sinθ
The torque (τ) of a force (F) about a point is given by τ = r x F, where r is the position vector from the point to the point where the force is applied. In the case of the normal force on a block, the magnitude of the torque depends on the magnitude of the normal force (N), the distance (r) from the axis of rotation to the line of action of the normal force, and the angle (θ) between r and N. If the normal force acts through the center, then the torque is zero.

Angular Momentum Conservation

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When no external torque acts on a system, its angular momentum remains constant. This means if the moment of inertia changes, the angular speed must also change to keep the angular momentum the same.
L = Iω, I₁ω₁ = I₂ω₂ (if net external torque is zero)
Angular momentum (L) is a measure of an object's rotation. For a rigid body, L = Iω, where I is the moment of inertia and ω is the angular velocity. The principle of conservation of angular momentum states that if the net external torque on a system is zero, the total angular momentum of the system remains constant. Therefore, if I changes, ω must change in such a way that Iω remains constant.

Rolling Motion

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When an object rolls without slipping, there is a relationship between its linear velocity and angular velocity. The speed of a point on the rim depends on both the linear speed of the center and the rotational speed.
v = rω
For rolling without slipping, v = rω, where v is the linear velocity of the center of mass, r is the radius, and ω is the angular velocity. The velocity of a point on the rim is the vector sum of the linear velocity of the center of mass and the tangential velocity due to rotation. At the point of contact with the ground, the velocity is zero (instantaneously). The velocity of the topmost point is 2v.

Rolling Down an Incline (Time)

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Objects with smaller moment of inertia roll down an incline faster, assuming rolling without slipping. When friction is not sufficient for pure rolling, the object with the least moment of inertia will reach the bottom first.
Acceleration a = (g sinθ) / (1 + I/MR²)
The time taken for an object to roll down an incline depends on its moment of inertia. For pure rolling, objects with smaller moment of inertia will accelerate more quickly and reach the bottom first. If friction is insufficient, the object that experiences the least amount of energy loss due to friction (i.e., the object with less rotational inertia relative to its mass and radius) will reach the bottom first. The order of moment of inertia for a solid sphere, hollow sphere, and disc (all having the same mass and radius) is: I_solid < I_disc < I_hollow.

Rolling Down an Incline (Kinetic Energy)

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If friction is present and prevents pure rolling during motion on the inclined plane, the object that covers more distance over the plane will have the smallest kinetic energy.
KE = 1/2 mv^2 + 1/2 Iω^2
The kinetic energy at the bottom of the incline depends on the amount of energy lost to friction. All the objects have the same potential energy initially, so the object with the smallest kinetic energy will have lost the most energy to friction. This implies more sliding and less rolling. Since the force of friction does negative work, we need to compare the energies and work done by friction to find the ordering.

String Wrapped Around a Cylinder

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When a string wrapped around a cylinder is pulled, and the cylinder rolls without slipping, the length of the string pulled is related to the distance the cylinder moves.
s = rθ
If a string is wrapped around a cylinder and the cylinder rolls without slipping when the string is pulled, the distance the cylinder moves is related to the length of the string unwound. For every revolution of the cylinder, a length of string equal to the circumference (2πr) is unwound. The man pulls the string a length that is twice the distance moved by the cylinder, i.e., 2l.

Axis of Rotation

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The axis of rotation of a rotating body may or may not pass through the center of mass but must pass through at least one particle of the body.
N/A
The axis of rotation is a line about which a body rotates. It may or may not pass through the center of mass. However, the axis must pass through at least one particle of the body.

Angular Momentum and Torque

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In inertial frames, both L=Iw and dL/dt=τ are valid. Angular momentum is equal to the moment of inertia times the angular velocity, and the rate of change of angular momentum is equal to the net external torque. But, in non-inertial frames, inertial forces must be considered when evaluating torques.
L = Iω, τ = dL/dt
In inertial frames, the relationship between angular momentum (L), moment of inertia (I), and angular velocity (ω) is given by L = Iω. Also, the net external torque (τ) is equal to the rate of change of angular momentum: τ = dL/dt. However, in non-inertial frames, fictitious forces (like Coriolis force) can exert torques, and these relationships need to be modified to account for those forces. Therefore, both L=Iw and dL/dt=G (where G is the external torque) are valid only in inertial frames.

