Rotational Motion MCQs with Answers and Detailed Explanations

Angular displacement describes how much an object rotates about an axis.

It is commonly written as \(\theta\) and measured in radians.

For a point at distance \(r\) from the axis, linear displacement is related to angular displacement by:

\[
s=r\theta
\]

Here, \(s\) is arc length. Therefore, angular displacement plays the same role in rotation that linear displacement plays in straight-line motion.

One complete revolution equals:

\[
2\pi\text{ radians}
\]

Therefore, the number of revolutions is:

\[
N=\frac{6\pi}{2\pi}
\]

\[
N=3
\]

The wheel completes 3 full revolutions.

Tangential speed is related to angular speed by:

\[
v=r\omega
\]

Substitute the values:

\[
v=0.50\times8
\]

\[
v=4\text{ m/s}
\]

The point moves along the circular path with speed \(4\text{ m/s}\).

For constant angular acceleration:

\[
\omega=\omega_0+\alpha t
\]

The wheel starts from rest, so:

\[
\omega_0=0
\]

Therefore:

\[
\omega=0+(4)(3)
\]

\[
\omega=12\text{ rad/s}
\]

Angular acceleration changes angular velocity in the same way that linear acceleration changes linear velocity.

Torque is the turning effect of a force.

\[
\tau=rF\sin\theta
\]

The force is perpendicular to the wrench, so:

\[
\theta=90^\circ
\]

and:

\[
\sin90^\circ=1
\]

Therefore:

\[
\tau=(0.30)(20)
\]

\[
\tau=6\text{ N}\cdot\text{m}
\]

A perpendicular force produces maximum torque.

Moment of inertia is the rotational equivalent of mass.

For point masses:

\[
I=\sum mr^2
\]

It depends on mass and distance from the rotation axis.

Changing the axis also changes distances \(r\), so it changes moment of inertia.

However, angular speed does not appear in the expression for \(I\). Therefore, the moment of inertia remains unchanged if only angular speed changes.

For point masses:

\[
I=\sum mr^2
\]

Each mass contributes:

\[
I_1=2(0.50)^2
\]

\[
I_1=0.50\text{ kg}\cdot\text{m}^2
\]

There are two identical masses:

\[
I=2(0.50)
\]

\[
I=1.0\text{ kg}\cdot\text{m}^2
\]

The rod is light, so its mass and moment of inertia are ignored.

Rotational equilibrium means angular acceleration is zero.

Using rotational dynamics:

\[
\tau_{\text{net}}=I\alpha
\]

If:

\[
\alpha=0
\]

then:

\[
\tau_{\text{net}}=0
\]

The object may still rotate with constant angular velocity. Therefore, rotational equilibrium does not always mean the object is at rest.

For a thin ring:

\[
I_{\text{ring}}=MR^2
\]

For a solid disc:

\[
I_{\text{disc}}=\frac{1}{2}MR^2
\]

The ring has more mass located farther from the axis.

Since moment of inertia depends strongly on distance squared, the thin ring has the greater moment of inertia.

Rotational kinetic energy is:

\[
K_{\text{rot}}=\frac{1}{2}I\omega^2
\]

Substitute the values:

\[
K_{\text{rot}}=\frac{1}{2}(4)(6^2)
\]

\[
K_{\text{rot}}=2\times36
\]

\[
K_{\text{rot}}=72\text{ J}
\]

This is the rotational version of linear kinetic energy:

\[
K=\frac{1}{2}mv^2
\]

For a solid disc about its central axis,

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

\[
I=\frac{1}{2}(2)(0.50)^2
\]

\[
I=0.25\text{ kg}\cdot\text{m}^2
\]

Rotational Newton’s second law is

\[
\tau=I\alpha
\]

Therefore,

\[
\alpha=\frac{\tau}{I}
\]

\[
\alpha=\frac{3}{0.25}
\]

\[
\alpha=12\text{ rad/s}^2
\]

The disc accelerates in the direction of the applied torque.

For rolling without slipping,

\[
a=\frac{g\sin\theta}{1+\frac{I}{MR^2}}
\]

For a thin ring,

\[
I=MR^2
\]

So,

\[
a_{\text{ring}}=\frac{g\sin\theta}{2}
\]

For a solid disc,

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

So,

\[
a_{\text{disc}}=\frac{g\sin\theta}{1+\frac{1}{2}}
\]

\[
a_{\text{disc}}=\frac{2}{3}g\sin\theta
\]

Since

\[
\frac{2}{3}g\sin\theta>\frac{1}{2}g\sin\theta
\]

the solid disc reaches the bottom first. It has less rotational inertia relative to its mass.

