501. A uniform disc of mass \(M\) and radius \(R\) starts from rest. In one trial it rotates about its central axis perpendicular to its plane. In another trial it rotates about a parallel tangent axis perpendicular to its plane. The same constant torque acts in both trials. If each disc turns through the same angular displacement \(\theta\), then
ⓐ. \(\displaystyle \frac{t_{\rm tangent}}{t_{\rm central}}=1,\quad \frac{\omega_{\rm tangent}}{\omega_{\rm central}}=1\)
ⓑ. \(\displaystyle \frac{t_{\rm tangent}}{t_{\rm central}}=3,\quad \frac{\omega_{\rm tangent}}{\omega_{\rm central}}=\frac13\)
ⓒ. \(\displaystyle \frac{t_{\rm tangent}}{t_{\rm central}}=\frac1{\sqrt3},\quad \frac{\omega_{\rm tangent}}{\omega_{\rm central}}=\sqrt3\)
ⓓ. \(\displaystyle \frac{t_{\rm tangent}}{t_{\rm central}}=\sqrt3,\quad \frac{\omega_{\rm tangent}}{\omega_{\rm central}}=\frac1{\sqrt3}\)
Correct Answer: \(\displaystyle \frac{t_{\rm tangent}}{t_{\rm central}}=\sqrt3,\quad \frac{\omega_{\rm tangent}}{\omega_{\rm central}}=\frac1{\sqrt3}\)
Explanation: For the central perpendicular axis,
\[
I_C=\frac12MR^2
\]
For the parallel tangent axis,
\[
I_T=I_C+MR^2
=\frac32MR^2
\]
Therefore,
\[
\frac{I_T}{I_C}=3
\]
With the same constant torque,
\[
\alpha=\frac{\tau}{I},
\]
so
\[
\alpha_T=\frac13\alpha_C
\]
Starting from rest,
\[
\theta=\frac12\alpha t^2
\]
For equal \(\theta\),
\[
\frac{t_T}{t_C}
=\sqrt{\frac{\alpha_C}{\alpha_T}}
=\sqrt3
\]
Also,
\[
\omega^2=2\alpha\theta
\]
Hence,
\[
\frac{\omega_T}{\omega_C}
=\sqrt{\frac{\alpha_T}{\alpha_C}}
=\frac1{\sqrt3}
\]
\(\textbf{Final answer:}\)
\[
\boxed{\frac{t_T}{t_C}=\sqrt3,\qquad
\frac{\omega_T}{\omega_C}=\frac1{\sqrt3}}
\]
502. A system initially has moment of inertia
\[
I_1=4\,\text{kg m}^2
\]
and angular speed
\[
\omega_1=2\,\text{rad s}^{-1}.
\]
A constant external torque \(+4\,\text{N m}\) acts for \(2\,\text{s}\). The system is then internally rearranged instantaneously so that its moment of inertia becomes \(2\,\text{kg m}^2\), with negligible external angular impulse during the rearrangement. Finally, a constant torque \(-2\,\text{N m}\) acts for \(3\,\text{s}\). The final angular momentum and angular speed are
ⓐ. \(6\,\text{kg m}^2\text{s}^{-1}\) and \(3\,\text{rad s}^{-1}\)
ⓑ. \(8\,\text{kg m}^2\text{s}^{-1}\) and \(4\,\text{rad s}^{-1}\)
ⓒ. \(10\,\text{kg m}^2\text{s}^{-1}\) and \(5\,\text{rad s}^{-1}\)
ⓓ. \(16\,\text{kg m}^2\text{s}^{-1}\) and \(8\,\text{rad s}^{-1}\)
Correct Answer: \(10\,\text{kg m}^2\text{s}^{-1}\) and \(5\,\text{rad s}^{-1}\)
Explanation: \(\textbf{Initial angular momentum:}\)
\[
L_1=I_1\omega_1
\]
Thus,
\[
L_1=(4)(2)=8\,\text{kg m}^2\text{s}^{-1}
\]
\(\textbf{First torque interval:}\)
Angular impulse gives
\[
\Delta L_1=\tau_1\Delta t_1
\]
Therefore,
\[
\Delta L_1=(4)(2)=8
\]
Hence,
\[
L_2=8+8=16\,\text{kg m}^2\text{s}^{-1}
\]
During the internal rearrangement the external angular impulse is negligible, so
\[
L_3=L_2=16
\]
With the new moment of inertia \(2\,\text{kg m}^2\),
\[
\omega_3=\frac{16}{2}=8\,\text{rad s}^{-1}
\]
\(\textbf{Second torque interval:}\)
\[
\Delta L_2=(-2)(3)=-6
\]
Thus,
\[
L_f=16-6=10\,\text{kg m}^2\text{s}^{-1}
\]
Finally,
\[
\omega_f=\frac{L_f}{I_f}
=\frac{10}{2}
=5\,\text{rad s}^{-1}
\]
\(\textbf{Final answer:}\)
\[
\boxed{L_f=10\,\text{kg m}^2\text{s}^{-1},\qquad
\omega_f=5\,\text{rad s}^{-1}}
\]
503. A rigid body starts from rest and rotates about a fixed axis under a constant non-zero net torque. At some instant, the magnitude of tangential acceleration of every off-axis point equals the magnitude of its radial acceleration. The angular displacement from the starting position at that instant is
ⓐ. \(\frac14\,\text{rad}\)
ⓑ. \(\frac12\,\text{rad}\)
ⓒ. \(1\,\text{rad}\)
ⓓ. \(2\,\text{rad}\)
Correct Answer: \(\frac12\,\text{rad}\)
Explanation: A constant torque acting on a rigid body of constant moment of inertia gives constant angular acceleration \(\alpha\).
