IB Physics Quiz: Understand Rigid Body Mechanics

20 questions · exam conditions

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Understand Rigid Body MechanicsQuestion 1 of 20

A force of 50 N is applied to the end of a wrench of length 20 cm. The force is applied at an angle of 60° to the handle of the wrench. What is the magnitude of the torque on the bolt?

5.0 N m

8.7 N m

10 N m

870 N m

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IB Physics Quiz

IB Physics Quiz: Understand Rigid Body Mechanics

Practice Understand 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 Understand 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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Question 1

A force of 50 N is applied to the end of a wrench of length 20 cm. The force is applied at an angle of 60° to the handle of the wrench. What is the magnitude of the torque on the bolt?

5.0 N m
8.7 N m (correct answer)
10 N m
870 N m

Explanation: The magnitude of the torque is given by the formula

\tau = Fr\sin\theta

, where

r

is the length of the lever arm and

\theta

is the angle between the force vector and the lever arm vector. First, convert the length to SI units:

r = 20 \text{ cm} = 0.20 \text{ m}

. Now, calculate the torque:

\tau = (50 \text{ N})(0.20 \text{ m})\sin(60°)

. Since

\sin(60°) = \frac{\sqrt{3}}{2} \approx 0.866

, the torque is

\tau = 10 \times 0.866 = 8.66 \text{ N m}

, which is approximately 8.7 N m.

Question 2

A flywheel with moment of inertia $I$ is accelerated from rest to a final angular velocity $\omega$ in a time $t$ by a constant torque. What is the total angle in radians through which the flywheel turns during this time?

$\omega t$
$\frac{1}{2}\omega t$ (correct answer)
$\frac{\omega}{t}$
$2\omega t$

Explanation: We can use the rotational kinematic equation

\theta = \frac{1}{2}(\omega_0 + \omega)t

. Since the flywheel starts from rest, the initial angular velocity

\omega_0 = 0

. Therefore, the angle turned is

\theta = \frac{1}{2}(0 + \omega)t = \frac{1}{2}\omega t

. Alternatively, one could find the constant angular acceleration

\alpha = \frac{\omega - \omega_0}{t} = \frac{\omega}{t}

and then use

\theta = \omega_0 t + \frac{1}{2}\alpha t^2 = 0 + \frac{1}{2}(\frac{\omega}{t})t^2 = \frac{1}{2}\omega t

Question 3

A solid sphere and a hollow sphere have the same mass and the same radius. They are released from rest at the top of an inclined plane and roll down without slipping. Which statement is correct?

Both spheres reach the bottom at the same time.
The solid sphere reaches the bottom first. (correct answer)
The hollow sphere reaches the bottom first.
Which sphere reaches first depends on their mass.

Explanation: For a rolling object, gravitational potential energy is converted into both translational and rotational kinetic energy. The moment of inertia of a solid sphere (

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

) is less than that of a hollow sphere (

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

). A smaller moment of inertia means that for a given amount of potential energy lost, a larger fraction is converted into translational kinetic energy (and less into rotational kinetic energy). A greater translational kinetic energy means a greater linear speed. Therefore, the solid sphere will have a greater linear acceleration down the ramp and will reach the bottom first.

Question 4

A yo-yo can be modelled as a solid cylinder of mass $m$ and radius $r$ with a moment of inertia $I=\frac{1}{2}mr^2$ . It is released from rest and unwinds from a vertical string without slipping. What is the downward linear acceleration of its centre of mass?

$g$
$\frac{1}{2}g$
$\frac{2}{3}g$ (correct answer)
$\frac{1}{3}g$

Explanation: Let

T

be the tension in the string and

a

be the linear acceleration. The linear motion is described by Newton's second law:

mg - T = ma

. The rotational motion is described by

\tau = I\alpha

, where the torque is provided by the tension:

Tr = I\alpha

. For rolling without slipping,

a = \alpha r

, so

\alpha = a/r

. Substituting this gives

Tr = I(a/r)

, so

T = Ia/r^2

. Now substitute this expression for

T

into the linear equation:

mg - \frac{Ia}{r^2} = ma

. Substitute

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

mg - \frac{(\frac{1}{2}mr^2)a}{r^2} = ma

, which simplifies to

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

. Solving for

a

mg = \frac{3}{2}ma

, which gives

a = \frac{2}{3}g

Question 5

A thin hoop and a solid disk have the same mass $M$ and outer radius $R$ . They are rotated about an axis passing through their centres and perpendicular to their planes. Which statement correctly compares their moments of inertia ( $I_{hoop}$ and $I_{disk}$ ) and the torques ( $\tau_{hoop}$ and $\tau_{disk}$ ) required to produce the same non-zero angular acceleration $\alpha$ ?

