Grade 11 Physics Rotational Motion Practice Quiz (Set 4)

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Grade 11 : Physics - Rotational Motion (Set 4)— Questions & Detailed Solutions

Setup: A uniform thin rod of length $L = 2m$ and mass $M = 3kg$ is pivoted at one of its ends so that it can rotate freely in a plane.
Calculate the moment of inertia of the rod about an axis passing through the pivot perpendicular to its length.

(A)

$2\text{ kg m}^2$

(B)

$4\text{ kg m}^2$

(C)

$6\text{ kg m}^2$

(D)

$12\text{ kg m}^2$

View Correct Answer & Detailed Solution

✓ Correct Option: (B)

Explanation:The moment of inertia of a uniform thin rod about an axis passing through one end perpendicular to its length is given by $I = \frac{1}{3} M L^2$. Substituting $M = 3kg$ and $L = 2m$ yields $I = \frac{1}{3} \times 3 \times 2^2 = 4\text{ kg m}^2$.

Object Axis of Rotation Moment of Inertia ($I$)
Object X Central axis perpendicular to disc plane $\frac{1}{2} M R^2$
Object Y Diameter axis $\frac{2}{5} M R^2$
Object Z Central axis perpendicular to ring plane $M R^2$
Based on the reference table, identify the geometry of objects X, Y, and Z respectively (all of mass $M$ and radius $R$).

Object	Axis of Rotation	Moment of Inertia ($I$)
Object X	Central axis perpendicular to disc plane	$\frac{1}{2} M R^2$
Object Y	Diameter axis	$\frac{2}{5} M R^2$
Object Z	Central axis perpendicular to ring plane	$M R^2$

(A)

Circular Disc, Solid Sphere, Thin Ring

(B)

Thin Ring, Solid Cylinder, Hollow Sphere

(C)

Solid Sphere, Circular Disc, Thin Ring

(D)

Circular Disc, Thin Ring, Solid Sphere

View Correct Answer & Detailed Solution

✓ Correct Option: (A)

Explanation:For Object X, $I = \frac{1}{2} M R^2$ corresponds to a uniform circular disc about its central axis. For Object Y, $I = \frac{2}{5} M R^2$ corresponds to a uniform solid sphere about its diameter. For Object Z, $I = M R^2$ corresponds to a thin circular ring about its central axis.

System Diagram: A circular ring of mass $M = 4kg$ and radius $R = 0.5m$ lies in the xy-plane. An axis is drawn tangent to the ring and parallel to its diameter in the plane.
Determine the moment of inertia of the ring about this tangential axis in its plane.

(A)

$0.5\text{ kg m}^2$

(B)

$1.0\text{ kg m}^2$

(C)

$1.5\text{ kg m}^2$

(D)

$2.0\text{ kg m}^2$

View Correct Answer & Detailed Solution

✓ Correct Option: (C)

Explanation:Moment of inertia of a ring about its diameter is $I_{dia} = \frac{1}{2} M R^2$. By the parallel axis theorem, for a tangent in the plane: $I_{tan} = I_{dia} + M R^2 = \frac{3}{2} M R^2$. Substituting values: $I_{tan} = \frac{3}{2}(4)(0.5)^2 = 1.5\text{ kg m}^2$.

A heavy flywheel starts from rest ($ \omega_0 = 0$) and undergoes constant angular acceleration of $\alpha = 4 rad/s^2for a duration oft = 5 s$.
What is the total angular displacement covered by the flywheel during this time interval?

(A)

20 rad

(B)

40 rad

(C)

50 rad

(D)

100 rad

View Correct Answer & Detailed Solution

✓ Correct Option: (C)

Explanation:Using the kinematic equation for rotational motion: $\theta = \omega_0 t + \frac{1}{2}\alpha t^2$. Since $\omega_0 = 0$, $\theta = \frac{1}{2}(4)(5)^2 = 50\text{ rad}$.

Vector Coordinates: Position vector of point of force application: $\vec{r} = (2\hat{i} + 1\hat{j})m$. Applied force vector: $\vec{F} = (3\hat{i} + 4\hat{j})\text{ N}$.
Find the torque vector $\vec{\tau}$ acting about the origin.

