401. A rolling body of mass \(M\) descends height \(h\) without slipping. If its speed at the bottom is \(v\), the total kinetic energy at the bottom is
ⓐ. \(\frac{1}{2}I_{\text{CM}}\omega^2\) only
ⓑ. \(\frac{1}{2}Mv^2\) only
ⓒ. zero because static friction acts
ⓓ. \(Mgh\)
Correct Answer: \(Mgh\)
Explanation: If rolling is without slipping and energy losses are neglected, mechanical energy is conserved. The decrease in gravitational potential energy is \(Mgh\). At the bottom, this energy appears as translational kinetic energy of the centre of mass plus rotational kinetic energy about the centre. Therefore, \(K_{\text{total}}=Mgh\). Static friction in ideal rolling need not dissipate mechanical energy, so it does not automatically make the final kinetic energy zero.
402. A wheel rolling without slipping is momentarily described as rotating about the contact point. This does not mean the contact point is a permanent fixed axle because
ⓐ. the centre of mass is fixed at the contact point
ⓑ. the angular velocity has become zero
ⓒ. the wheel has stopped translating
ⓓ. a different wheel point touches the ground next
Correct Answer: a different wheel point touches the ground next
Explanation: In pure rolling, the point of the wheel touching the ground is instantaneously at rest relative to the ground. This allows the wheel's motion at that instant to be described as rotation about the contact point. However, as the wheel moves forward, a different material point reaches the ground. The instantaneous axis is therefore not a permanent axle fixed in the wheel. The centre of mass continues moving forward while the wheel rotates.
403. A solid cylinder rolls without slipping down an incline. If \(I_{\text{CM}}=\frac{1}{2}MR^2\), the static friction magnitude is
ⓐ. \(\frac{1}{3}Mg\sin\theta\)
ⓑ. \(\frac{1}{2}Mg\sin\theta\)
ⓒ. \(\frac{2}{3}Mg\sin\theta\)
ⓓ. \(Mg\sin\theta\)
Correct Answer: \(\frac{1}{3}Mg\sin\theta\)
Explanation: \( \textbf{For rolling down the incline:} \) The translational equation along the plane is
\[
Mg\sin\theta-f=Ma
\]
\( \textbf{Torque about the centre:} \)
\[
fR=I_{\text{CM}}\alpha
\]
\( \textbf{No-slip condition:} \)
\[
a=R\alpha
\]
\( \textbf{Use the cylinder inertia:} \)
\[
fR=\left(\frac{1}{2}MR^2\right)\left(\frac{a}{R}\right)
\]
\[
f=\frac{1}{2}Ma
\]
\( \textbf{Known acceleration for a solid cylinder:} \)
\[
a=\frac{2}{3}g\sin\theta
\]
\( \textbf{Substitute into friction:} \)
\[
f=\frac{1}{2}M\left(\frac{2}{3}g\sin\theta\right)
\]
\[
f=\frac{1}{3}Mg\sin\theta
\]
\( \textbf{Final answer:} \) The static friction magnitude is \(\frac{1}{3}Mg\sin\theta\).
404. In ideal rolling without slipping down an incline, the work done by static friction on the rolling body is zero because
ⓐ. gravity does no work during rolling
ⓑ. the body has no rotational kinetic energy
ⓒ. static friction is zero in every rolling case
ⓓ. the contact point has no displacement then
Correct Answer: the contact point has no displacement then
Explanation: In pure rolling, the contact point of the body is instantaneously at rest relative to the surface. Static friction acts at this contact point. Since the point of application has no instantaneous displacement relative to the ground, static friction does no work in the ideal rolling model. Static friction may still be non-zero and may provide the torque needed for rotation. The energy conversion is mainly from gravitational potential energy into translational and rotational kinetic energy.