$I_{hoop} > I_{disk}$ and $\tau_{hoop} > \tau_{disk}$ (correct answer)
$I_{hoop} < I_{disk}$ and $\tau_{hoop} < \tau_{disk}$
$I_{hoop} > I_{disk}$ and $\tau_{hoop} < \tau_{disk}$
$I_{hoop} = I_{disk}$ and $\tau_{hoop} = \tau_{disk}$

Explanation: The moment of inertia depends on the distribution of mass relative to the axis of rotation. For a hoop, all the mass is at radius

R

, so

I_{hoop} = MR^2

. For a solid disk, the mass is distributed from the centre to the radius

R

, so its moment of inertia is smaller:

I_{disk} = \frac{1}{2}MR^2

. Therefore,

I_{hoop} > I_{disk}

. From Newton's second law for rotation,

\tau = I\alpha

. To produce the same angular acceleration

\alpha

, the object with the greater moment of inertia requires the greater torque. Since

I_{hoop} > I_{disk}

, it follows that

\tau_{hoop} > \tau_{disk}

Question 6

A student sits at rest on a rotating stool holding a spinning bicycle wheel with its axis vertical. The wheel is spinning clockwise when viewed from above. The student then inverts the wheel so its axis is again vertical, but it spins counter-clockwise when viewed from above. What is the subsequent motion of the student and the stool?

They remain at rest.
They rotate counter-clockwise.
They rotate clockwise. (correct answer)
They rotate briefly then stop.

Explanation: The system consists of the student, the stool, and the wheel. The action of flipping the wheel is an internal torque, so the total angular momentum of the system must be conserved. Let the upward direction be positive. Initially, the wheel's angular momentum is negative (

-L_{wheel}

) and the student is at rest (

L_{student}=0

), so the total initial angular momentum is

L_{initial} = -L_{wheel}

. After inversion, the wheel's angular momentum is positive (

+L_{wheel}

). To conserve total angular momentum, the student and stool must acquire angular momentum

L_{final, student}

such that

L_{final, student} + (+L_{wheel}) = -L_{wheel}

. This gives

L_{final, student} = -2L_{wheel}

. The negative sign indicates that the student and stool will rotate in the original direction of the wheel's spin, which was clockwise.

Question 7

A wheel of radius $R$ rolls without slipping on a horizontal surface. The centre of the wheel moves with a constant linear speed $v$ . What is the instantaneous speed, relative to the surface, of a point on the very top of the wheel?

0
$v$
$2v$ (correct answer)
$v\sqrt{2}$

Explanation: The motion of any point on the wheel is the superposition of the translational motion of the centre of mass and the rotational motion about the centre of mass. The translational speed of every point is

v

. The tangential speed of a point on the rim due to rotation is

v_{tan} = \omega R

. The condition for rolling without slipping is

v = \omega R

, so

v_{tan} = v

. At the very top of the wheel, the translational velocity vector and the tangential velocity vector both point in the same direction (forward). Therefore, the total speed relative to the surface is the sum of these speeds:

v_{total} = v + v_{tan} = v + v = 2v

Question 8

A uniform block of height H and square base of side length W is placed on a plane. The plane is slowly tilted by an angle θ. Assuming the block does not slide, what is tan(θ) when the block is on the verge of toppling?

$\frac{W}{H}$ (correct answer)
$\frac{H}{W}$
$\frac{W}{2H}$
$\frac{H}{2W}$

Explanation: The block is on the verge of toppling when its centre of mass is vertically directly above the pivot point, which is the lower edge of its base of support. The centre of mass of a uniform block is at its geometric centre, at a height of

H/2

from the base and a horizontal distance of

W/2

from the edge. When the plane is tilted by an angle

\theta

, the angle between the vertical line through the centre of mass and the line perpendicular to the base is also

\theta

. A right-angled triangle is formed with the adjacent side

H/2

and the opposite side

W/2

. From trigonometry,

\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{W/2}{H/2} = \frac{W}{H}

Question 9

A horizontal turntable with moment of inertia $I$ rotates freely with angular velocity $\omega$ . A piece of clay of mass $m$ is dropped from rest and sticks to the turntable at a distance $r$ from the axis. After the clay sticks, a brake applies a constant external torque to stop the rotation. How does the angular momentum of the turntable-clay system change, first during the collision with the clay, and then during the application of the brake?