For rolling without slipping,

\[
v_{\text{cm}}=R\omega
\]

Therefore,

\[
\omega=\frac{v_{\text{cm}}}{R}
\]

\[
\omega=\frac{5}{0.20}
\]

\[
\omega=25\text{ rad/s}
\]

The rolling condition connects translational motion and rotational motion. It is valid only when the contact point does not slide relative to the surface.

For a constant torque acting through angular displacement,

\[
W=\tau\theta
\]

Here,

\[
\tau=10\text{ N}\cdot\text{m}
\]

and

\[
\theta=4\text{ rad}
\]

Therefore,

\[
W=10\times4
\]

\[
W=40\text{ J}
\]

Radians are dimensionless in this formula. The work becomes rotational kinetic energy if no other energy transfer occurs.

Rotational power is

\[
P=\tau\omega
\]

Substitute the given values:

\[
P=8\times15
\]

\[
P=120\text{ W}
\]

This is the rotational form of the linear power relation

\[
P=Fv
\]

A larger torque or a larger angular speed produces greater power.

For a rolling body,

\[
a=\frac{g\sin\theta}{1+\frac{I}{MR^2}}
\]

For a solid sphere,

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

Therefore,

\[
a=\frac{g\sin\theta}{1+\frac{2}{5}}
\]

\[
a=\frac{g\sin\theta}{\frac{7}{5}}
\]

\[
a=\frac{5}{7}g\sin\theta
\]

Static friction provides the torque needed for rolling. It does not necessarily do mechanical work because the contact point is instantaneously at rest.

For the falling mass,

\[
mg-T=ma
\]

The pulley rotates because of the string tension.

\[
TR=I\alpha
\]

Since the string does not slip,

\[
a=R\alpha
\]

Thus,

\[
T=\frac{Ia}{R^2}
\]

Substitute into the mass equation:

\[
mg-\frac{Ia}{R^2}=ma
\]

\[
mg=a\left(m+\frac{I}{R^2}\right)
\]

Therefore,

\[
a=\frac{mg}{m+\frac{I}{R^2}}
\]

The pulley’s rotational inertia reduces the acceleration below \(g\).

If external torque is negligible, angular momentum is conserved.

\[
I_i\omega_i=I_f\omega_f
\]

Therefore,

\[
\omega_f=\frac{I_i\omega_i}{I_f}
\]

\[
\omega_f=\frac{4\times3}{1}
\]

\[
\omega_f=12\text{ rad/s}
\]

The skater spins faster because her mass moves closer to the rotation axis, reducing moment of inertia.

For a uniform rod about its center,

\[
I_{\text{center}}=\frac{1}{12}ML^2
\]

The axis through one end is a distance

\[
d=\frac{L}{2}
\]

from the center.

Using the parallel-axis theorem,

\[
I_{\text{end}}=I_{\text{center}}+Md^2
\]

\[
I_{\text{end}}=\frac{1}{12}ML^2+M\left(\frac{L}{2}\right)^2
\]

\[
I_{\text{end}}=\frac{1}{12}ML^2+\frac{1}{4}ML^2
\]

\[
I_{\text{end}}=\frac{1}{3}ML^2
\]

For a solid cylinder,

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

Rolling without slipping gives

\[
\omega=\frac{v}{R}
\]

Rotational kinetic energy is

\[
K_{\text{rot}}=\frac{1}{2}I\omega^2
\]

\[
K_{\text{rot}}=
\frac{1}{2}
\left(\frac{1}{2}MR^2\right)
\left(\frac{v}{R}\right)^2
\]

\[
K_{\text{rot}}=\frac{1}{4}Mv^2
\]

Translational kinetic energy is

\[
K_{\text{trans}}=\frac{1}{2}Mv^2
\]

Thus, total kinetic energy is

\[
K_{\text{total}}=\frac{3}{4}Mv^2
\]

The required fraction is

\[
\frac{K_{\text{rot}}}{K_{\text{total}}}
=
\frac{\frac{1}{4}Mv^2}{\frac{3}{4}Mv^2}
\]

\[
=\frac{1}{3}
\]

Gravity acts at the rod’s center of mass, located at a distance \(\dfrac{L}{2}\) from the hinge.