For a point at distance \(r\),
\[
a_t=\alpha r
\]
Its radial acceleration is
\[
a_r=\omega^2r
\]
The stated equality gives
\[
\alpha r=\omega^2r
\]
For any off-axis point,
\[
\omega^2=\alpha
\]
Because the body starts from rest and \(\alpha\) is constant,
\[
\omega^2=2\alpha\theta
\]
Combining the two relations,
\[
\alpha=2\alpha\theta
\]
Since \(\alpha\ne0\),
\[
\theta=\frac12\,\text{rad}
\]
The radius cancels, so the equality occurs simultaneously for every off-axis point of the rigid body.
\(\textbf{Final answer:}\)
\[
\boxed{\theta=\frac12\,\text{rad}}
\]
504. A system of fixed total mass rotates about an axis with no external torque. Due to internal redistribution, its angular speed decreases from \(6\,\text{rad s}^{-1}\) to \(3\,\text{rad s}^{-1}\). If its initial moment of inertia and radius of gyration are \(I_1\) and \(k_1\), respectively, the final values are
ⓐ. \(I_2=2I_1,\quad k_2=\sqrt2k_1\)
ⓑ. \(I_2=\frac12I_1,\quad k_2=\frac{k_1}{\sqrt2}\)
ⓒ. \(I_2=4I_1,\quad k_2=2k_1\)
ⓓ. \(I_2=2I_1,\quad k_2=2k_1\)
Correct Answer: \(I_2=2I_1,\quad k_2=\sqrt2k_1\)
Explanation: With zero external torque, angular momentum is conserved:
\[
I_1\omega_1=I_2\omega_2
\]
Substituting,
\[
I_1(6)=I_2(3)
\]
Therefore,
\[
I_2=2I_1
\]
For a fixed total mass \(M\),
\[
I=Mk^2
\]
Hence,
\[
\frac{I_2}{I_1}
=\frac{k_2^2}{k_1^2}
\]
Using
\[
\frac{I_2}{I_1}=2,
\]
we get
\[
\frac{k_2}{k_1}=\sqrt2
\]
Therefore,
\[
k_2=\sqrt2k_1
\]
The decrease in angular speed is caused by the increase in rotational inertia required by angular-momentum conservation.
\(\textbf{Final answer:}\)
\[
\boxed{I_2=2I_1,\qquad k_2=\sqrt2k_1}
\]
505. A particle of mass \(M\), initially at rest, receives a linear impulse \(J\). Separately, a rigid body of the same mass \(M\) and radius of gyration \(k\), initially at rest, receives an angular impulse of magnitude \(Jk\) about its rotation axis. If the resulting speeds are \(v\) and \(\omega\), respectively, then
ⓐ. \(v=k^2\omega\)
ⓑ. \(v=\omega/k\)
ⓒ. \(v=k\omega\)
ⓓ. \(v=2k\omega\)
Correct Answer: \(v=k\omega\)
Explanation: For the particle, linear impulse equals change in linear momentum:
\[
J=Mv
\]
Therefore,
\[
v=\frac{J}{M}
\]
For the rigid body,
\[
I=Mk^2
\]
Angular impulse equals change in angular momentum:
\[
Jk=I\omega
\]
Thus,
\[
Jk=Mk^2\omega
\]
Hence,
\[
\omega=\frac{J}{Mk}
\]
Multiplying by \(k\),
\[
k\omega=\frac{J}{M}
\]
But this is exactly the particle speed \(v\).
Therefore,
\[
v=k\omega
\]
The relation is a direct consequence of the linear–rotational impulse analogy together with \(I=Mk^2\).
506. Consider the following statements.
Statement I: A system can have zero resultant external force but still fail to be in equilibrium because a non-zero external couple may act.
Statement II: If the net external torque about a chosen fixed axis is zero, angular momentum about that axis is conserved.
Statement III: In the parallel-axis theorem, the shifted axis must be parallel to the corresponding axis through the centre of mass.