405. A rolling body descends an incline without slipping. If the angle of the incline is increased while the body and surface condition remain suitable for pure rolling, its acceleration
ⓐ. decreases because \(\sin\theta\) decreases
ⓑ. increases because \(g\sin\theta\) increases
ⓒ. remains independent of the incline angle
ⓓ. becomes equal to zero
Correct Answer: increases because \(g\sin\theta\) increases
Explanation: For rolling without slipping, the acceleration is
\[
a=\frac{g\sin\theta}{1+k}
\]
where \(I_{\text{CM}}=kMR^2\). For a given body, \(k\) is fixed. Increasing the incline angle increases \(\sin\theta\). Since the numerator increases while the denominator remains the same, the centre-of-mass acceleration increases. This conclusion assumes rolling without slipping continues to hold on the steeper incline.
406. A thin ring and a solid cylinder roll without slipping down the same incline. The ratio of their accelerations \(a_{\text{ring}}:a_{\text{cylinder}}\) is
ⓐ. \(3:4\)
ⓑ. \(1:1\)
ⓒ. \(2:3\)
ⓓ. \(4:3\)
Correct Answer: \(3:4\)
Explanation: \( \textbf{Rolling acceleration formula:} \)
\[
a=\frac{g\sin\theta}{1+k}
\]
\( \textbf{For a thin ring:} \)
\[
k_{\text{ring}}=1
\]
\[
a_{\text{ring}}=\frac{g\sin\theta}{2}
\]
\( \textbf{For a solid cylinder:} \)
\[
k_{\text{cylinder}}=\frac{1}{2}
\]
\[
a_{\text{cylinder}}=\frac{g\sin\theta}{1+\frac{1}{2}}=\frac{2}{3}g\sin\theta
\]
\( \textbf{Ratio:} \)
\[
a_{\text{ring}}:a_{\text{cylinder}}=\frac{1}{2}g\sin\theta:\frac{2}{3}g\sin\theta
\]
\[
a_{\text{ring}}:a_{\text{cylinder}}=\frac{1}{2}:\frac{2}{3}=3:4
\]
\( \textbf{Final answer:} \) The ratio is \(3:4\).
407. Use the arrangement described below. A wheel rolls to the right without slipping. Point P is at the top, point Q is at the centre, point R is at the contact point, and point S is at the front of the rim.
The point having zero instantaneous speed relative to the ground is
ⓐ. Point S
ⓑ. Point R
ⓒ. Point P
ⓓ. Point Q
Correct Answer: Point R
Explanation: For pure rolling, the velocity of a rim point is found by adding the translational velocity of the centre and the rotational velocity about the centre. At the contact point, the forward translational velocity \(v_{\text{CM}}\) is exactly cancelled by the backward rotational velocity \(R\omega\). Since \(v_{\text{CM}}=R\omega\), the contact point has zero instantaneous speed relative to the ground. The centre has speed \(v_{\text{CM}}\), the top point has speed \(2v_{\text{CM}}\), and the front point has speed \(\sqrt{2}v_{\text{CM}}\). The zero-speed result belongs only to the instantaneous point of contact.
408. A disc rolls without slipping with centre speed \(6\,\text{m s}^{-1}\). The speed of its topmost point relative to the ground is
ⓐ. \(8.5\,\text{m s}^{-1}\)
ⓑ. \(12\,\text{m s}^{-1}\)
ⓒ. \(6\,\text{m s}^{-1}\)
ⓓ. \(0\,\text{m s}^{-1}\)
Correct Answer: \(12\,\text{m s}^{-1}\)
Explanation: \( \textbf{Given data:} \) Centre speed \(v_{\text{CM}}=6\,\text{m s}^{-1}\).
\( \textbf{Pure rolling condition:} \)
\[
v_{\text{CM}}=R\omega
\]
\( \textbf{Velocity of top point:} \) At the top, translational velocity and rotational velocity are in the same forward direction.
\( \textbf{Rotational speed at rim:} \)
\[
R\omega=v_{\text{CM}}=6\,\text{m s}^{-1}
\]
\( \textbf{Total top-point speed:} \)
\[
v_{\text{top}}=v_{\text{CM}}+R\omega
\]
\[
v_{\text{top}}=6+6=12\,\text{m s}^{-1}
\]
\( \textbf{Final answer:} \) The topmost point has speed \(12\,\text{m s}^{-1}\).