Decreases during collision; Decreases during braking
Is conserved during collision; Is conserved during braking
Decreases during collision; Is conserved during braking
Is conserved during collision; Decreases during braking (correct answer)

Explanation: During the collision, the forces between the clay and the turntable are internal to the system. Assuming no external torques act on the system during this brief period, the total angular momentum of the turntable-clay system is conserved. After the collision, the brake applies an external torque. An external torque causes a change in the system's angular momentum, according to

\Delta L = \tau_{ext} \Delta t

. Since the brake opposes the motion, the torque is retarding, and the angular momentum of the system decreases until it becomes zero.

Question 10

A solid ball of mass $M$ rolls without slipping along a horizontal surface with speed $v$ . It makes a perfectly inelastic collision with a stationary block of mass $M$ . After the collision, the combined object slides without rotating. What is the speed of the combined object?

$v$
$v/2$ (correct answer)
$v/4$
$7v/10$

Explanation: In this collision, we should consider the conservation of linear momentum for the system (ball + block) in the horizontal direction, as there are no net external horizontal forces. Mechanical energy is not conserved because the collision is perfectly inelastic. The fact that the ball was initially rolling is relevant for its energy, but not for its linear momentum. The initial linear momentum of the system is just that of the ball,

P_i = Mv

. The block is stationary. After the collision, the two objects stick together, forming a combined mass of (2M) moving with a final speed

v_f

. The final linear momentum is

P_f = (2M)v_f

. By conservation of linear momentum,

P_i = P_f

, so

Mv = (2M)v_f

. Solving for

v_f

gives

v_f = \frac{Mv}{2M} = v/2

Question 11

A uniform ladder of mass $M$ and length $L$ leans against a frictionless vertical wall at an angle $\theta$ to the horizontal ground. The ground is rough. Which equation represents the condition for rotational equilibrium about the point where the ladder touches the ground?

$Mg\frac{L}{2}\cos\theta = N_{wall}L\sin\theta$ (correct answer)
$Mg\frac{L}{2}\sin\theta = N_{wall}L\cos\theta$
$MgL\cos\theta = N_{wall}L\sin\theta$
$Mg\frac{L}{2}\cos\theta = N_{wall}L\cos\theta$

Explanation: For rotational equilibrium, the net torque about any point must be zero. Choosing the pivot at the base of the ladder simplifies the calculation, as the normal force from the ground and the friction force exert no torque. The weight of the ladder

Mg

acts downwards at its centre of mass, at a distance

L/2

from the base. The horizontal distance from the pivot to the line of action of the weight is the lever arm, which is

(\frac{L}{2})\cos\theta

. This creates a clockwise torque of

\tau_{cw} = Mg\frac{L}{2}\cos\theta

. The normal force from the wall,

N_{wall}

, acts horizontally at the top of the ladder. Its lever arm is the vertical distance from the pivot, which is

L\sin\theta

. This creates a counter-clockwise torque of

\tau_{ccw} = N_{wall}L\sin\theta

. Setting clockwise torques equal to counter-clockwise torques gives

Mg\frac{L}{2}\cos\theta = N_{wall}L\sin\theta

Question 12

A uniform rod of length $L$ and mass $M$ is pivoted at its centre. A small object of mass $m$ is placed on the rod at a distance $x$ from the pivot. The rod is released from a horizontal position. What is the initial angular acceleration of the rod?

$\frac{mgx}{\frac{1}{12}ML^2}$
$\frac{mgx}{\frac{1}{3}ML^2 + mx^2}$
$\frac{mgx}{\frac{1}{12}ML^2 + mx^2}$ (correct answer)
$\frac{(M+m)gx}{\frac{1}{12}ML^2 + mx^2}$

Explanation: The initial torque on the system about the pivot is caused by the weight of the small object

m

, so

\tau = mgx

. The weight of the rod

Mg

acts at the pivot, so it produces no torque. The total moment of inertia of the system is the sum of the moment of inertia of the rod about its centre (

I_{rod} = \frac{1}{12}ML^2

) and the moment of inertia of the point mass

m

(

I_{mass} = mx^2

). So,

I_{total} = \frac{1}{12}ML^2 + mx^2

. Using Newton's second law for rotation,

\tau = I\alpha

, the initial angular acceleration is

\alpha = \frac{\tau}{I_{total}} = \frac{mgx}{\frac{1}{12}ML^2 + mx^2}

Question 13

A solid sphere of mass $m$ and radius $r$ rolls on a horizontal surface with a linear speed $v$ . It then rolls up an incline without slipping. What is the maximum vertical height $h$ it reaches? The moment of inertia of a solid sphere is $I = \frac{2}{5}mr^2$ .