Statement IV: In fixed-axis rigid-body rotation, all points share the same angular velocity, although their linear speeds may differ.
Which combination is correct?
ⓐ. I and II only
ⓑ. II, III and IV only
ⓒ. I, III and IV only
ⓓ. I, II, III and IV
Correct Answer: I, II, III and IV
Explanation: Statement I is correct because translational equilibrium alone is not sufficient for a rigid body.
A pure couple has zero resultant force but produces a non-zero torque.
Statement II expresses the angular-momentum conservation condition about the specified axis.
If the corresponding net external torque vanishes,
\[
\frac{dL}{dt}=0.
\]
Statement III gives an essential geometrical condition of the parallel-axis theorem.
The theorem does not connect arbitrary non-parallel axes.
Statement IV is a defining kinematic feature of a rigid body rotating about a fixed axis.
All its points share the same \(\omega\) at an instant.
However,
\[
v=\omega r,
\]
so points at different perpendicular distances from the axis generally have different linear speeds.
Therefore all four statements are correct.
507. Match the situations in List I with the consequences in List II.
| List I | Situation | List II | Consequence |
|---|
| P | No external torque while a rotating system decreases its moment of inertia | 1 | Zero resultant force but non-zero resultant torque |
| Q | An axis is shifted parallel away from the centre-of-mass axis | 2 | Angular speed increases for non-zero conserved angular momentum |
| R | A pure couple acts on a rigid body | 3 | Moment of inertia increases by \(Md^2\) |
| S | Two points of a rigid body rotate about the same fixed axis at different radii | 4 | Same angular speed but generally different linear speeds |
ⓐ. P-2, Q-1, R-3, S-4
ⓑ. P-3, Q-2, R-1, S-4
ⓒ. P-2, Q-3, R-1, S-4
ⓓ. P-4, Q-3, R-2, S-1
Correct Answer: P-2, Q-3, R-1, S-4
Explanation: For P, zero external torque conserves angular momentum:
\[
I\omega=\text{constant}.
\]
If \(I\) decreases while angular momentum is non-zero, \(\omega\) must increase, so P-2.
For Q, the parallel-axis theorem gives
\[
I=I_{\rm CM}+Md^2,
\]
so shifting away from the centre-of-mass axis increases the moment of inertia by \(Md^2\); hence Q-3.
For R, a pure couple consists of equal and opposite forces.
Its resultant force is zero, but its torque is non-zero, giving R-1.
For S, all points of a rigid body share the same angular speed.
Their linear speeds satisfy
\[
v=\omega r,
\]
so different radii generally give different speeds; hence S-4.
The correct matching is P-2, Q-3, R-1, S-4.
508. A thin ring of mass \(M\) and radius \(R\) rotates with angular velocity \(+\omega\) about its central axis. A coaxial uniform disc of the same mass and radius rotates with angular velocity \(-\omega\). They are brought gently into contact and eventually rotate together. External torque about the axis is negligible. The final angular velocity and the fraction of initial rotational kinetic energy remaining are
ⓐ. \(\displaystyle +\frac{\omega}{3}\) and \(\displaystyle \frac19\)
ⓑ. \(\displaystyle +\frac{\omega}{2}\) and \(\displaystyle \frac14\)
ⓒ. \(0\) and \(0\)
ⓓ. \(\displaystyle -\frac{\omega}{3}\) and \(\displaystyle \frac19\)
Correct Answer: \(\displaystyle +\frac{\omega}{3}\) and \(\displaystyle \frac19\)
Explanation: For the ring,
\[
I_r=MR^2
\]
For the disc,
\[
I_d=\frac12MR^2
\]
\(\textbf{Conserve angular momentum:}\)
The initial angular momentum is
\[
L_i=MR^2\omega-\frac12MR^2\omega
\]
Thus,
\[
L_i=\frac12MR^2\omega
\]
After they rotate together,
\[
I_f=I_r+I_d=\frac32MR^2
\]
Hence,
\[
\frac12MR^2\omega
=\frac32MR^2\omega_f
\]
Therefore,
\[
\omega_f=\frac{\omega}{3}
\]
\(\textbf{Compare kinetic energies:}\)
Initially,
\[
K_i=\frac12MR^2\omega^2
+\frac12\left(\frac12MR^2\right)\omega^2
=\frac34MR^2\omega^2
\]
Finally,
\[
K_f
=\frac12\left(\frac32MR^2\right)\left(\frac{\omega}{3}\right)^2
=\frac1{12}MR^2\omega^2
\]
Therefore,
\[
\frac{K_f}{K_i}
=\frac{1/12}{3/4}
=\frac19
\]
Angular momentum is conserved, but rotational kinetic energy is not conserved during the dissipative coupling.