Initially, the rod is horizontal, so gravity is perpendicular to the rod. The gravitational torque is:

\[
\tau=Mg\left(\frac{L}{2}\right)
\]

For a uniform rod about one end:

\[
I=\frac{1}{3}ML^2
\]

Using rotational Newton’s second law:

\[
\tau=I\alpha
\]

\[
Mg\left(\frac{L}{2}\right)
=
\left(\frac{1}{3}ML^2\right)\alpha
\]

Therefore:

\[
\alpha=\frac{3g}{2L}
\]

The angular acceleration is maximum at release because the torque is maximum when the rod is horizontal.

The moment of inertia of the disc is:

\[
I_{\text{disc}}=\frac{1}{2}MR^2
\]

The attached small mass behaves as a point mass at distance \(R\):

\[
I_m=mR^2
\]

Therefore, the total moment of inertia is:

\[
I_{\text{total}}=I_{\text{disc}}+I_m
\]

\[
I_{\text{total}}=\frac{1}{2}MR^2+mR^2
\]

\[
I_{\text{total}}=
\left(\frac{M}{2}+m\right)R^2
\]

Moment of inertia adds directly when all parts rotate about the same axis.

Gravity pulls the cylinder’s center of mass down the incline. However, gravity acts through the center and cannot create torque about the center.

The cylinder needs clockwise angular acceleration to roll downhill.

Static friction acts up the incline. This friction produces the required torque that increases angular speed.

For rolling without slipping, the friction direction is determined by the tendency of the contact point to slide. It is not always opposite to the object’s center-of-mass motion.

During the short collision, angular momentum about the hinge is conserved because the hinge force passes through the hinge and produces zero torque about that point.

Initial angular momentum of the particle is:

\[
L_i=mvL
\]

After sticking, the total moment of inertia about the hinge is:

\[
I=
\frac{1}{3}ML^2+mL^2
\]

\[
I=
\left(\frac{M}{3}+m\right)L^2
\]

Using angular momentum conservation:

\[
mvL=I\omega
\]

\[
mvL=
\left(\frac{M}{3}+m\right)L^2\omega
\]

Therefore:

\[
\omega=
\frac{mv}{\left(\frac{M}{3}+m\right)L}
\]

Kinetic energy is not conserved because the particle sticks to the rod.

Angular impulse equals the integral of torque over time:

\[
J_{\text{angular}}=\int\tau\,dt
\]

Given:

\[
\tau=4t
\]

Therefore:

\[
J_{\text{angular}}=
\int_0^3 4t\,dt
\]

\[
J_{\text{angular}}=
\left[2t^2\right]_0^3
\]

\[
J_{\text{angular}}=2(9)
\]

\[
J_{\text{angular}}=18\text{ N}\cdot\text{m}\cdot\text{s}
\]

Angular impulse equals the change in angular momentum.

For a small angular displacement \(\theta\), the restoring torque is approximately:

\[
\tau=-Mgd\theta
\]

Rotational equation of motion is:

\[
I\frac{d^2\theta}{dt^2}=-Mgd\theta
\]

Rearranging gives:

\[
\frac{d^2\theta}{dt^2}
+
\frac{Mgd}{I}\theta
=
0
\]

Compare this with the SHM form:

\[
\frac{d^2\theta}{dt^2}+\omega^2\theta=0
\]

Therefore:

\[
\omega=\sqrt{\frac{Mgd}{I}}
\]

This formula applies only for small oscillations.

Take rightward translation and clockwise rotation as positive.

For translation:

\[
F+f=Ma
\]

For rotation about the center:

\[
Fr-fR=I\alpha
\]

For rolling without slipping:

\[
a=R\alpha
\]

Therefore:

\[
Fr-fR=\frac{I}{R}a
\]

From translation:

\[
f=Ma-F
\]

Substitute into the torque equation:

\[
Fr-(Ma-F)R=\frac{I}{R}a
\]

\[
F(r+R)-MaR=\frac{I}{R}a
\]

\[
F(r+R)=a\left(MR+\frac{I}{R}\right)
\]

Therefore:

\[
a=
\frac{F(R+r)}{MR+\frac{I}{R}}
\]

The spool moves in the direction of the pull for this string arrangement.