(A)

$3\hat{k}\text{ N m}$

(B)

$5\hat{k}\text{ N m}$

(C)

$8\hat{k}\text{ N m}$

(D)

$11\hat{k}\text{ N m}$

View Correct Answer & Detailed Solution

✓ Correct Option: (B)

Explanation:Torque is given by the cross product $\vec{\tau} = \vec{r} \times \vec{F} = (2\hat{i} + \hat{j}) \times (3\hat{i} + 4\hat{j})$. $\vec{\tau} = (2 \times 4 - 1 \times 3)\hat{k} = 5\hat{k}\text{ N m}$.

A uniform solid spherical star contracts symmetrically under its own gravitational attraction to half its original radius ($R_2 = R_1 / 2$) without losing any mass ($M$ remains constant).
Assuming no external torque acts on the star, what is the ratio of its new angular velocity $\omega_2$ to its initial angular velocity $\omega_1$?

(A)

(B)

(C)

(D)

0.25

View Correct Answer & Detailed Solution

✓ Correct Option: (B)

Explanation:By conservation of angular momentum, $I_1 \omega_1 = I_2 \omega_2$. For a solid sphere, $I = \frac{2}{5} M R^2$. Therefore, $I_2 = \frac{2}{5} M (R/2)^2 = \frac{1}{4} I_1$. Thus, $\omega_2 / \omega_1 = I_1 / I_2 = 4$.

Track Geometry: Two particles A and B move along concentric circular paths of radius $R_A = 1m$ and $R_B = 2m$ respectively, fixed on a rotating rigid platform turning at angular speed $\omega = 3\text{ rad/s}$.
What is the ratio of linear speed of particle B to that of particle A ($v_B / v_A$)?

(A)

0.5

(B)

1.0

(C)

2.0

(D)

4.0

View Correct Answer & Detailed Solution

✓ Correct Option: (C)

Explanation:Linear velocity is related to angular velocity by $v = r \omega$. Since both particles share the same rigid rotation $\omega$, $v_A = R_A \omega$ and $v_B = R_B \omega$. The ratio $v_B / v_A = R_B / R_A = 2 / 1 = 2$.

A solid cylinder of mass $M = 2kg$ and radius $R = 0.2m$ rolls without slipping on a flat horizontal floor at center of mass speed $v = 4m/s$.
Calculate the total kinetic energy of the rolling cylinder.

(A)

16 J

(B)

24 J

(C)

32 J

(D)

48 J

View Correct Answer & Detailed Solution

✓ Correct Option: (B)

Explanation:Total kinetic energy of rolling without slipping is $K_{total} = K_{trans} + K_{rot} = \frac{1}{2} M v^2 \left(1 + \frac{k^2}{R^2}\right)$. For a solid cylinder, $\frac{k^2}{R^2} = \frac{1}{2}$. Thus, $K_{total} = \frac{1}{2}(2)(4)^2 \left(1 + 0.5\right) = 16 \times 1.5 = 24\text{ J}$.

[ Turntable 1: I₁ = 2 kg m², ω₁ = 10 rad/s ] + [ Stationary Disc 2: I₂ = 3 kg m², ω₂ = 0 ] ---> [ Combined System: I₁ + I₂, ω = ? ]
A horizontal turntable of moment of inertia $I_1 = 2\text{ kg m}^2$ rotates freely at $\omega_1 = 10\text{ rad/s}$. A non-rotating disc of moment of inertia $I_2 = 3\text{ kg m}^2$ is gently dropped coaxially onto it. What is their final common angular speed?

(A)

2 rad/s

(B)

4 rad/s

(C)

5 rad/s

(D)

6 rad/s

View Correct Answer & Detailed Solution

✓ Correct Option: (B)

Explanation:Since no external torque acts on the system, angular momentum is conserved: $I_1 \omega_1 = (I_1 + I_2) \omega$. Solving for $\omega$: $\omega = \frac{I_1 \omega_1}{I_1 + I_2} = \frac{2 \times 10}{2 + 3} = \frac{20}{5} = 4\text{ rad/s}$.