\(\textbf{Final answer:}\)
\[
\boxed{\omega_f=\frac{\omega}{3},\qquad \frac{K_f}{K_i}=\frac19}
\]
509. A thin ring and a uniform disc have the same mass \(M\) and radius \(R\). The ring rotates about an axis perpendicular to its plane and tangent to its rim, while the disc rotates about its central perpendicular axis. Starting from rest, each receives the same angular impulse \(J\). The ratios of final angular momenta and angular speeds are
ⓐ. \(\displaystyle \frac{L_d}{L_r}=\frac14,\quad \frac{\omega_d}{\omega_r}=4\)
ⓑ. \(\displaystyle \frac{L_d}{L_r}=4,\quad \frac{\omega_d}{\omega_r}=1\)
ⓒ. \(\displaystyle \frac{L_d}{L_r}=1,\quad \frac{\omega_d}{\omega_r}=2\)
ⓓ. \(\displaystyle \frac{L_d}{L_r}=1,\quad \frac{\omega_d}{\omega_r}=4\)
Correct Answer: \(\displaystyle \frac{L_d}{L_r}=1,\quad \frac{\omega_d}{\omega_r}=4\)
Explanation: Each body starts from rest and receives the same angular impulse.
Therefore,
\[
L_d=L_r=J
\]
For the thin ring about its central perpendicular axis,
\[
I_C=MR^2
\]
The tangent perpendicular axis is parallel and at distance \(R\), so
\[
I_r=MR^2+MR^2=2MR^2
\]
For the disc about its central perpendicular axis,
\[
I_d=\frac12MR^2
\]
Using
\[
L=I\omega,
\]
we have
\[
\frac{\omega_d}{\omega_r}
=\frac{I_r}{I_d}
\]
Therefore,
\[
\frac{\omega_d}{\omega_r}
=\frac{2}{1/2}=4
\]
The angular momenta are equal because the impulses are equal, while the angular speeds differ because the moments of inertia differ.
\(\textbf{Final answer:}\)
\[
\boxed{\frac{L_d}{L_r}=1,\qquad
\frac{\omega_d}{\omega_r}=4}
\]
510. A thin ring and a solid sphere have the same mass \(M\) and radius \(R\). Each starts from rest and is acted upon by the same constant torque through the same angular displacement \(\theta\), about the ring's central perpendicular axis and a sphere diameter, respectively. Which statement is correct?
ⓐ. \(K_{\rm ring}=K_{\rm sphere},\qquad \dfrac{\omega_{\rm sphere}}{\omega_{\rm ring}}=\sqrt{\frac52}\)
ⓑ. \(K_{\rm ring}\gt K_{\rm sphere},\qquad \dfrac{\omega_{\rm sphere}}{\omega_{\rm ring}}=1\)
ⓒ. \(K_{\rm ring}\lt K_{\rm sphere},\qquad \dfrac{\omega_{\rm sphere}}{\omega_{\rm ring}}=\sqrt{\frac25}\)
ⓓ. \(K_{\rm ring}=K_{\rm sphere},\qquad \dfrac{\omega_{\rm sphere}}{\omega_{\rm ring}}=1\)
Correct Answer: \(K_{\rm ring}=K_{\rm sphere},\qquad \dfrac{\omega_{\rm sphere}}{\omega_{\rm ring}}=\sqrt{\frac52}\)
Explanation: The work done by a constant torque through angular displacement \(\theta\) is
\[
W=\tau\theta
\]
Both bodies experience the same torque through the same angular displacement.
Therefore the same work is done on each.
Since both start from rest,
\[
K_{\rm ring}=K_{\rm sphere}=\tau\theta
\]
For the ring,
\[
I_r=MR^2
\]
For the solid sphere,
\[
I_s=\frac25MR^2
\]
Using
\[
K=\frac12I\omega^2,
\]
equal final kinetic energies require
\[
I_r\omega_r^2=I_s\omega_s^2
\]
Thus,
\[
\frac{\omega_s}{\omega_r}
=\sqrt{\frac{I_r}{I_s}}
\]
Hence,
\[
\frac{\omega_s}{\omega_r}
=\sqrt{\frac{1}{2/5}}
=\sqrt{\frac52}
\]
Equal work does not imply equal angular momentum or equal angular speed when the moments of inertia differ.
\(\textbf{Final answer:}\)
\[
K_{\rm sphere}=K_{\rm ring},\qquad
\frac{\omega_{\rm sphere}}{\omega_{\rm ring}}=\sqrt{\frac52}
\]
511. A free rigid body of mass \(M\) has radius of gyration \(k\) about an axis through its centre of mass. A force \(F\) is applied perpendicular to a radius drawn from the centre of mass to its line of action, the perpendicular distance being exactly \(k\). At the instant the force is applied, compare the centre-of-mass acceleration \(a_{\rm CM}\) with the tangential acceleration \(a_t\), relative to the centre of mass, of a point situated at distance \(k\) from the same rotation axis.