For a solid sphere:

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

Rolling condition gives:

\[
\omega=\frac{v}{R}
\]

Initial total kinetic energy is:

\[
K=
\frac{1}{2}Mv^2
+
\frac{1}{2}I\omega^2
\]

\[
K=
\frac{1}{2}Mv^2
+
\frac{1}{2}
\left(\frac{2}{5}MR^2\right)
\left(\frac{v}{R}\right)^2
\]

\[
K=
\frac{1}{2}Mv^2+\frac{1}{5}Mv^2
\]

\[
K=\frac{7}{10}Mv^2
\]

At maximum height, the sphere stops momentarily. Therefore:

\[
Mgh=\frac{7}{10}Mv^2
\]

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

Static friction maintains rolling but does no net work in ideal rolling.

For rotational equilibrium, net torque about the hinge must be zero.

The rod’s weight acts at its center of mass, located at:

\[
\frac{L}{2}
\]

from the hinge.

Torque due to weight is:

\[
\tau_W=Mg\left(\frac{L}{2}\right)
\]

The applied force acts at distance \(L\), producing opposite torque:

\[
\tau_F=FL
\]

Set torques equal:

\[
FL=Mg\left(\frac{L}{2}\right)
\]

Therefore:

\[
F=\frac{Mg}{2}
\]

The hinge also provides vertical and horizontal reaction forces, but those forces produce no torque about the hinge.

The clay is dropped vertically, so it has no initial angular momentum about the disc’s vertical rotation axis.

External torque about the axis is negligible during contact. Therefore, angular momentum is conserved.

Initial angular momentum:

\[
L_i=I\omega
\]

After sticking, total moment of inertia becomes:

\[
I_f=I+mR^2
\]

Thus:

\[
I\omega=(I+mR^2)\omega_f
\]

Therefore:

\[
\omega_f=
\frac{I\omega}{I+mR^2}
\]

The angular speed decreases because the system’s moment of inertia increases.

During the very short collision, linear momentum is not conserved because the hinge gives an external impulse.

However, angular momentum about the hinge is conserved because the hinge force has zero moment arm about the hinge.

The incoming particle has angular momentum:

\[
L_i=mvL
\]

The rod has moment of inertia about the hinge:

\[
I_{\text{rod}}=\frac{1}{3}ML^2
\]

The particle sticks at distance \(L\), so:

\[
I_{\text{particle}}=mL^2
\]

Therefore:

\[
I_{\text{total}}=
\left(\frac{M}{3}+m\right)L^2
\]

Using angular momentum conservation:

\[
mvL=
\left(\frac{M}{3}+m\right)L^2\omega
\]

Hence:

\[
\omega=
\frac{mv}{\left(\frac{M}{3}+m\right)L}
\]

The collision is perfectly inelastic, so kinetic energy is not conserved.

On a rough incline, static friction can provide torque and maintain rolling without slipping.

However, the incline is smooth, so friction is zero.

Gravity has a component down the incline, which slows the center of mass while the sphere moves upward.

Since gravity acts through the center of mass, it produces no torque about the center.

Therefore:

\[
\tau_{\text{cm}}=0
\]

and:

\[
\alpha=0
\]

The angular speed remains initially constant, while translational speed decreases.

As a result, the rolling condition:

\[
v=R\omega
\]

is no longer maintained. The sphere begins to slip.

The particle has zero initial angular momentum about the disc axis.

Angular momentum is conserved:

\[
I_i\omega_0=I_f\omega_f
\]

For the disc:

\[
I_i=\frac{1}{2}MR^2
\]

After the particle sticks:

\[
I_f=\frac{1}{2}MR^2+mR^2
\]

\[
I_f=
\left(\frac{M}{2}+m\right)R^2
\]

Therefore:

\[
\omega_f=
\frac{\frac{1}{2}MR^2}{\left(\frac{M}{2}+m\right)R^2}\omega_0
\]

\[
\omega_f=
\frac{M}{M+2m}\omega_0
\]

Initial rotational kinetic energy:

\[
K_i=\frac{1}{2}I_i\omega_0^2
\]

Final rotational kinetic energy:

\[
K_f=\frac{1}{2}I_f\omega_f^2
\]

Using angular momentum conservation, the ratio becomes:

\[
\frac{K_f}{K_i}
=
\frac{I_i}{I_f}
\]

\[
\frac{K_f}{K_i}
=
\frac{\frac{1}{2}M}{\frac{1}{2}M+m}
\]

\[
\frac{K_f}{K_i}
=
\frac{M}{M+2m}
\]

Some kinetic energy becomes internal energy during sticking.