Angular Momentum of a Particle

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The angular momentum of a particle moving with uniform velocity is constant about any given point.
L = r x p
The angular momentum of a particle is given by L = r x p, where r is the position vector from the reference point to the particle and p is the linear momentum of the particle (p = mv). If the particle moves with uniform velocity, its linear momentum is constant. If the particle moves along a straight line, the perpendicular distance from the reference point to the line of motion remains constant. Thus, the magnitude and direction of the angular momentum about any point remains constant.

Conservation Laws for Non-Rigid Bodies

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When no external force is acting on a non-rigid body, its linear momentum and angular momentum are conserved. Kinetic energy and moment of inertia need not be constant because the body can deform.
N/A
If no external force acts on a non-rigid body, the total linear momentum and the total angular momentum of the body must remain constant. However, the kinetic energy and moment of inertia are not necessarily constant because the body can deform, and its internal energy can change.

Parallel Axis Theorem

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The moment of inertia about an axis not passing through the center of mass is always greater than the moment of inertia about a parallel axis passing through the center of mass.
I = I_cm + Md²
The parallel axis theorem states that the moment of inertia (I) of a body about any axis is equal to the moment of inertia about a parallel axis through the center of mass (I_cm) plus the product of the mass (M) of the body and the square of the distance (d) between the two axes: I = I_cm + Md². Thus, the moment of inertia is always least about an axis passing through the center of mass. If the axes are not parallel, there is no general ordering possible between I_A and I_B.

Rotation of a Sphere

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During the rotation of a sphere, points on the surface have acceleration, but particles on the diameter of rotation have zero acceleration.
a_c = v^2/r = rω^2
When a sphere is rotating about a diameter, the particles on the surface of the sphere undergo circular motion and thus have centripetal acceleration (and potentially tangential acceleration if the angular velocity is changing). However, the particles on the diameter about which the sphere is rotating are on the axis of rotation and do not have any linear acceleration (assuming the axis is fixed).

Rolling Motion

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Conditions for rolling, factors affecting rolling motion on inclined planes, and the effect of friction on rolling objects.
None explicitly mentioned in the text (but concepts imply use of τ = Iα, F = ma, etc., depending on the problem context).
  • Rolling on a Smooth Surface: A sphere cannot roll on a smooth surface (horizontal or inclined) because there is no friction to provide the torque necessary for rotation.
  • Rolling on a Rough Surface: A sphere can roll on a rough surface (horizontal or inclined) due to the presence of friction.
  • Rear-Wheel Drive Cars: In rear-wheel drive cars accelerating on a horizontal road, friction on the rear wheels is in the forward direction, propelling the car forward.
  • Rolling on an Inclined Plane: A sphere can roll on a surface inclined at an angle θ if the friction coefficient is sufficiently high.
  • Effect of Friction on Rolling: Friction tries to decrease the linear velocity and increase the angular velocity of a sphere rolling on a rough horizontal surface until pure rolling is achieved.
  • Pure Rolling on an Inclined Plane in an Accelerating Car: If a sphere is set in pure rolling on an inclined plane fixed in a car accelerating with a = g tanθ, it will continue pure rolling.

Angular Motion

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Basic concepts of angular motion, including angular acceleration, angular velocity, and their relationship to linear motion.
While no explicit formulas are listed, the text implies use of ω = ω₀ + αt, θ = ω₀t + (1/2)αt², and v = rω in relevant problems.
  • Angular Acceleration: Uniform angular acceleration is considered in the context of a wheel making revolutions about its axis.
  • Relationship between Linear and Angular Speed: All particles on the surface of a rolling object do not have the same linear speed.

Rotational Inertia and Distribution of Mass

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How the distribution of mass affects rotational motion and the concept of the moment of inertia.
Again, no explicit formulas, but the content strongly implies using I = ∫r²dm (for moment of inertia) and considerations of kinetic energy (KE = (1/2)Iω² + (1/2)mv²).
  • Density Variation in a Rod: When a rod with gradually decreasing density from one end is pivoted, the angular acceleration, angular velocity, angular momentum, and torque depend on the distribution of mass.
  • Rolling Motion on Inclines: Solid spheres and hollow spheres with the same mass and radii roll down a rough inclined plane differently. The solid sphere reaches the bottom with greater speed and kinetic energy.

Velocity in Rolling Motion

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Understanding the velocity of different points on a rolling wheel.
v = rω
  • Velocity of Points on a Rolling Wheel: The speed of the particle at the point of contact (A) is zero. The speed of the topmost point (C) is 2v₀. The speed of the center of the wheel (O) is v₀.

