IB Physics Quiz: Apply Rigid Body Mechanics

20 questions · exam conditions

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Apply Rigid Body MechanicsQuestion 1 of 20

A rigid disk of radius $R$ rotates with constant angular velocity $\omega$ . A point P is at a distance $R/2$ from the center, and a point Q is on the rim at distance $R$ . What is the ratio of the centripetal acceleration of Q to that of P ( $a_Q / a_P$ )?

1/2

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IB Physics Quiz: Apply Rigid Body Mechanics

Practice Apply Rigid Body Mechanics in IB Physics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Apply Rigid Body Mechanics, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Physics.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

1/2
1
2 (correct answer)
4

Explanation: Centripetal acceleration is given by

a = \omega^2 r

. Since the disk is rigid, the angular velocity

\omega

is the same for all points. For point Q,

a_Q = \omega^2 R

. For point P,

a_P = \omega^2 (R/2)

. The ratio is

\frac{a_Q}{a_P} = \frac{\omega^2 R}{\omega^2 (R/2)} = 2

Question 2

Two point masses, $m_1 = 3.0 \text{ kg}$ and $m_2 = 1.0 \text{ kg}$ , are fixed to the ends of a massless rod of length 2.0 m. What is the moment of inertia of the system about an axis of rotation perpendicular to the rod and located 0.50 m from mass $m_2$ ?

3.0 kg m²
4.0 kg m²
5.0 kg m²
7.0 kg m² (correct answer)

Explanation: The moment of inertia of a system of point masses is

I = \sum m_i r_i^2

. The distance of

m_2

from the axis is

r_2 = 0.50 \text{ m}

. The distance of

m_1

from the axis is

r_1 = 2.0 \text{ m} - 0.50 \text{ m} = 1.5 \text{ m}

. Therefore,

I = m_1 r_1^2 + m_2 r_2^2 = (3.0)(1.5)^2 + (1.0)(0.50)^2 = (3.0)(2.25) + (1.0)(0.25) = 6.75 + 0.25 = 7.0 \text{ kg m}^2

Question 3

A solid sphere and a hollow sphere have the same mass $M$ and radius $R$ . Both are initially at rest. The same constant net torque is applied to each, causing them to rotate through the same total angular displacement. Which statement correctly compares their final states?

The solid sphere has a greater final angular velocity. (correct answer)
The hollow sphere has a greater final angular velocity.
Both spheres have the same final angular momentum.
Both spheres have the same final angular velocity.

Explanation: The work done by the torque is

W = \tau \theta

. Since both

\tau

and

\theta

are the same for both spheres, the work done on each is identical. By the work-energy theorem, they have the same final rotational kinetic energy (

KE_{rot}

). The formula for rotational kinetic energy is

KE_{rot} = \frac{1}{2}I\omega^2

. The moment of inertia of a solid sphere (

I_s = \frac{2}{5}MR^2

) is less than that of a hollow sphere (

I_h = \frac{2}{3}MR^2

). Since

KE_{rot}

is the same, the sphere with the smaller moment of inertia (the solid sphere) must have the greater angular velocity

\omega

Question 4

A block of mass $m$ hangs from a light string wrapped around a solid cylindrical pulley of mass $M$ and radius $R$ . The block is released from rest and accelerates downwards. What is the tension $T$ in the string? The moment of inertia of the pulley is $I = \frac{1}{2}MR^2$ .

$mg$
$mg \left( \frac{M}{2m+M} \right)$ (correct answer)
$mg \left( \frac{M}{m+M} \right)$
$mg \left( \frac{m}{m+M} \right)$

Explanation: For the hanging mass:

mg - T = ma

. For the pulley:

\tau = TR = I\alpha

. With

a = \alpha R

and

I = \frac{1}{2}MR^2

, we get

TR = (\frac{1}{2}MR^2)(\frac{a}{R})

, which simplifies to

T = \frac{1}{2}Ma

. Substituting this expression for T into the first equation:

mg - \frac{1}{2}Ma = ma

. Solving for acceleration

a

mg = a(m + \frac{M}{2}) \implies a = \frac{mg}{m + M/2}

. Now substitute this

a

back into the equation for tension:

T = \frac{1}{2}Ma = \frac{1}{2}M \left( \frac{mg}{m+M/2} \right) = mg \frac{M/2}{m+M/2} = mg \left( \frac{M}{2m+M} \right)

Question 5

A uniform rod of mass 2.0 kg and length 1.2 m rotates about its center with an angular velocity of 5.0 rad s⁻¹. What is its rotational kinetic energy? The moment of inertia of a rod about its center is $I = \frac{1}{12}ML^2$ .