Q10

Rod Orientations: Axis 1 passes through center of mass perpendicular to length ($I_{cm}$). Axis 2 passes through one end perpendicular to length ($I_{end}$).
For a uniform thin rod, what is the ratio of its moment of inertia $I_{cm}$ to $I_{end}$?

(A)

$1 : 4$

(B)

$1 : 3$

(C)

$1 : 2$

(D)

$1 : 12$

View Correct Answer & Detailed Solution

✓ Correct Option: (A)

Explanation:For a thin uniform rod of mass $M$ and length $L$, $I_{cm} = \frac{1}{12} M L^2$ and $I_{end} = \frac{1}{3} M L^2$. The ratio is $\frac{I_{cm}}{I_{end}} = \frac{\frac{1}{12} M L^2}{\frac{1}{3} M L^2} = \frac{3}{12} = \frac{1}{4}$.

Q11

Consider a solid uniform sphere of radius $R$. We wish to determine its radius of gyration $k$ about a axis tangent to its outer surface.
Express the radius of gyration $k$ in terms of $R$.

(A)

$\sqrt{\frac{2}{5}} R$

(B)

$\sqrt{\frac{3}{5}} R$

(C)

$\sqrt{\frac{7}{5}} R$

(D)

$\frac{7}{5} R$

View Correct Answer & Detailed Solution

✓ Correct Option: (C)

Explanation:$I_{cm} = \frac{2}{5} M R^2$. By parallel axis theorem, $I_{tangent} = \frac{2}{5} M R^2 + M R^2 = \frac{7}{5} M R^2$. Setting $M k^2 = \frac{7}{5} M R^2$ gives $k = \sqrt{\frac{7}{5}} R$.

Q12

Timing Interval: Initial angular velocity $\omega_0 = 10\text{ rad/s}$. Final angular velocity $\omega = 0\text{ rad/s}$. Stopping time $t = 5\text{ s}$.
A rotating wheel is brought to rest by a uniform braking torque in 5 s. What is the magnitude of its angular acceleration?

(A)

$1\text{ rad/s}^2$

(B)

$2\text{ rad/s}^2$

(C)

$4\text{ rad/s}^2$

(D)

$5\text{ rad/s}^2$

View Correct Answer & Detailed Solution

✓ Correct Option: (B)

Explanation:Using $\omega = \omega_0 + \alpha t$, we get $0 = 10 + \alpha (5)$, which yields $\alpha = -2\text{ rad/s}^2$. The magnitude of angular deceleration is $2\text{ rad/s}^2$.

Q13

Body Shape Factor ($k^2/R^2$) Acceleration Formula
Ring 1.0 $a = \frac{g \sin\theta}{1 + k^2/R^2}$
Disc 0.5 $a = \frac{g \sin\theta}{1 + k^2/R^2}$
Solid Sphere 0.4 $a = \frac{g \sin\theta}{1 + k^2/R^2}$
Three bodies (a ring, a disc, and a solid sphere) of equal mass and radius are released simultaneously from rest from the top of an inclined plane. Which object reaches the bottom first?

Body	Shape Factor ($k^2/R^2$)	Acceleration Formula
Ring	1.0	$a = \frac{g \sin\theta}{1 + k^2/R^2}$
Disc	0.5	$a = \frac{g \sin\theta}{1 + k^2/R^2}$
Solid Sphere	0.4	$a = \frac{g \sin\theta}{1 + k^2/R^2}$

(A)

Ring

(B)

Disc

(C)

Solid Sphere

(D)

All reach at the same time

View Correct Answer & Detailed Solution

✓ Correct Option: (C)

Explanation:Linear acceleration down the incline is $a = \frac{g \sin\theta}{1 + k^2/R^2}$. The body with the smallest $k^2/R^2$ ratio has the greatest acceleration. Since $0.4 < 0.5 < 1.0$, the solid sphere has the maximum acceleration and arrives first.

Q14

A constant torque of $\tau = 20\text{ N m}$ is applied to rotate a flywheel through an angular displacement of 5 full revolutions.
Calculate the total work done by the applied torque.