ⓐ. \(a_t=a_{\rm CM}\)
ⓑ. \(a_t=2a_{\rm CM}\)
ⓒ. \(a_t=\frac12a_{\rm CM}\)
ⓓ. \(a_t=k\,a_{\rm CM}\)
Correct Answer: \(a_t=a_{\rm CM}\)
Explanation: The resultant external force controls the acceleration of the centre of mass:
\[
a_{\rm CM}=\frac{F}{M}
\]
The same force also produces a torque about the centre of mass.
Because its perpendicular moment arm is \(k\),
\[
\tau_{\rm CM}=Fk
\]
The moment of inertia about the stated axis is
\[
I=Mk^2
\]
Therefore the angular acceleration is
\[
\alpha=\frac{\tau_{\rm CM}}{I}
\]
Substituting,
\[
\alpha=\frac{Fk}{Mk^2}
=\frac{F}{Mk}
\]
A point at distance \(k\) from the axis has tangential acceleration relative to the centre of mass
\[
a_t=\alpha k
\]
Hence,
\[
a_t=\frac{F}{Mk}\,k
=\frac{F}{M}
\]
Therefore,
\[
a_t=a_{\rm CM}
\]
The equality follows from the special choice of moment arm and point radius equal to the body's radius of gyration.
\(\textbf{Final answer:}\)
\[
a_t=a_{\rm CM}
\]
512. Consider the following statements about the centre of mass and centre of gravity of a rigid body.
Statement I: In a sufficiently uniform gravitational field, the centre of gravity coincides with the centre of mass.
Statement II: Under the same condition, the body's weight may be treated as acting through this common point in equilibrium calculations.
Statement III: For every non-uniform rigid body, the centre of gravity must coincide with its geometric centre.
Statement IV: Changing the point about which torques are calculated physically shifts the centre of gravity of the body.
Which combination is correct?
ⓐ. II and IV only
ⓑ. I, II and III only
ⓒ. I and II only
ⓓ. I, III and IV only
Correct Answer: I and II only
Explanation: The centre of mass is determined by the body's mass distribution.
The centre of gravity is the point through which the resultant gravitational force may be considered to act.
In a sufficiently uniform gravitational field, all parts of an ordinary-sized rigid body experience essentially the same gravitational acceleration.
Under that condition, the gravitational weighting of the mass elements is proportional to their masses, so the centre of gravity coincides with the centre of mass.
Therefore Statements I and II are correct.
A non-uniform body need not have its centre of mass, and hence its centre of gravity in uniform gravity, at its geometric centre.
Thus Statement III is incorrect.
The choice of pivot changes the calculated moment arms of forces but does not physically relocate the body's mass distribution or centre of gravity.
Therefore Statement IV is also incorrect.
513. A uniform rod of length \(4a\) and mass \(3M\) lies horizontally from \(x=0\) to \(x=4a\). A point mass \(M\) is attached at its right end. The composite body is supported vertically at A, \(x=a\), and B, \(x=4a\). After equilibrium is established, the point mass is removed without moving the rod or supports. Which set gives the support reactions before and after removal?
ⓐ. Initially \(R_A=\frac32Mg,\ R_B=\frac52Mg\); after removal \(R_A'=\frac32Mg,\ R_B'=\frac32Mg\)
ⓑ. Initially \(R_A=Mg,\ R_B=3Mg\); after removal \(R_A'=2Mg,\ R_B'=Mg\)
ⓒ. Initially \(R_A=2Mg,\ R_B=2Mg\); after removal \(R_A'=Mg,\ R_B'=2Mg\)
ⓓ. Initially \(R_A=2Mg,\ R_B=2Mg\); after removal \(R_A'=2Mg,\ R_B'=Mg\)
Correct Answer: Initially \(R_A=2Mg,\ R_B=2Mg\); after removal \(R_A'=2Mg,\ R_B'=Mg\)
Explanation: The problem first requires the centre of mass of the composite body and then a separate equilibrium calculation after the mass distribution changes.
\(\textbf{Composite centre of mass before removal:}\)
The rod's centre is at
\[
x_r=2a
\]
Hence,
\[
x_{\rm CM}
=\frac{3M(2a)+M(4a)}{4M}
\]
Therefore,
\[
x_{\rm CM}=\frac{10a}{4}=\frac{5a}{2}
\]
The total weight is
\[
4Mg
\]
\(\textbf{Initial moment balance about A:}\)
The centre of mass is
\[
\frac{5a}{2}-a=\frac{3a}{2}
\]
to the right of A, while B is \(3a\) from A.
Thus,
\[
R_B(3a)=4Mg\left(\frac{3a}{2}\right)
\]
Hence,
\[
R_B=2Mg
\]
Vertical force balance gives
\[
R_A+R_B=4Mg,
\]
so
\[
R_A=2Mg
\]
\(\textbf{After removing the point mass:}\)
Only the uniform rod remains, with weight \(3Mg\) acting at \(x=2a\).