3.0 J (correct answer)
6.0 J
9.0 J
12 J

Explanation: First, calculate the moment of inertia:

I = \frac{1}{12}ML^2 = \frac{1}{12}(2.0 \text{ kg})(1.2 \text{ m})^2 = \frac{1}{12}(2.0)(1.44) = 0.24 \text{ kg m}^2

. Next, calculate the rotational kinetic energy using

KE_{rot} = \frac{1}{2}I\omega^2

KE_{rot} = \frac{1}{2}(0.24 \text{ kg m}^2)(5.0 \text{ rad s}^{-1})^2 = \frac{1}{2}(0.24)(25) = 3.0 \text{ J}

Question 6

A solid cylinder, a hollow cylinder, and a solid sphere, all having the same mass M and radius R, are released from rest at the top of the same incline. They all roll without slipping. In which order do they reach the bottom?

Solid sphere, solid cylinder, hollow cylinder (correct answer)
Hollow cylinder, solid cylinder, solid sphere
Solid cylinder, solid sphere, hollow cylinder
They all reach the bottom at the same time

Explanation: The object with the greatest translational acceleration will reach the bottom first. The acceleration depends on the moment of inertia, I. A smaller I means less of the initial potential energy is converted to rotational kinetic energy and more is converted to translational kinetic energy, resulting in a higher final speed and faster time. The moments of inertia are:

I_{sphere} = \frac{2}{5}MR^2

I_{solid cyl} = \frac{1}{2}MR^2

I_{hollow cyl} = MR^2

. Since

I_{sphere} < I_{solid cyl} < I_{hollow cyl}

, the solid sphere will be the fastest, followed by the solid cylinder, and then the hollow cylinder.

Question 7

A disk is rotating with angular velocity $\omega$ . A constant frictional torque $\tau$ is applied, bringing the disk to rest. Which expression represents the total angular displacement $\Delta\theta$ of the disk before it stops? Let $I$ be the moment of inertia of the disk.

$\frac{I\omega^2}{2\tau}$ (correct answer)
$\frac{I\omega}{\tau}$
$\frac{2\tau}{I\omega^2}$
$\frac{\tau}{I\omega}$

Explanation: We can use the work-energy theorem for rotation:

W = \Delta KE_{rot}

. The work done by the frictional torque is

W = -\tau \Delta\theta

(negative because it opposes the motion). The change in kinetic energy is

\Delta KE_{rot} = KE_f - KE_i = 0 - \frac{1}{2}I\omega^2

. Setting them equal:

-\tau \Delta\theta = -\frac{1}{2}I\omega^2

. Solving for

\Delta\theta

gives

\Delta\theta = \frac{I\omega^2}{2\tau}

Question 8

A constant torque of 12 N m is applied to a flywheel, causing its angular velocity to increase from 20 rad s⁻¹ to 50 rad s⁻¹. The flywheel has a moment of inertia of 6.0 kg m². What is the duration of the applied torque?

2.5 s
5.0 s
15 s (correct answer)
35 s

Explanation: The relationship between torque, moment of inertia, and angular acceleration is

\tau = I\alpha

. The angular acceleration can be found from

\alpha = \frac{\Delta\omega}{\Delta t}

. Substituting this into the torque equation gives

\tau = I \frac{\Delta\omega}{\Delta t}

. This is the angular impulse equation

\tau \Delta t = I \Delta \omega = \Delta L

. Rearranging to solve for

\Delta t

\Delta t = \frac{I \Delta \omega}{\tau} = \frac{(6.0 \text{ kg m}^2)(50 \text{ rad s}^{-1} - 20 \text{ rad s}^{-1})}{12 \text{ N m}} = \frac{(6.0)(30)}{12} = \frac{180}{12} = 15 \text{ s}

Question 9

A torque $\tau$ applied to an object for a time $\Delta t$ causes its angular momentum to change by $\Delta L$ . If the same torque is applied for a time $2\Delta t$ , what is the change in the object's rotational kinetic energy? Assume the object starts from rest.

It doubles.
It triples.
It quadruples. (correct answer)
It increases by a factor of $\sqrt{2}$ .