(A)

100 J

(B)

$100\pi\text{ J}$

(C)

$200\pi\text{ J}$

(D)

$400\pi\text{ J}$

View Correct Answer & Detailed Solution

✓ Correct Option: (C)

Explanation:Angular displacement $\theta = 5 \text{ rev} \times 2\pi \text{ rad/rev} = 10\pi\text{ rad}$. Work done by torque is $W = \tau \theta = 20 \times 10\pi = 200\pi\text{ J}$.

Q15

Plane Lamina Setup: A thin flat lamina lies in the xy-plane. Its moments of inertia about the $x$-axis and $y$-axis are $I_x = 4\text{ kg m}^2$ and $I_y = 6\text{ kg m}^2$.
Using the Perpendicular Axes Theorem, calculate its moment of inertia $I_z$ about the $z$-axis perpendicular to the lamina plane.

(A)

$2\text{ kg m}^2$

(B)

$5\text{ kg m}^2$

(C)

$10\text{ kg m}^2$

(D)

$24\text{ kg m}^2$

View Correct Answer & Detailed Solution

✓ Correct Option: (C)

Explanation:By the Perpendicular Axes Theorem for a flat planar lamina, $I_z = I_x + I_y$. Substituting the given values gives $I_z = 4 + 6 = 10\text{ kg m}^2$.

Q16

A particle of mass $m = 2kg$ moves with a constant velocity $v = 5m/s$ along the horizontal line $y = 3m$ in the xy-plane.
Find the magnitude of the angular momentum of the particle about the origin.

(A)

$10\text{ kg m}^2/\text{s}$

(B)

$15\text{ kg m}^2/\text{s}$

(C)

$30\text{ kg m}^2/\text{s}$

(D)

$60\text{ kg m}^2/\text{s}$

View Correct Answer & Detailed Solution

✓ Correct Option: (C)

Explanation:Angular momentum magnitude for a particle moving in a straight line is $L = m v r_{\perp}$, where $r_{\perp}$ is the perpendicular distance from origin to the line of motion. Here $r_{\perp} = 3m$. Thus $L = 2 \times 5 \times 3 = 30\text{ kg m}^2/\text{s}$.

Q17

Thin Cylindrical Shell (Hollow): I_hollow = M R² vs Solid Cylinder: I_solid = (1/2) M R²
What is the ratio of moment of inertia of a thin hollow cylinder to that of a solid cylinder of the same mass $M$ and radius $R$ about their central symmetry axis?

(A)

$1 : 2$

(B)

$2 : 1$

(C)

$1 : 1$

(D)

$4 : 1$

View Correct Answer & Detailed Solution

✓ Correct Option: (B)

Explanation:For a thin hollow cylinder, $I_{hollow} = M R^2$. For a solid cylinder, $I_{solid} = \frac{1}{2} M R^2$. The ratio $I_{hollow} / I_{solid} = \frac{M R^2}{\frac{1}{2} M R^2} = 2$, which gives $2 : 1$.

Q18

A body undergoing pure rotation possesses rotational kinetic energy $E = 100\text{ J}$ and angular momentum $L = 20\text{ J s}$.
Determine the moment of inertia of the body about its axis of rotation.

(A)

$1\text{ kg m}^2$

(B)

$2\text{ kg m}^2$

(C)

$4\text{ kg m}^2$

(D)

$5\text{ kg m}^2$

View Correct Answer & Detailed Solution

✓ Correct Option: (B)

Explanation:Rotational kinetic energy is related to angular momentum by $E = \frac{L^2}{2 I}$. Rearranging gives $I = \frac{L^2}{2 E} = \frac{20^2}{2 \times 100} = \frac{400}{200} = 2\text{ kg m}^2$.

Q19

Time Interval Initial Speed Acceleration Displacement Formula
1st second ($t=0$ to 1 s) $\omega_0 = 0$ $\alpha$ $\theta_1 = \frac{1}{2} \alpha (1)^2$
2nd second ($t=1$ to 2 s) $\omega_1 = \alpha$ $\alpha$ $\theta_2 = \frac{1}{2} \alpha (2)^2 - \theta_1$
A wheel starts from rest and rotates with uniform angular acceleration. Find the ratio of angular displacements $\theta_1 : \theta_2$ covered in the 1st second and the 2nd second respectively.