Taking moments about A,
\[
R_B'(3a)=3Mg(a)
\]
Therefore,
\[
R_B'=Mg
\]
Then,
\[
R_A'+R_B'=3Mg,
\]
so
\[
R_A'=2Mg
\]
The removal changes both the total weight and its line of action, so the reaction change cannot be inferred from total weight alone.
\(\textbf{Final answer:}\)
\[
(R_A,R_B)=(2Mg,2Mg),
\qquad
(R_A',R_B')=(2Mg,Mg)
\]
514. Assembly A consists of two point masses \(m\) fixed at \(x=+a\) and \(x=-a\). Assembly B consists of two point masses \(m\) fixed at \(x=+2a\) and \(x=-2a\). Each assembly rotates about a \(z\)-axis through the origin. Starting from rest, the same constant torque acts on each for the same time \(t\). Which pair is correct?
ⓐ. \(\displaystyle \frac{L_B}{L_A}=1,\qquad \frac{\omega_B}{\omega_A}=\frac14\)
ⓑ. \(\displaystyle \frac{L_B}{L_A}=4,\qquad \frac{\omega_B}{\omega_A}=1\)
ⓒ. \(\displaystyle \frac{L_B}{L_A}=1,\qquad \frac{\omega_B}{\omega_A}=4\)
ⓓ. \(\displaystyle \frac{L_B}{L_A}=\frac14,\qquad \frac{\omega_B}{\omega_A}=1\)
Correct Answer: \(\displaystyle \frac{L_B}{L_A}=1,\qquad \frac{\omega_B}{\omega_A}=\frac14\)
Explanation: Both assemblies have the same total mass and the same centre of mass at the origin.
However, equal centre-of-mass positions do not imply equal moments of inertia.
\(\textbf{Moment of inertia of A:}\)
\[
I_A=m a^2+m a^2=2ma^2
\]
\(\textbf{Moment of inertia of B:}\)
\[
I_B=m(2a)^2+m(2a)^2
\]
Thus,
\[
I_B=8ma^2=4I_A
\]
The same torque acts for the same time, so both receive the same angular impulse:
\[
\Delta L=\tau t
\]
Since both start from rest,
\[
L_A=L_B=\tau t
\]
Therefore,
\[
\frac{L_B}{L_A}=1
\]
Using
\[
L=I\omega,
\]
we obtain
\[
\frac{\omega_B}{\omega_A}
=\frac{I_A}{I_B}
=\frac14
\]
The question distinguishes centre-of-mass information from rotational inertia: the same centre of mass can coexist with very different mass distributions about the axis.
\(\textbf{Final answer:}\)
\[
\boxed{\frac{L_B}{L_A}=1,\qquad
\frac{\omega_B}{\omega_A}=\frac14}
\]
515.
The same constant torque \(12\,\text{N m}\) is applied separately to a rigid body about several mutually parallel axes. If \(d\) is the perpendicular distance of an axis from the parallel centre-of-mass axis, a graph of \(1/\alpha\) against \(d^2\) is a straight line. Its intercept on the \(1/\alpha\) axis is \(0.25\,\text{s}^2\), and its slope is \(0.50\,\text{s}^2\text{m}^{-2}\).
The mass of the body, its moment of inertia about the centre-of-mass axis, and its angular acceleration for \(d=1\,\text{m}\) are
ⓐ. \(3\,\text{kg},\ 6\,\text{kg m}^2,\ \frac43\,\text{rad s}^{-2}\)
ⓑ. \(6\,\text{kg},\ 3\,\text{kg m}^2,\ \frac43\,\text{rad s}^{-2}\)
ⓒ. \(6\,\text{kg},\ 3\,\text{kg m}^2,\ 4\,\text{rad s}^{-2}\)
ⓓ. \(3\,\text{kg},\ 3\,\text{kg m}^2,\ 2\,\text{rad s}^{-2}\)
Correct Answer: \(6\,\text{kg},\ 3\,\text{kg m}^2,\ \frac43\,\text{rad s}^{-2}\)
Explanation: The graph combines the parallel-axis theorem with rotational dynamics.