$\frac{v^2}{2g}$
$\frac{3v^2}{5g}$
$\frac{v^2}{g}$
$\frac{7v^2}{10g}$ (correct answer)

Explanation: By conservation of mechanical energy, the initial total kinetic energy on the horizontal surface will be equal to the final gravitational potential energy at the maximum height. The initial total kinetic energy is the sum of translational and rotational kinetic energy:

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

. For rolling without slipping,

\omega = v/r

. Substituting for

I

and

\omega

E_k = \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

. The final potential energy is

E_p = mgh

. Setting

E_k = E_p

, we have

\frac{7}{10}mv^2 = mgh

. Solving for

h

gives

h = \frac{7v^2}{10g}

Question 14

Four identical point masses $m$ are placed at the corners of a square with side length $a$ . What is the moment of inertia of the system about an axis perpendicular to the square and passing through its centre?

$ma^2$
$\sqrt{2}ma^2$
$4ma^2$
$2ma^2$ (correct answer)

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

I = \sum m_i r_i^2

, where

r_i

is the perpendicular distance of each mass from the axis of rotation. The centre of the square is the origin (0,0). The corners are at distances

x=\pm a/2

and

y=\pm a/2

. The distance

r

from the centre to any corner is the same. Using the Pythagorean theorem,

r^2 = (a/2)^2 + (a/2)^2 = a^2/4 + a^2/4 = a^2/2

. Since all four masses are identical and at the same distance from the axis, the total moment of inertia is

I = 4 \times mr^2 = 4m(a^2/2) = 2ma^2

Question 15

A solid disk of mass $M$ and radius $R$ is free to rotate about its fixed central axis. The moment of inertia of the disk is $I = \frac{1}{2}MR^2$ . A string is wrapped around its rim and pulled with a constant tangential force $F$ . What is the angular acceleration of the disk?

$\frac{F}{MR}$
$\frac{2F}{MR}$ (correct answer)
$\frac{F}{MR^2}$
$\frac{2F}{M}$

Explanation: The torque applied to the disk is

\tau = F \times R

since the force is tangential. According to Newton's second law for rotation,

\tau = I\alpha

. Substituting the expressions for torque and moment of inertia, we get

FR = (\frac{1}{2}MR^2)\alpha

. Solving for the angular acceleration

\alpha

, we find

\alpha = \frac{FR}{\frac{1}{2}MR^2} = \frac{2F}{MR}

Question 16

A bicycle wheel with moment of inertia 0.50 kg m² is spinning at an initial angular velocity of 10 rad s⁻¹. A constant braking torque is applied, bringing the wheel to rest in 2.5 s. What is the magnitude of the angular impulse applied to the wheel?

2.0 N m s
4.0 N m s
12.5 N m s
5.0 N m s (correct answer)

Explanation: Angular impulse is equal to the change in angular momentum,

\Delta L

. The initial angular momentum is

L_i = I \omega_i = (0.50 \text{ kg m}^2)(10 \text{ rad s}^{-1}) = 5.0 \text{ kg m}^2 \text{s}^{-1}

. The final angular momentum is

L_f = I \omega_f = (0.50)(0) = 0

. The change in angular momentum is

\Delta L = L_f - L_i = 0 - 5.0 = -5.0 \text{ kg m}^2 \text{s}^{-1}

. The magnitude of the angular impulse is therefore 5.0 N m s. The time taken to stop is not needed to find the angular impulse if initial and final states are known, but could be used to find the torque:

\tau = \Delta L / \Delta t = -5.0 / 2.5 = -2.0 \text{ N m}

Question 17

A disk is spinning about a horizontal axle oriented along an east-west axis. Viewed from the east, the disk spins clockwise. The axle is supported at both ends. The eastern support is then removed. What is the initial motion of the free eastern end of the axle due to precession?

It moves vertically downwards.
It moves horizontally towards the north. (correct answer)
It moves horizontally towards the south.
It remains stationary horizontally but starts to fall.