Time Interval	Initial Speed	Acceleration	Displacement Formula
1st second ($t=0$ to 1 s)	$\omega_0 = 0$	$\alpha$	$\theta_1 = \frac{1}{2} \alpha (1)^2$
2nd second ($t=1$ to 2 s)	$\omega_1 = \alpha$	$\alpha$	$\theta_2 = \frac{1}{2} \alpha (2)^2 - \theta_1$

(A)

$1 : 2$

(B)

$1 : 3$

(C)

$1 : 4$

(D)

$2 : 3$

View Correct Answer & Detailed Solution

✓ Correct Option: (B)

Explanation:Displacement in total $t=1\text{ s}$ is $\theta(1) = \frac{1}{2}\alpha(1)^2 = \frac{1}{2}\alpha$. Displacement in total $t=2\text{ s}$ is $\theta(2) = \frac{1}{2}\alpha(2)^2 = 2\alpha$. Displacement in 2nd second is $\theta_2 = 2\alpha - \frac{1}{2}\alpha = \frac{3}{2}\alpha$. Thus, $\theta_1 : \theta_2 = \frac{1}{2}\alpha : \frac{3}{2}\alpha = 1 : 3$.

Q20

Cutout Geometry: A uniform disc of mass $M = 6kg$ and radius $R = 1m$. A circular hole of radius $r = 0.5m$ is punched out such that its center is at distance $d = 0.5m$ from the disc's center.
Calculate the moment of inertia of the remaining portion about the axis through the original disc's center perpendicular to its plane.

(A)

$1.5\text{ kg m}^2$

(B)

$2.4375\text{ kg m}^2$

(C)

$2.75\text{ kg m}^2$

(D)

$3.0\text{ kg m}^2$

View Correct Answer & Detailed Solution

✓ Correct Option: (B)

Explanation:Original disc mass $M = 6kg$, $I_0 = \frac{1}{2} M R^2 = 3\text{ kg m}^2$. Mass of cut portion $m = M \frac{\pi r^2}{\pi R^2} = 6 \times \frac{0.25}{1} = 1.5kg$. Moment of inertia of removed disc about original center (by parallel axis theorem): $I_{cut} = \frac{1}{2} m r^2 + m d^2 = \frac{1}{2}(1.5)(0.25) + 1.5(0.25) = 0.1875 + 0.375 = 0.5625\text{ kg m}^2$. Remaining $I = 3 - 0.5625 = 2.4375\text{ kg m}^2$.

Grade 11 : Physics - Rotational Motion (Set 4)

Grade 11 : Physics - Rotational Motion (Set 4)— Questions & Detailed Solutions

Setup: A uniform thin rod of length $L = 2m$ and mass $M = 3kg$ is pivoted at one of its ends so that it can rotate freely in a plane.
Calculate the moment of inertia of the rod about an axis passing through the pivot perpendicular to its length.

System Diagram: A circular ring of mass $M = 4kg$ and radius $R = 0.5m$ lies in the xy-plane. An axis is drawn tangent to the ring and parallel to its diameter in the plane.
Determine the moment of inertia of the ring about this tangential axis in its plane.

A heavy flywheel starts from rest ($ \omega_0 = 0$) and undergoes constant angular acceleration of $\alpha = 4 rad/s^2for a duration oft = 5 s$.
What is the total angular displacement covered by the flywheel during this time interval?

Vector Coordinates: Position vector of point of force application: $\vec{r} = (2\hat{i} + 1\hat{j})m$. Applied force vector: $\vec{F} = (3\hat{i} + 4\hat{j})\text{ N}$.
Find the torque vector $\vec{\tau}$ acting about the origin.

Track Geometry: Two particles A and B move along concentric circular paths of radius $R_A = 1m$ and $R_B = 2m$ respectively, fixed on a rotating rigid platform turning at angular speed $\omega = 3\text{ rad/s}$.
What is the ratio of linear speed of particle B to that of particle A ($v_B / v_A$)?