409. A uniform solid sphere rolls without slipping from height \(h\), while an identical solid sphere slides without friction from the same height. The rolling sphere reaches the bottom with smaller speed because
ⓐ. some energy becomes rotational kinetic energy
ⓑ. gravity is weaker on the rolling sphere
ⓒ. static friction always removes all energy as heat
ⓓ. the rolling sphere has zero translational kinetic energy
Correct Answer: some energy becomes rotational kinetic energy
Explanation: The sliding sphere converts its gravitational potential energy into translational kinetic energy only, giving a larger centre-of-mass speed. The rolling sphere converts the same potential energy into both translational and rotational kinetic energy. Since part of the energy is stored in rotation, less is available for translational motion of the centre. For a solid sphere, the rolling speed from height \(h\) is \(\sqrt{\frac{10gh}{7}}\), which is less than \(\sqrt{2gh}\). The difference comes from energy sharing, not from a weaker gravitational force.
410. A body rolls without slipping from rest down a height \(h\). If its value of \(k\) in \(I_{\text{CM}}=kMR^2\) is \(0\), the speed formula reduces to
ⓐ. \(v=\sqrt{gh}\)
ⓑ. \(v=0\)
ⓒ. \(v=\sqrt{\frac{gh}{2}}\)
ⓓ. \(v=\sqrt{2gh}\)
Correct Answer: \(v=\sqrt{2gh}\)
Explanation: The general rolling speed from height \(h\) is
\[
v=\sqrt{\frac{2gh}{1+k}}
\]
If \(k=0\), the denominator becomes \(1\). Therefore,
\[
v=\sqrt{2gh}
\]
This is the same expression as for a particle or frictionless sliding body when rotational kinetic energy is absent. Real rolling bodies have \(k\gt0\), so their translational speed is less than this limiting value. The limiting case helps show how rotational inertia reduces translational speed.
411. A hollow spherical shell and a solid sphere roll without slipping down the same incline. The hollow shell has smaller acceleration because
ⓐ. its radius must be larger
ⓑ. its mass must be smaller
ⓒ. gravity cannot act at its centre
ⓓ. its \(k\) value is larger
Correct Answer: its \(k\) value is larger
Explanation: Rolling acceleration down an incline is \(a=\frac{g\sin\theta}{1+k}\). A hollow spherical shell has \(k=\frac{2}{3}\), while a solid sphere has \(k=\frac{2}{5}\). The hollow shell has more mass farther from the axis, so its rotational inertia factor is larger. A larger \(k\) gives a larger denominator and hence a smaller acceleration. The comparison is controlled by mass distribution, not by mass or radius alone.
412. A rigid body rolls without slipping with \(v_{\text{CM}}=R\omega\). If its centre has acceleration \(a_{\text{CM}}\) but the angular acceleration is not equal to \(\frac{a_{\text{CM}}}{R}\), the motion
ⓐ. must still satisfy pure rolling throughout the motion
ⓑ. has no translational motion
ⓒ. cannot satisfy no-slip acceleration condition
ⓓ. has zero angular speed
Correct Answer: cannot satisfy no-slip acceleration condition
Explanation: For rolling without slipping, the speed relation is \(v_{\text{CM}}=R\omega\). If the no-slip condition holds continuously during accelerated motion, differentiating gives \(a_{\text{CM}}=R\alpha\). Therefore, \(\alpha=\frac{a_{\text{CM}}}{R}\). If the given angular acceleration does not satisfy this relation, the motion is not consistent with pure rolling throughout that interval. The body may still translate and rotate, but slipping or another constraint change must be present.
413. Study the table and select the row that correctly compares sliding and rolling from the same height \(h\), neglecting losses.
| Row | Motion | Energy at bottom |
|---|
| P | Frictionless sliding | Only translational kinetic energy |
| Q | Pure rolling | Only rotational kinetic energy |
| R | Pure rolling | No kinetic energy |
| S | Frictionless sliding | Only rotational kinetic energy |
ⓐ. Row P
ⓑ. Row Q
ⓒ. Row R
ⓓ. Row S
Correct Answer: Row P
Explanation: A frictionless sliding body does not rotate due to contact torque, so its gravitational potential energy becomes translational kinetic energy. A rolling body has both translational kinetic energy of the centre and rotational kinetic energy about the centre. Therefore, row P is suitable. Row Q is incomplete because rolling includes translational motion as well as rotation. Row R is impossible without losses being larger than the available energy, and row S assigns rotational energy to a frictionless sliding body without a rolling constraint.