Torque

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Torque is a twisting force that causes rotation. It depends on the force applied and the distance from the axis of rotation.
τ = r × F = rFsinθ, where τ is the torque, r is the position vector from the axis of rotation to the point where the force is applied, F is the force vector, and θ is the angle between r and F.
Torque: Torque is a measure of the turning force on an object about an axis of rotation. It is a vector quantity. The magnitude of the torque is given by the product of the magnitude of the force and the perpendicular distance from the axis of rotation to the line of action of the force. This perpendicular distance is also known as the lever arm or moment arm.

Angular Acceleration

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Angular acceleration is the rate of change of angular velocity.
α = dω/dt
Angular Acceleration: Angular acceleration (α) is the rate of change of angular velocity (ω) with respect to time (t). If the angular velocity changes from ω₁ to ω₂ in time Δt, then angular acceleration α= (ω₂ - ω₁) / Δt.

Angular Velocity

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Angular velocity is the rate at which an object rotates.
ω = dθ/dt
Angular Velocity: Angular velocity (ω) is the rate of change of angular displacement (θ) with respect to time (t).

Linear Speed and Angular Speed

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Relationship between linear speed and angular speed
v = rω
The linear speed (v) of a point on a rotating object is related to the angular speed (ω) by the equation v = rω, where r is the distance of the point from the axis of rotation.

Radial and Tangential Acceleration

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Components of acceleration in circular motion.
a_r = rω^2, a_t = rα
Radial (Centripetal) Acceleration: a_r = v^2/r = rω^2, directed towards the center of the circle. Tangential Acceleration: a_t = rα, directed along the tangent to the circle.

Moment of Inertia

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Moment of inertia is a measure of an object's resistance to rotational motion.
I = Σ mᵢrᵢ² (for discrete particles), I = ∫ r² dm (for continuous bodies)
Moment of Inertia: The moment of inertia (I) of a body about a given axis is the sum of the products of the masses of all the particles in the body and the squares of their distances from the axis of rotation. For a continuous body, the sum is replaced by an integral. Moment of Inertia depends on the mass and the distribution of mass relative to the axis of rotation. Unit is kg m².

Radius of Gyration

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Radius of gyration is the distance from the axis of rotation at which the entire mass of the object can be assumed to be concentrated to give the same moment of inertia.
k = √(I/M)
Radius of Gyration: The radius of gyration (k) is defined as the distance from the axis of rotation to a point where the entire mass of the object could be concentrated without changing its moment of inertia. If M is the total mass, then I = Mk². Therefore, k = √(I/M).

Rotational Kinetic Energy

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Energy due to rotation.
KE = (1/2)Iω²
The kinetic energy of a rotating object is given by KE = (1/2)Iω², where I is the moment of inertia and ω is the angular velocity.

Angular Momentum

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Measure of an object's rotational motion.
L = Iω
Angular momentum (L) is the product of the moment of inertia (I) and the angular velocity (ω). L = Iω. It is a vector quantity. For a single particle, L = r × p = rmvsinθ, where r is the position vector from the origin to the particle, p is the linear momentum, m is the mass, v is the velocity, and θ is the angle between r and v.

Torque

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Torque is a twisting force that causes rotation. It depends on the force applied and the distance from the axis of rotation.
τ = r × F, τ = rFsinθ
Torque is a vector quantity that measures the tendency of a force to cause rotation about an axis. It is calculated as the cross product of the position vector (from the axis of rotation to the point where the force is applied) and the force vector. The magnitude of the torque is given by τ = rFsinθ, where r is the magnitude of the position vector, F is the magnitude of the force, and θ is the angle between the position and force vectors.

Angular Deceleration

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Angular deceleration is the rate at which the angular velocity of an object decreases over time.
α = Δω / Δt
Angular deceleration (α) is the rate of decrease of angular velocity (ω) with respect to time (t). It is the rotational equivalent of linear deceleration and is measured in radians per second squared (rad/s²). A constant torque can cause angular deceleration.

Moment of Inertia

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Moment of inertia is a measure of an object's resistance to changes in its rotational motion. It depends on the mass and the distribution of mass relative to the axis of rotation.
Varies depending on the object's shape and axis of rotation. Examples include I = MR² for a hoop about its central axis and I = (1/2)MR² for a solid cylinder about its central axis.
Moment of inertia (I) is the rotational analog of mass in linear motion. It represents an object's resistance to changes in its rotational velocity. The moment of inertia depends on both the mass of the object and the distribution of that mass relative to the axis of rotation. Objects with mass farther from the axis of rotation have a larger moment of inertia.