Explanation: The angular impulse is

\tau \Delta t = \Delta L

. If the time is doubled to

2\Delta t

, the new change in angular momentum is

\Delta L' = \tau (2\Delta t) = 2(\tau \Delta t) = 2\Delta L

. Since the object starts from rest, its final angular momentum is

L' = 2L

. Rotational kinetic energy is related to angular momentum by

KE = \frac{L^2}{2I}

. The initial KE was

KE = \frac{L^2}{2I}

. The new KE is

KE' = \frac{(L')^2}{2I} = \frac{(2L)^2}{2I} = \frac{4L^2}{2I} = 4 \times KE

. Therefore, the kinetic energy quadruples.

Question 10

A child of mass 40 kg stands at the edge of a merry-go-round of radius 2.0 m and moment of inertia 160 kg m², which is rotating at 1.0 rad s⁻¹. The child then walks to the center of the merry-go-round. What is the final angular velocity of the system? Treat the child as a point mass.

1.0 rad s⁻¹
1.5 rad s⁻¹
2.0 rad s⁻¹ (correct answer)
2.5 rad s⁻¹

Explanation: Angular momentum is conserved. Initially, the total moment of inertia is

I_i = I_{mgr} + I_{child} = 160 + mr^2 = 160 + (40)(2.0)^2 = 160 + 160 = 320 \text{ kg m}^2

. The initial angular momentum is

L_i = I_i \omega_i = (320)(1.0) = 320 \text{ kg m}^2 \text{s}^{-1}

. When the child is at the center, their distance from the axis is r=0, so their moment of inertia is zero. The final moment of inertia is just that of the merry-go-round,

I_f = 160 \text{ kg m}^2

. By conservation of angular momentum,

L_f = L_i

, so

I_f \omega_f = 320

. The final angular velocity is

\omega_f = \frac{320}{160} = 2.0 \text{ rad s}^{-1}

Question 11

A figure skater is spinning with an initial angular velocity $\omega_0$ and moment of inertia $I_0$ . She pulls her arms in, reducing her moment of inertia to $I_0/3$ . What is the ratio of her final rotational kinetic energy to her initial rotational kinetic energy?

1/3
1
3 (correct answer)
9

Explanation: By conservation of angular momentum,

L_i = L_f

, so

I_0\omega_0 = I_f\omega_f

. With

I_f = I_0/3

, her final angular velocity is

\omega_f = \frac{I_0\omega_0}{I_0/3} = 3\omega_0

. The initial kinetic energy is

KE_i = \frac{1}{2}I_0\omega_0^2

. The final kinetic energy is

KE_f = \frac{1}{2}I_f\omega_f^2 = \frac{1}{2}(\frac{I_0}{3})(3\omega_0)^2 = \frac{1}{2}\frac{I_0}{3}(9\omega_0^2) = 3(\frac{1}{2}I_0\omega_0^2) = 3KE_i

. The ratio

KE_f/KE_i

is 3. The extra energy comes from the work the skater does to pull her arms in.

Question 12

A constant torque of 5.0 N m is applied to a grinding wheel with a moment of inertia of 0.20 kg m², initially at rest. What is the rotational kinetic energy of the wheel after it has completed 10 revolutions?

50 J
100 J
157 J
314 J (correct answer)

Explanation: The work-energy theorem for rotation states that the work done by the net torque equals the change in rotational kinetic energy,

W = \Delta KE_{rot}

. The work done is

W = \tau \theta

. First, the angular displacement

\theta

must be in radians:

\theta = 10 \text{ rev} \times 2\pi \text{ rad/rev} = 20\pi \text{ rad}

. Then,

W = (5.0 \text{ N m})(20\pi \text{ rad}) = 100\pi \text{ J}

. Since the wheel starts from rest, its change in kinetic energy is equal to its final kinetic energy.

KE_{final} = 100\pi \approx 314 \text{ J}

Question 13

A uniform rod of length 2.0 m is pivoted at its center. A force of 10 N is applied 0.50 m from the pivot at an angle of 30° to the rod, tending to cause a clockwise rotation. Another force of 5.0 N is applied at one end of the rod, perpendicular to it, tending to cause a counter-clockwise rotation. What is the magnitude of the net torque about the pivot?

0 N m
2.5 N m (correct answer)
4.3 N m
7.5 N m

Explanation: Torque is calculated as

\tau = Fr \sin \theta

. The counter-clockwise torque from the 5.0 N force is

\tau_{ccw} = (5.0 \text{ N})(1.0 \text{ m})\sin(90^\circ) = 5.0 \text{ N m}

. The clockwise torque from the 10 N force is

\tau_{cw} = (10 \text{ N})(0.50 \text{ m})\sin(30^\circ) = (10)(0.50)(0.5) = 2.5 \text{ N m}

. The net torque is the difference between the two, as they act in opposite directions:

\tau_{net} = \tau_{ccw} - \tau_{cw} = 5.0 - 2.5 = 2.5 \text{ N m}

Question 14

A spinning solid sphere has a mass of 5.0 kg, a radius of 0.10 m, and an angular velocity of 20 rad s⁻¹. What is the magnitude of its angular momentum? The moment of inertia of a solid sphere is $I = \frac{2}{5}MR^2$ .