414. A solid cylinder rolling without slipping has total kinetic energy \(\frac{3}{4}Mv^2\). The ratio of its translational kinetic energy to rotational kinetic energy is
ⓐ. \(1:1\)
ⓑ. \(3:1\)
ⓒ. \(2:1\)
ⓓ. \(1:2\)
Correct Answer: \(2:1\)
Explanation: \( \textbf{Translational kinetic energy:} \)
\[
K_{\text{trans}}=\frac{1}{2}Mv^2
\]
\( \textbf{For a solid cylinder:} \)
\[
I_{\text{CM}}=\frac{1}{2}MR^2
\]
\( \textbf{Rotational kinetic energy in rolling:} \)
\[
K_{\text{rot}}=\frac{1}{2}I_{\text{CM}}\omega^2
\]
Using \(\omega=\frac{v}{R}\),
\[
K_{\text{rot}}=\frac{1}{2}\left(\frac{1}{2}MR^2\right)\left(\frac{v}{R}\right)^2
\]
\[
K_{\text{rot}}=\frac{1}{4}Mv^2
\]
\( \textbf{Ratio:} \)
\[
K_{\text{trans}}:K_{\text{rot}}=\frac{1}{2}Mv^2:\frac{1}{4}Mv^2=2:1
\]
\( \textbf{Final answer:} \) The ratio is \(2:1\).
415. A thin ring rolling without slipping has translational kinetic energy \(12\,\text{J}\). Its rotational kinetic energy is
ⓐ. \(6\,\text{J}\)
ⓑ. \(12\,\text{J}\)
ⓒ. \(18\,\text{J}\)
ⓓ. \(24\,\text{J}\)
Correct Answer: \(12\,\text{J}\)
Explanation: For a thin ring,
\[
I_{\text{CM}}=MR^2
\]
In rolling without slipping,
\[
\omega=\frac{v}{R}
\]
The rotational kinetic energy is
\[
K_{\text{rot}}=\frac{1}{2}I_{\text{CM}}\omega^2
\]
Substitute \(I_{\text{CM}}=MR^2\):
\[
K_{\text{rot}}=\frac{1}{2}MR^2\left(\frac{v}{R}\right)^2
\]
\[
K_{\text{rot}}=\frac{1}{2}Mv^2
\]
This equals the translational kinetic energy of the ring. Since \(K_{\text{trans}}=12\,\text{J}\), \(K_{\text{rot}}=12\,\text{J}\).
416. A wheel rolling without slipping covers \(12\,\text{m}\) in \(3\,\text{s}\). Its radius is \(0.50\,\text{m}\). Its angular speed is
ⓐ. \(4\,\text{rad s}^{-1}\)
ⓑ. \(8\,\text{rad s}^{-1}\)
ⓒ. \(6\,\text{rad s}^{-1}\)
ⓓ. \(12\,\text{rad s}^{-1}\)
Correct Answer: \(8\,\text{rad s}^{-1}\)
Explanation: \( \textbf{Distance covered:} \) \(s=12\,\text{m}\).
\( \textbf{Time taken:} \) \(t=3\,\text{s}\).
\( \textbf{Centre-of-mass speed:} \)
\[
v_{\text{CM}}=\frac{s}{t}=\frac{12}{3}=4\,\text{m s}^{-1}
\]
\( \textbf{Rolling condition:} \)
\[
v_{\text{CM}}=R\omega
\]
\( \textbf{Given radius:} \)
\[
R=0.50\,\text{m}
\]
\( \textbf{Solve for angular speed:} \)
\[
\omega=\frac{v_{\text{CM}}}{R}
\]
\[
\omega=\frac{4}{0.50}=8\,\text{rad s}^{-1}
\]
\( \textbf{Final answer:} \) The angular speed is \(8\,\text{rad s}^{-1}\).