For an axis at distance \(d\),
\[
I=I_{\rm CM}+Md^2
\]
Under the same torque,
\[
\alpha=\frac{\tau}{I}
\]
Therefore,
\[
\frac1\alpha=\frac{I_{\rm CM}+Md^2}{\tau}
\]
Rearranging,
\[
\frac1\alpha
=\frac{I_{\rm CM}}{\tau}
+\frac{M}{\tau}d^2
\]
\(\textbf{Use the graph intercept:}\)
\[
\frac{I_{\rm CM}}{12}=0.25
\]
Hence,
\[
I_{\rm CM}=3\,\text{kg m}^2
\]
\(\textbf{Use the graph slope:}\)
\[
\frac{M}{12}=0.50
\]
Therefore,
\[
M=6\,\text{kg}
\]
\(\textbf{Evaluate the shifted axis:}\)
\[
I=3+6(1)^2=9\,\text{kg m}^2
\]
Thus,
\[
\alpha=\frac{12}{9}
=\frac43\,\text{rad s}^{-2}
\]
The linearity of \(1/\alpha\) with \(d^2\), rather than \(\alpha\) itself, follows directly from the inverse dependence of angular acceleration on moment of inertia.
\(\textbf{Final answer:}\)
\[
\boxed{M=6\,\text{kg},\qquad
I_{\rm CM}=3\,\text{kg m}^2,\qquad
\alpha=\frac43\,\text{rad s}^{-2}}
\]
516. A plane lamina has centre of mass O. Axes \(x\) and \(y\) are mutually perpendicular in its plane through O, while \(z\) is perpendicular to the lamina through O. The same torque magnitude \(\tau\) is applied separately about these axes and produces
\[
\alpha_x=6a,\qquad \alpha_y=3a.
\]
A fourth axis \(z'\) is parallel to \(z\) and at distance \(d\) from O, where
\[
Md^2=\frac{\tau}{6a}.
\]
The angular accelerations \(\alpha_z\) and \(\alpha_{z'}\), under the same torque magnitude \(\tau\), are
ⓐ. \(3a\) and \(2a\)
ⓑ. \(2a\) and \(\frac32a\)
ⓒ. \(2a\) and \(3a\)
ⓓ. \(\frac32a\) and \(2a\)
Correct Answer: \(2a\) and \(\frac32a\)
Explanation: The measured angular accelerations first reveal the moments of inertia about the two in-plane axes.
\(\textbf{Moments inferred dynamically:}\)
\[
I_x=\frac{\tau}{\alpha_x}
=\frac{\tau}{6a},
\]
\[
I_y=\frac{\tau}{\alpha_y}
=\frac{\tau}{3a}
\]
For the plane lamina, the perpendicular-axis theorem gives
\[
I_z=I_x+I_y
\]
Therefore,
\[
I_z
=\frac{\tau}{6a}+\frac{\tau}{3a}
=\frac{\tau}{2a}
\]
Hence,
\[
\alpha_z=\frac{\tau}{I_z}=2a
\]
\(\textbf{Shift to the parallel axis:}\)
The parallel-axis theorem gives
\[
I_{z'}=I_z+Md^2
\]
Using the stated condition,
\[
I_{z'}
=\frac{\tau}{2a}+\frac{\tau}{6a}
\]
Thus,
\[
I_{z'}=\frac{2\tau}{3a}
\]
Therefore,
\[
\alpha_{z'}
=\frac{\tau}{I_{z'}}
=\frac{3a}{2}
\]
This experiment requires both axis theorems in sequence, with the dynamical measurements supplying the required moments of inertia.
\(\textbf{Final answer:}\)
\[
\boxed{\alpha_z=2a,\qquad
\alpha_{z'}=\frac32a}
\]
517. A uniform thin rod of mass \(M\) and length \(L\) lies in a horizontal plane and rotates about a frictionless vertical axis through a point \(L/4\) from its left end. A force of magnitude \(F\) is applied at the right end in the horizontal plane, making an angle \(60^\circ\) with the rod. At the instant considered, the angular acceleration produced by this force is
ⓐ. \(\displaystyle \frac{18\sqrt3}{7}\frac{F}{ML}\)
ⓑ. \(\displaystyle \frac{12\sqrt3}{7}\frac{F}{ML}\)
ⓒ. \(\displaystyle \frac{9\sqrt3}{7}\frac{F}{ML}\)
ⓓ. \(\displaystyle \frac{18}{7}\frac{F}{ML}\)
Correct Answer: \(\displaystyle \frac{18\sqrt3}{7}\frac{F}{ML}\)
Explanation: The pivot is not at the rod's centre, so both the moment arm and the shifted-axis moment of inertia must be determined.
\(\textbf{Torque of the applied force:}\)
The distance from the pivot to the right end is
\[
r=L-\frac L4=\frac{3L}{4}
\]
Hence,
\[
\tau=rF\sin60^\circ
\]
Therefore,
\[
\tau
=\frac{3L}{4}F\left(\frac{\sqrt3}{2}\right)
=\frac{3\sqrt3}{8}FL
\]
\(\textbf{Moment of inertia about the pivot:}\)
For the rod about its midpoint,
\[
I_{\rm CM}=\frac1{12}ML^2
\]
The pivot is a distance
\[
d=\frac L2-\frac L4=\frac L4
\]
from the midpoint.