0.40 kg m² s⁻¹ (correct answer)
0.50 kg m² s⁻¹
1.0 kg m² s⁻¹
10 kg m² s⁻¹

Explanation: First, calculate the moment of inertia:

I = \frac{2}{5}MR^2 = \frac{2}{5}(5.0 \text{ kg})(0.10 \text{ m})^2 = (2.0)(0.01) = 0.020 \text{ kg m}^2

. Then, calculate the angular momentum using

L = I\omega

L = (0.020 \text{ kg m}^2)(20 \text{ rad s}^{-1}) = 0.40 \text{ kg m}^2 \text{s}^{-1}

Question 15

A horizontal turntable with moment of inertia $I$ rotates freely with angular velocity $\omega_0$ . A small piece of putty of mass $m$ is dropped from rest and sticks to the turntable at a distance $r$ from the axis of rotation. What is the new angular velocity of the system?

$\frac{I \omega_0}{I+mr^2}$ (correct answer)
$\frac{(I+mr^2)\omega_0}{I}$
$\frac{I \omega_0}{I-mr^2}$
$\omega_0$

Explanation: Since there are no external torques acting on the turntable-putty system, its total angular momentum is conserved. The initial angular momentum is

L_i = I\omega_0

(the putty has zero initial angular momentum). The final moment of inertia of the system is

I_f = I + mr^2

, where

mr^2

is the moment of inertia of the putty. The final angular momentum is

L_f = I_f \omega_f = (I+mr^2)\omega_f

. Setting

L_i = L_f

, we get

I\omega_0 = (I+mr^2)\omega_f

. Solving for the final angular velocity gives

\omega_f = \frac{I\omega_0}{I+mr^2}

Question 16

A solid cylinder of mass $M$ and radius $R$ has a moment of inertia $I = \frac{1}{2}MR^2$ . It is rotating with angular velocity $\omega$ . What is the ratio of its rotational kinetic energy to its translational kinetic energy if it were moving with a translational speed $v = \omega R$ ?

1/4
1/2 (correct answer)
1
2

Explanation: The rotational kinetic energy is

KE_{rot} = \frac{1}{2}I\omega^2 = \frac{1}{2}(\frac{1}{2}MR^2)\omega^2 = \frac{1}{4}MR^2\omega^2

. The translational kinetic energy is

KE_{trans} = \frac{1}{2}Mv^2

. Substituting

v = \omega R

, we get

KE_{trans} = \frac{1}{2}M(\omega R)^2 = \frac{1}{2}MR^2\omega^2

. The ratio is

\frac{KE_{rot}}{KE_{trans}} = \frac{\frac{1}{4}MR^2\omega^2}{\frac{1}{2}MR^2\omega^2} = \frac{1/4}{1/2} = 1/2

Question 17

A solid sphere starts from rest at the top of an incline of vertical height $h$ and rolls without slipping to the bottom. What is its translational speed at the bottom? The moment of inertia of a solid sphere is $I = \frac{2}{5}mr^2$ .

$\sqrt{2gh}$
$\sqrt{gh}$
$\sqrt{\frac{4}{3}gh}$
$\sqrt{\frac{10}{7}gh}$ (correct answer)

Explanation: By conservation of energy, the initial potential energy

mgh

is converted into translational and rotational kinetic energy:

mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2

. For rolling without slipping,

\omega = v/r

. Substituting this and

I = \frac{2}{5}mr^2

mgh = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mr^2)(\frac{v}{r})^2 = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2

. Solving for

v

, we get

v^2 = \frac{10}{7}gh

, so

v = \sqrt{\frac{10}{7}gh}

Question 18

A solid cylinder of mass $M$ and radius $R$ is initially spinning with angular velocity $\omega_0$ about its central axis while moving with linear velocity $v_0$ along a rough horizontal surface. The coefficient of kinetic friction is $\mu$ . After what time will the cylinder begin to roll without slipping?