A solid cylinder of mass $M = 2kg$ and radius $R = 0.2m$ rolls without slipping on a flat horizontal floor at center of mass speed $v = 4m/s$.
Calculate the total kinetic energy of the rolling cylinder.

Rod Orientations: Axis 1 passes through center of mass perpendicular to length ($I_{cm}$). Axis 2 passes through one end perpendicular to length ($I_{end}$).
For a uniform thin rod, what is the ratio of its moment of inertia $I_{cm}$ to $I_{end}$?

Consider a solid uniform sphere of radius $R$. We wish to determine its radius of gyration $k$ about a axis tangent to its outer surface.
Express the radius of gyration $k$ in terms of $R$.

Timing Interval: Initial angular velocity $\omega_0 = 10\text{ rad/s}$. Final angular velocity $\omega = 0\text{ rad/s}$. Stopping time $t = 5\text{ s}$.
A rotating wheel is brought to rest by a uniform braking torque in 5 s. What is the magnitude of its angular acceleration?

A constant torque of $\tau = 20\text{ N m}$ is applied to rotate a flywheel through an angular displacement of 5 full revolutions.
Calculate the total work done by the applied torque.

A particle of mass $m = 2kg$ moves with a constant velocity $v = 5m/s$ along the horizontal line $y = 3m$ in the xy-plane.
Find the magnitude of the angular momentum of the particle about the origin.

Thin Cylindrical Shell (Hollow): I_hollow = M R² vs Solid Cylinder: I_solid = (1/2) M R²
What is the ratio of moment of inertia of a thin hollow cylinder to that of a solid cylinder of the same mass $M$ and radius $R$ about their central symmetry axis?

A body undergoing pure rotation possesses rotational kinetic energy $E = 100\text{ J}$ and angular momentum $L = 20\text{ J s}$.
Determine the moment of inertia of the body about its axis of rotation.

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Grade 11 : Physics - Rotational Motion (Set 4)

Grade 11 : Physics - Rotational Motion (Set 4)— Questions & Detailed Solutions

Setup: A uniform thin rod of length $L = 2m$ and mass $M = 3kg$ is pivoted at one of its ends so that it can rotate freely in a plane.Calculate the moment of inertia of the rod about an axis passing through the pivot perpendicular to its length.

System Diagram: A circular ring of mass $M = 4kg$ and radius $R = 0.5m$ lies in the xy-plane. An axis is drawn tangent to the ring and parallel to its diameter in the plane.Determine the moment of inertia of the ring about this tangential axis in its plane.

A heavy flywheel starts from rest ($ \omega_0 = 0$) and undergoes constant angular acceleration of $\alpha = 4 rad/s^2for a duration oft = 5 s$.What is the total angular displacement covered by the flywheel during this time interval?

Vector Coordinates: Position vector of point of force application: $\vec{r} = (2\hat{i} + 1\hat{j})m$. Applied force vector: $\vec{F} = (3\hat{i} + 4\hat{j})\text{ N}$.Find the torque vector $\vec{\tau}$ acting about the origin.

Track Geometry: Two particles A and B move along concentric circular paths of radius $R_A = 1m$ and $R_B = 2m$ respectively, fixed on a rotating rigid platform turning at angular speed $\omega = 3\text{ rad/s}$.What is the ratio of linear speed of particle B to that of particle A ($v_B / v_A$)?

A solid cylinder of mass $M = 2kg$ and radius $R = 0.2m$ rolls without slipping on a flat horizontal floor at center of mass speed $v = 4m/s$.Calculate the total kinetic energy of the rolling cylinder.

Rod Orientations: Axis 1 passes through center of mass perpendicular to length ($I_{cm}$). Axis 2 passes through one end perpendicular to length ($I_{end}$).For a uniform thin rod, what is the ratio of its moment of inertia $I_{cm}$ to $I_{end}$?

Consider a solid uniform sphere of radius $R$. We wish to determine its radius of gyration $k$ about a axis tangent to its outer surface.Express the radius of gyration $k$ in terms of $R$.

Timing Interval: Initial angular velocity $\omega_0 = 10\text{ rad/s}$. Final angular velocity $\omega = 0\text{ rad/s}$. Stopping time $t = 5\text{ s}$.A rotating wheel is brought to rest by a uniform braking torque in 5 s. What is the magnitude of its angular acceleration?