Relationship between Torque and Angular Acceleration

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Torque is proportional to angular acceleration, with the moment of inertia being the constant of proportionality.
τ = Iα
The net torque acting on an object is directly proportional to its angular acceleration. The constant of proportionality is the moment of inertia (I) of the object. This relationship is analogous to Newton's second law of motion (F = ma) in linear motion.

Rotational Kinetic Energy

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Rotational kinetic energy is the kinetic energy due to the rotation of an object.
KE_rotational = (1/2)Iω²
A rotating object possesses kinetic energy due to its rotation. This rotational kinetic energy depends on the object's moment of inertia (I) and its angular velocity (ω).

Angular Momentum

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Angular momentum is a measure of an object's rotation. It depends on the object's moment of inertia and angular velocity.
L = I * ω, where L is angular momentum, I is moment of inertia, and ω is angular velocity.
Angular momentum is a vector quantity that describes the amount of rotational motion an object has. For a single particle, it is the cross product of the position vector and the linear momentum. For a rigid body, it is the product of the moment of inertia and the angular velocity. Conservation of angular momentum is a fundamental principle in physics, stating that the total angular momentum of a closed system remains constant if no external torque acts on it.

Moment of Inertia

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Moment of inertia is a body's resistance to angular acceleration. It depends on the mass and its distribution relative to the axis of rotation.
I = Σ mr², where m is the mass of each particle and r is the distance from the axis of rotation. Reduced mass: m = (m1*m2) / (m1 + m2)
Moment of inertia (I) is the rotational analog of mass in linear motion. It quantifies an object's resistance to changes in its rotational motion. The moment of inertia depends not only on the mass of the object but also on how that mass is distributed relative to the axis of rotation. Different shapes and axes of rotation will result in different moment of inertia values. The parallel axis theorem is used to calculate the moment of inertia about any axis parallel to an axis through the center of mass.

Torque

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Torque is the rotational equivalent of force. It causes an object to rotate.
τ = r × F, where τ is torque, r is the position vector, and F is the force.
Torque is a twisting force that tends to cause rotation. It is a vector quantity equal to the cross product of the position vector and the force vector. The magnitude of the torque depends on the magnitude of the force, the distance from the axis of rotation, and the angle between the force and position vectors.

Kinetic Energy (Rotational)

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Rotational kinetic energy is the energy an object has due to its rotation.
KE = (1/2) * I * ω², where KE is kinetic energy, I is moment of inertia, and ω is angular speed.
Rotational kinetic energy is the kinetic energy due to the rotation of an object and is part of its total kinetic energy. It depends on the object's moment of inertia and angular speed.

Impulse and Angular Impulse

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A force applied for a short time interval (impulse) changes the linear momentum of an object. Similarly, a torque applied for a short time interval (angular impulse) changes the angular momentum of an object.
['Impulse = ∫F dt = Change in linear momentum', 'Angular Impulse = ∫τ dt = Change in angular momentum']
When a force F is applied for a short time interval dt, it imparts an impulse which changes the linear momentum. Similarly, when a torque acts for a short duration, it imparts an angular impulse which changes the angular momentum.

Kinetic Energy of a Rotating Body

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The kinetic energy of a rotating body is related to its moment of inertia and angular speed.
['Kinetic Energy (Rotational) = (1/2) * I * ω^2']
A rotating body possesses kinetic energy due to its rotation, which depends on the body's moment of inertia (I) and its angular speed (ω).

Angular Momentum

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Angular momentum is a measure of the amount of rotational motion an object has. It depends on the object's moment of inertia and angular velocity.
['Angular Momentum (L) = I * ω']
Angular momentum is a vector quantity that describes the rotational inertia and rotational velocity of an object. For a rigid body rotating about a fixed axis, it's the product of moment of inertia and angular velocity.

Rolling Motion

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Rolling motion is a combination of translational and rotational motion. The speed of a point on a rolling object depends on its position relative to the center of mass.
['v = rω (for pure rolling, where v is the velocity of the center of mass, r is the radius, and ω is the angular velocity)', 'Speed of highest point = 2v (in pure rolling)']
In rolling motion, an object simultaneously translates and rotates. For pure rolling (without slipping), the velocity of the point of contact with the surface is zero. The velocity of the highest point is twice the velocity of the center of mass.