$\frac{v_0 - \omega_0 R}{3\mu g}$ (correct answer)
$\frac{v_0 + \omega_0 R}{3\mu g}$
$\frac{2(v_0 - \omega_0 R)}{3\mu g}$
$\frac{v_0 - \omega_0 R}{2\mu g}$

Explanation: The friction force

f = \mu Mg

opposes the relative motion at the contact point. Initially, the contact point moves with velocity

v_0 - \omega_0 R

. If this is positive, friction acts backward on translation and forward on rotation. The linear acceleration is

a = -\mu g

and angular acceleration is

\alpha = \frac{\mu MgR}{I} = \frac{\mu MgR}{\frac{1}{2}MR^2} = \frac{2\mu g}{R}

. The velocities become:

v(t) = v_0 - \mu gt

and

\omega(t) = \omega_0 + \frac{2\mu gt}{R}

. Rolling without slipping occurs when

v(t) = \omega(t)R

v_0 - \mu gt = (\omega_0 + \frac{2\mu gt}{R})R = \omega_0 R + 2\mu gt

. Solving:

v_0 - \omega_0 R = 3\mu gt

, so

t = \frac{v_0 - \omega_0 R}{3\mu g}

Question 19

A yo-yo consists of two uniform disks of mass $M$ and radius $R$ connected by a thin axle of mass $m$ and radius $r$ . The yo-yo is released from rest with the string wound around the axle. As it falls and unwinds, what is the ratio of its rotational kinetic energy to its translational kinetic energy?

$\frac{MR^2 + mr^2}{(M + m)r^2}$
$\frac{2MR^2 + mr^2}{2(M + m)r^2}$ (correct answer)
$\frac{MR^2}{(M + m)r^2}$
$\frac{MR^2 + \frac{mr^2}{2}}{(M + m)r^2}$

Explanation: The total moment of inertia about the center is

I = 2 \cdot \frac{1}{2}MR^2 + \frac{1}{2}mr^2 = MR^2 + \frac{mr^2}{2}

. For a yo-yo, the constraint is

v = r\omega

(the string unwinds at radius

r

). The rotational kinetic energy is

KE_{rot} = \frac{1}{2}I\omega^2 = \frac{1}{2}\left(MR^2 + \frac{mr^2}{2}\right)\omega^2

. The translational kinetic energy is

KE_{trans} = \frac{1}{2}(M + m)v^2 = \frac{1}{2}(M + m)r^2\omega^2

. The ratio is

\frac{KE_{rot}}{KE_{trans}} = \frac{\frac{1}{2}\left(MR^2 + \frac{mr^2}{2}\right)\omega^2}{\frac{1}{2}(M + m)r^2\omega^2} = \frac{MR^2 + \frac{mr^2}{2}}{(M + m)r^2} = \frac{2MR^2 + mr^2}{2(M + m)r^2}

Question 20

A uniform rod of mass $M$ and length $L$ lies on a frictionless horizontal surface. A bullet of mass $m$ moving with velocity $v$ strikes the rod at a distance $\frac{L}{3}$ from one end, perpendicular to the rod's length, and embeds in it. What is the velocity of the center of mass of the rod immediately after the collision?

$\frac{3mv}{2(M + m)}$
$\frac{mv}{M + m} \cdot \frac{L}{3}$
$\frac{mv}{M + m}$ (correct answer)
$\frac{mv}{M + m} \cdot \frac{2L}{3}$

Explanation: When you see a collision problem involving embedding (a perfectly inelastic collision), the key principle is conservation of momentum. However, you need to carefully distinguish between linear momentum of the entire system and the motion of specific parts. Since the bullet embeds in the rod, this is a perfectly inelastic collision where linear momentum is conserved for the entire system. Before collision, only the bullet has momentum:

mv

. After collision, both the bullet and rod move together as a single system with total mass

(M + m)

. Using conservation of linear momentum:

mv = (M + m)v_{cm}

, where

v_{cm}

is the velocity of the center of mass of the entire system. Solving:

v_{cm} = \frac{mv}{M + m}

. This is answer C. The wrong answers represent common misconceptions about this type of problem. Answer A (

\frac{3mv}{2(M + m)}

) incorrectly attempts to account for the collision location by multiplying by a factor, but the center of mass velocity doesn't depend on where the collision occurs. Answer B (

\frac{mv}{M + m} \cdot \frac{L}{3}

) and Answer D (

\frac{mv}{M + m} \cdot \frac{2L}{3}

) make the dimensional error of multiplying a velocity by a length, which gives units of

\frac{length^2}{time}

, not velocity. These also incorrectly assume the collision location affects the center of mass velocity. Remember: In collision problems, the velocity of the center of mass depends only on total momentum and total mass, never on where the collision occurs on the object. The collision location affects rotation, not translation.