The force produces translation of the center of mass and rotation about the center of mass.

Torque about the center is:

\[
\tau=F\left(\frac{L}{2}\right)
\]

For a uniform rod about its center:

\[
I_{\text{cm}}=\frac{1}{12}ML^2
\]

Using:

\[
\tau=I_{\text{cm}}\alpha
\]

\[
F\left(\frac{L}{2}\right)
=
\left(\frac{1}{12}ML^2\right)\alpha
\]

Therefore:

\[
\alpha=
\frac{F(L/2)}{ML^2/12}
\]

\[
\alpha=
\frac{6F}{ML}
\]

The rod also has center-of-mass acceleration:

\[
a_{\text{cm}}=\frac{F}{M}
\]

Initially:

\[
v_{\text{cm}}=2v
\]

and:

\[
\omega=\frac{v}{R}
\]

For rolling without slipping, the final condition is:

\[
V=R\Omega
\]

Friction is an external force, so linear momentum is not conserved.

However, angular momentum about the contact point with the ground is conserved because friction and normal force act through that point.

For a hoop:

\[
I_{\text{cm}}=MR^2
\]

Initial angular momentum about the contact point is:

\[
L_i=
MR(2v)+I_{\text{cm}}\left(\frac{v}{R}\right)
\]

\[
L_i=2MRv+MRv
\]

\[
L_i=3MRv
\]

After rolling begins:

\[
L_f=
MRV+MR^2\left(\frac{V}{R}\right)
\]

\[
L_f=2MRV
\]

Therefore:

\[
3MRv=2MRV
\]

\[
V=\frac{3v}{2}
\]

The final speed is between the original rotational-equivalent speed \(v\) and the initial translational speed \(2v\).

Immediately before breaking, the particle has tangential speed:

\[
v=\omega L
\]

Breaking the rod removes the constraint force, but it cannot instantly change the particle’s velocity.

Therefore, immediately after the break:

\[
v_{\text{after}}=v_{\text{before}}
\]

\[
v_{\text{after}}=\omega L
\]

The particle then follows projectile motion under gravity.

At the highest point, its velocity is horizontal. The direction is tangent to the circular path.

The disc begins with gravitational potential energy:

\[
U_i=Mgh
\]

It rolls without slipping, so static friction does no net work.

At the bottom, its energy is divided into translational and rotational kinetic energy:

\[
Mgh=
\frac{1}{2}Mv^2+\frac{1}{2}I\omega^2
\]

However, total mechanical energy remains equal to \(Mgh\).

The disc then climbs the second rough incline without slipping. Static friction again maintains rolling and does no net work.

At the maximum height, both translational and rotational speeds become zero.

Therefore:

\[
Mgh=Mgh_{\text{max}}
\]

\[
h_{\text{max}}=h
\]

The disc reaches the same vertical height because mechanical energy is conserved.

The center of mass falls through height:

\[
\frac{L}{2}
\]

Using energy conservation:

\[
Mg\left(\frac{L}{2}\right)
=
\frac{1}{2}I\omega^2
\]

For a rod about one end:

\[
I=\frac{1}{3}ML^2
\]

Thus:

\[
Mg\left(\frac{L}{2}\right)
=
\frac{1}{2}
\left(\frac{1}{3}ML^2\right)\omega^2
\]

\[
\omega^2=\frac{3g}{L}
\]

At the lowest point, the center of mass has centripetal acceleration:

\[
a_c=\omega^2\left(\frac{L}{2}\right)
\]

\[
a_c=
\frac{3g}{L}\left(\frac{L}{2}\right)
\]

\[
a_c=\frac{3g}{2}
\]

Apply vertical force equation toward the pivot:

\[
R-Mg=M\left(\frac{3g}{2}\right)
\]

Therefore:

\[
R=
Mg+\frac{3Mg}{2}
\]

\[
R=\frac{5Mg}{2}
\]

The pivot force is upward at this instant.

Initially, the person is at the center, so their contribution to moment of inertia is approximately zero.

Initial angular momentum is:

\[
L_i=I\omega_0
\]

After walking outward, total moment of inertia becomes:

\[
I_f=I+mR^2
\]

No external torque acts on the platform-person system, so:

\[
I\omega_0=(I+mR^2)\omega_f
\]

Therefore:

\[
\omega_f=
\frac{I}{I+mR^2}\omega_0
\]

The angular speed decreases because moving mass outward increases moment of inertia.