Explanation: This is a problem of gyroscopic precession. First, determine the direction of the angular momentum vector

\vec{L}

. Using the right-hand rule for a clockwise spin viewed from the east, the

\vec{L}

vector points to the west. When the eastern support is removed, the force of gravity acts downwards at the center of mass, creating a torque

\vec{\tau}

about the western support (the pivot). The torque vector is given by

\vec{\tau} = \vec{r} \times \vec{F}

, where

\vec{r}

points east (from pivot to center of mass) and

\vec{F}

points down. Using the right-hand rule for the cross product,

\vec{\tau}

points horizontally to the north. The torque causes a change in angular momentum,

\Delta \vec{L}

, in the same direction as the torque. So,

\Delta \vec{L}

points north. The new angular momentum is

\vec{L}_{new} = \vec{L}_{old} + \Delta \vec{L}

. The axle will precess in the direction of the torque, so the eastern end moves horizontally to the north.

Question 18

A figure skater is spinning with angular velocity $\omega$ and has a moment of inertia $I$ . Their rotational kinetic energy is $E_k$ . They pull their arms in, reducing their moment of inertia to $I/3$ . What is the new rotational kinetic energy in terms of $E_k$ ?

$E_k/3$
$E_k$
$3E_k$ (correct answer)
$9E_k$

Explanation: Since there is no external torque, angular momentum is conserved. Let the initial state be 1 and final state be 2.

L_1 = L_2

, so

I_1\omega_1 = I_2\omega_2

. We are given

I_1 = I

\omega_1 = \omega

, and

I_2 = I/3

. Therefore,

I\omega = (I/3)\omega_2

, which gives

\omega_2 = 3\omega

. The initial kinetic energy is

E_k = E_{k1} = \frac{1}{2}I\omega^2

. The final kinetic energy is

E_{k2} = \frac{1}{2}I_2\omega_2^2 = \frac{1}{2}(\frac{I}{3})(3\omega)^2 = \frac{1}{2}(\frac{I}{3})(9\omega^2) = 3(\frac{1}{2}I\omega^2) = 3E_k

Question 19

A uniform solid cylinder rolls down a ramp and then moves across a horizontal surface with coefficient of kinetic friction $\mu$ . If the cylinder has initial translational velocity $v_0$ when it reaches the horizontal surface, what distance will it travel before coming to rest?

$\frac{v_0^2}{2\mu g}$
$\frac{3v_0^2}{4\mu g}$ (correct answer)
$\frac{v_0^2}{\mu g}$
$\frac{2v_0^2}{3\mu g}$

Explanation: For a rolling cylinder, the total kinetic energy is

KE = \frac{1}{2}mv_0^2 + \frac{1}{2}I\omega_0^2 = \frac{1}{2}mv_0^2 + \frac{1}{2} \cdot \frac{1}{2}mR^2 \cdot \frac{v_0^2}{R^2} = \frac{1}{2}mv_0^2 + \frac{1}{4}mv_0^2 = \frac{3}{4}mv_0^2

. The work done by friction is

W = \mu mg \cdot d

, where

d

is the distance traveled. By the work-energy theorem:

\mu mgd = \frac{3}{4}mv_0^2

. Solving for

d

d = \frac{3v_0^2}{4\mu g}

Question 20

Two identical uniform rods, each of mass $m$ and length $L$ , are connected at their ends to form an L-shape. The system rotates about an axis through the corner, perpendicular to the plane containing both rods. If the system starts from rest and a constant torque $\tau_0$ is applied, what is the angular velocity after the system has rotated through angle $\theta$ ?

$\sqrt{\frac{3\tau_0\theta}{mL^2}}$ (correct answer)
$\sqrt{\frac{6\tau_0\theta}{mL^2}}$
$\sqrt{\frac{\tau_0\theta}{mL^2}}$
$\sqrt{\frac{9\tau_0\theta}{2mL^2}}$

Explanation: The moment of inertia of each rod about the corner is

I_{rod} = \frac{1}{3}mL^2

(about one end). The total moment of inertia is

I_{total} = 2 \times \frac{1}{3}mL^2 = \frac{2}{3}mL^2

. Using the work-energy theorem:

W = \Delta KE

, where

W = \tau_0 \theta

and

\Delta KE = \frac{1}{2}I_{total}\omega^2 - 0 = \frac{1}{2} \cdot \frac{2}{3}mL^2 \cdot \omega^2 = \frac{1}{3}mL^2\omega^2

. Therefore:

\tau_0\theta = \frac{1}{3}mL^2\omega^2

, which gives

\omega = \sqrt{\frac{3\tau_0\theta}{mL^2}}