417. The angular momentum of a rolling body about its centre of mass is
ⓐ. \(MR\omega^2\)
ⓑ. \(I_{\text{CM}}\omega\)
ⓒ. \(Mv_{\text{CM}}\)
ⓓ. \(\frac{1}{2}I_{\text{CM}}\omega^2\)
Correct Answer: \(I_{\text{CM}}\omega\)
Explanation: The angular momentum of a rigid body about its centre of mass during rotation is \(I_{\text{CM}}\omega\). Rolling motion also has translational motion of the centre of mass, but that is not part of the angular momentum about the centre itself. The expression \(Mv_{\text{CM}}\) is linear momentum. The expression \(\frac{1}{2}I_{\text{CM}}\omega^2\) is rotational kinetic energy. The reference point must be clear whenever angular momentum is discussed.
418. In pure rolling, the total kinetic energy may be written as \(\frac{1}{2}I_{\text{contact}}\omega^2\). For a solid sphere, \(I_{\text{CM}}=\frac{2}{5}MR^2\). Then \(I_{\text{contact}}\) is
ⓐ. \(2MR^2\)
ⓑ. \(\frac{2}{5}MR^2\)
ⓒ. \(\frac{5}{7}MR^2\)
ⓓ. \(\frac{7}{5}MR^2\)
Correct Answer: \(\frac{7}{5}MR^2\)
Explanation: \( \textbf{Moment of inertia about centre:} \)
\[
I_{\text{CM}}=\frac{2}{5}MR^2
\]
\( \textbf{Instantaneous contact point:} \) In pure rolling, the contact point is instantaneously at rest.
\( \textbf{Use parallel-axis theorem:} \)
\[
I_{\text{contact}}=I_{\text{CM}}+MR^2
\]
\( \textbf{Substitution:} \)
\[
I_{\text{contact}}=\frac{2}{5}MR^2+MR^2
\]
\[
I_{\text{contact}}=\frac{2}{5}MR^2+\frac{5}{5}MR^2
\]
\[
I_{\text{contact}}=\frac{7}{5}MR^2
\]
\( \textbf{Final answer:} \) \(I_{\text{contact}}=\frac{7}{5}MR^2\).
419. A rolling body has total kinetic energy \(K=\frac{1}{2}Mv^2(1+k)\). For a thin ring, the fraction of total kinetic energy that is translational is
ⓐ. \(1\)
ⓑ. \(\frac{1}{2}\)
ⓒ. \(\frac{1}{3}\)
ⓓ. \(\frac{2}{3}\)
Correct Answer: \(\frac{1}{2}\)
Explanation: The translational kinetic energy is
\[
K_{\text{trans}}=\frac{1}{2}Mv^2
\]
The total rolling kinetic energy is
\[
K=\frac{1}{2}Mv^2(1+k)
\]
For a thin ring,
\[
k=1
\]
So the total kinetic energy is
\[
K=\frac{1}{2}Mv^2(2)=Mv^2
\]
The translational fraction is
\[
\frac{K_{\text{trans}}}{K}=\frac{\frac{1}{2}Mv^2}{Mv^2}=\frac{1}{2}
\]
\( \textbf{Final answer:} \) The translational fraction is \(\frac{1}{2}\).
420. A body rolls without slipping down an incline from rest. The acceleration formula \(a=\frac{g\sin\theta}{1+k}\) is smaller than \(g\sin\theta\) because
ⓐ. the normal reaction acts down the incline
ⓑ. static friction always does negative work down the incline
ⓒ. the body's mass cancels to zero
ⓓ. gravity also produces angular acceleration
Correct Answer: gravity also produces angular acceleration
Explanation: If a body simply slides without friction, its acceleration down the incline is \(g\sin\theta\). In rolling without slipping, the body must also rotate while its centre of mass accelerates. Static friction supplies the torque needed for angular acceleration. As a result, the translational acceleration is reduced to \(a=\frac{g\sin\theta}{1+k}\). The reduction reflects the sharing of motion between translation and rotation, not an automatic loss of mechanical energy.