Rolling Motion

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Rolling motion is a combination of translational and rotational motion. Understanding the relationship between linear and angular speed/acceleration is key. Friction plays a crucial role in enabling pure rolling.
['Kinetic Energy (Rolling): KE = (1/2)mv^2 + (1/2)Iω^2, where m is mass, v is linear speed, I is moment of inertia, and ω is angular speed.', 'Condition for Pure Rolling: v = Rω, where v is linear speed, R is radius, and ω is angular speed.', 'Relationship between linear and angular acceleration: a=Rα, where a is linear acceleration, R is the radius and α is angular acceleration']
Rolling motion can be analyzed by considering both translational and rotational kinetic energy. Pure rolling occurs when there is no slipping at the point of contact, implying a specific relationship between linear and angular velocity (v = rω). Friction provides the necessary torque for angular acceleration and can affect the linear acceleration as well. The moment of inertia of the rolling object is important in calculating its rotational kinetic energy.

Rotational Motion Definitions and Formulas

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This section likely contains definitions and formulas related to rotational motion, as it provides answers to objective questions and exercises on the topic.
['Angular velocity (ω)', 'Angular acceleration (α)', 'Torque (τ)', 'Moment of inertia (I)', 'Rotational kinetic energy (KE_rot = 1/2 * I * ω^2)', 'Angular momentum (L = I * ω)', 'Relationship between torque and angular acceleration (τ = I * α)', 'v = rω', 'a = rα', 'Conservation of Angular Momentum (I1ω1 = I2ω2)']
Based on the provided information, the content covers various aspects of rotational mechanics including: * **Rotational Kinematics:** Angular displacement, angular velocity, angular acceleration. * **Torque:** The rotational equivalent of force. * **Moment of Inertia:** The resistance of an object to changes in its rotational motion. * **Rotational Kinetic Energy:** The kinetic energy due to the rotation of an object. * **Angular Momentum:** A measure of the amount of rotation an object has. * **Relationship between Linear and Rotational Motion:** Connecting linear velocity, acceleration, and displacement to their angular counterparts (v = rω, a = rα, etc.). * **Conservation of Angular Momentum:** The total angular momentum of a closed system remains constant if no external torque acts on it. It's also implied that problems involving pulleys, rolling motion, and collisions with rotating objects are discussed.

Rolling Motion

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This section likely contains formulas and concepts related to rolling motion, potentially including kinetic energy, conservation of energy, and forces involved in rolling with and without slipping.
['v = ωR (for rolling without slipping)', 'K.E. = 1/2 mv^2 + 1/2 Iω^2 (Total kinetic energy of rolling body)', 'I = Moment of Inertia (depends on the shape and axis of rotation)', 'Conservation of Energy: Potential Energy + Kinetic Energy (translational + rotational) = constant']
Based on the provided answers, the problems cover topics such as: * **Conservation of Energy in Rolling:** Problems involve calculating velocities, heights, and forces using conservation of energy principles, considering both translational and rotational kinetic energy. * **Rolling without Slipping:** Conditions for rolling without slipping are likely used, relating linear velocity to angular velocity (v = ωR). * **Forces in Rolling:** Problems involve calculating forces like friction required for rolling and normal reaction. * **Moment of Inertia:** The moment of inertia of various objects (like rods, spheres, cylinders) is implicitly used in calculating rotational kinetic energy. * **Collisions involving Rotation:** Some problems might involve collisions where rotational motion is affected.

Rotational Dynamics

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Calculations for angular velocity, and relationships between forces and angles in rotational systems.
['τ = Iα (Torque = Moment of Inertia * Angular Acceleration)', 'ω = v/R (Angular velocity = linear velocity / radius)', 'Conditions for Equilibrium: ΣF = 0, Στ = 0']
Based on the provided answers, the problems cover topics such as: * **Rotational Kinematics:** Problems involving angular velocity (ω). * **Torque and Angular Acceleration:** Problems might implicitly involve the relationship between torque and angular acceleration (τ = Iα). * **Equilibrium:** Problems likely involve static equilibrium where net force and net torque are zero.

Rotational Motion

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The motion of objects around an axis.
Torque = Iα; Angular Momentum = Iω; Kinetic Energy (Rotational) = (1/2)Iω^2
Includes angular velocity, angular acceleration, torque, angular momentum, and moment of inertia.