A constant torque of $\tau = 20\text{ N m}$ is applied to rotate a flywheel through an angular displacement of 5 full revolutions.Calculate the total work done by the applied torque.

A particle of mass $m = 2kg$ moves with a constant velocity $v = 5m/s$ along the horizontal line $y = 3m$ in the xy-plane.Find the magnitude of the angular momentum of the particle about the origin.

Thin Cylindrical Shell (Hollow): I_hollow = M R² vs Solid Cylinder: I_solid = (1/2) M R²What is the ratio of moment of inertia of a thin hollow cylinder to that of a solid cylinder of the same mass $M$ and radius $R$ about their central symmetry axis?

A body undergoing pure rotation possesses rotational kinetic energy $E = 100\text{ J}$ and angular momentum $L = 20\text{ J s}$.Determine the moment of inertia of the body about its axis of rotation.

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Setup: A uniform thin rod of length $L = 2m$ and mass $M = 3kg$ is pivoted at one of its ends so that it can rotate freely in a plane.
Calculate the moment of inertia of the rod about an axis passing through the pivot perpendicular to its length.

System Diagram: A circular ring of mass $M = 4kg$ and radius $R = 0.5m$ lies in the xy-plane. An axis is drawn tangent to the ring and parallel to its diameter in the plane.
Determine the moment of inertia of the ring about this tangential axis in its plane.

A heavy flywheel starts from rest ($ \omega_0 = 0$) and undergoes constant angular acceleration of $\alpha = 4 rad/s^2for a duration oft = 5 s$.
What is the total angular displacement covered by the flywheel during this time interval?

Vector Coordinates: Position vector of point of force application: $\vec{r} = (2\hat{i} + 1\hat{j})m$. Applied force vector: $\vec{F} = (3\hat{i} + 4\hat{j})\text{ N}$.
Find the torque vector $\vec{\tau}$ acting about the origin.

Track Geometry: Two particles A and B move along concentric circular paths of radius $R_A = 1m$ and $R_B = 2m$ respectively, fixed on a rotating rigid platform turning at angular speed $\omega = 3\text{ rad/s}$.
What is the ratio of linear speed of particle B to that of particle A ($v_B / v_A$)?

A solid cylinder of mass $M = 2kg$ and radius $R = 0.2m$ rolls without slipping on a flat horizontal floor at center of mass speed $v = 4m/s$.
Calculate the total kinetic energy of the rolling cylinder.

Rod Orientations: Axis 1 passes through center of mass perpendicular to length ($I_{cm}$). Axis 2 passes through one end perpendicular to length ($I_{end}$).
For a uniform thin rod, what is the ratio of its moment of inertia $I_{cm}$ to $I_{end}$?

Consider a solid uniform sphere of radius $R$. We wish to determine its radius of gyration $k$ about a axis tangent to its outer surface.
Express the radius of gyration $k$ in terms of $R$.

Timing Interval: Initial angular velocity $\omega_0 = 10\text{ rad/s}$. Final angular velocity $\omega = 0\text{ rad/s}$. Stopping time $t = 5\text{ s}$.
A rotating wheel is brought to rest by a uniform braking torque in 5 s. What is the magnitude of its angular acceleration?

A constant torque of $\tau = 20\text{ N m}$ is applied to rotate a flywheel through an angular displacement of 5 full revolutions.
Calculate the total work done by the applied torque.

A particle of mass $m = 2kg$ moves with a constant velocity $v = 5m/s$ along the horizontal line $y = 3m$ in the xy-plane.
Find the magnitude of the angular momentum of the particle about the origin.

Thin Cylindrical Shell (Hollow): I_hollow = M R² vs Solid Cylinder: I_solid = (1/2) M R²
What is the ratio of moment of inertia of a thin hollow cylinder to that of a solid cylinder of the same mass $M$ and radius $R$ about their central symmetry axis?

A body undergoing pure rotation possesses rotational kinetic energy $E = 100\text{ J}$ and angular momentum $L = 20\text{ J s}$.
Determine the moment of inertia of the body about its axis of rotation.