Thus,
\[
I_P=I_{\rm CM}+Md^2
\]
Hence,
\[
I_P
=\frac1{12}ML^2+M\left(\frac L4\right)^2
\]
Therefore,
\[
I_P
=\left(\frac4{48}+\frac3{48}\right)ML^2
=\frac7{48}ML^2
\]
\(\textbf{Rotational response:}\)
\[
\tau=I_P\alpha
\]
Thus,
\[
\alpha
=\frac{(3\sqrt3/8)FL}{(7/48)ML^2}
\]
Therefore,
\[
\alpha
=\frac{18\sqrt3}{7}\frac{F}{ML}
\]
\(\textbf{Final answer:}\)
\[
\boxed{\alpha=\frac{18\sqrt3}{7}\frac{F}{ML}}
\]
518. A thin ring, a uniform disc and a solid sphere have the same mass \(M\) and radius \(R\). Each starts from rest about its standard central axis, and the same amount of rotational work \(W\) is done on each. Which pair of orderings is correct?
ⓐ. \(\omega_{\rm ring}\gt \omega_{\rm disc}\gt \omega_{\rm sphere}\) and \(L_{\rm sphere}\gt L_{\rm disc}\gt L_{\rm ring}\)
ⓑ. \(\omega_{\rm sphere}\gt \omega_{\rm disc}\gt \omega_{\rm ring}\) and \(L_{\rm ring}\gt L_{\rm disc}\gt L_{\rm sphere}\)
ⓒ. \(\omega_{\rm sphere}\gt \omega_{\rm ring}\gt \omega_{\rm disc}\) and \(L_{\rm disc}\gt L_{\rm ring}\gt L_{\rm sphere}\)
ⓓ. All three acquire the same \(\omega\) and the same \(L\)
Correct Answer: \(\omega_{\rm sphere}\gt \omega_{\rm disc}\gt \omega_{\rm ring}\) and \(L_{\rm ring}\gt L_{\rm disc}\gt L_{\rm sphere}\)
Explanation: Equal rotational work from rest gives equal final rotational kinetic energy:
\[
W=\frac12I\omega^2.
\]
The standard moments of inertia are
\[
I_{\rm ring}=MR^2,
\]
\[
I_{\rm disc}=\frac12MR^2,
\]
\[
I_{\rm sphere}=\frac25MR^2.
\]
For fixed \(W\),
\[
\omega=\sqrt{\frac{2W}{I}}.
\]
Thus smaller moment of inertia gives larger angular speed:
\[
\omega_{\rm sphere}\gt \omega_{\rm disc}\gt \omega_{\rm ring}.
\]
Angular momentum is
\[
L=I\omega.
\]
Substituting the equal-work expression,
\[
L=\sqrt{2IW}.
\]
Therefore, for fixed \(W\), angular momentum increases with \(\sqrt I\).
Hence,
\[
L_{\rm ring}\gt L_{\rm disc}\gt L_{\rm sphere}.
\]
Equal work therefore produces equal kinetic energies but neither equal angular speeds nor equal angular momenta.
\(\textbf{Final answer:}\)
\[
\omega_{\rm sphere}\gt \omega_{\rm disc}\gt \omega_{\rm ring},
\qquad
L_{\rm ring}\gt L_{\rm disc}\gt L_{\rm sphere}.
\]
519. A rigid body of constant moment of inertia rotates about a fixed axis. A constant torque \(-\tau_0\) acts while the initial angular velocity is positive. The torque continues to act after the body stops and reverses its rotational sense.
Statement I: Before the body stops, the instantaneous rotational power is negative.
Statement II: At the instant the angular velocity is zero, the instantaneous rotational power is zero even though the angular acceleration is non-zero.
Statement III: After reversal, while both \(\tau\) and \(\omega\) are negative, the instantaneous rotational power is positive.
Statement IV: After reversal, the magnitude of angular velocity must continue to decrease.
Which combination is correct?
ⓐ. I and II only
ⓑ. I, II, III and IV
ⓒ. II, III and IV only
ⓓ. I, II and III only
Correct Answer: I, II and III only
Explanation: Instantaneous rotational power about the fixed axis is
\[
P=\tau\omega.
\]
Before the body stops,
\[
\tau\lt 0,\qquad \omega\gt 0.
\]
Therefore,
\[
P\lt 0,
\]
so rotational kinetic energy is decreasing and Statement I is correct.
At the stopping instant,
\[
\omega=0,
\]
so
\[
P=\tau\omega=0.
\]
However,
\[
\alpha=\frac{\tau}{I}
\]
is still non-zero because the torque is still acting.
Thus Statement II is correct.
After reversal,
\[
\tau\lt 0,\qquad \omega\lt 0,
\]
so
\[
P\gt 0.
\]
The rotational kinetic energy then increases, making Statement III correct.
Because torque and angular velocity now have the same sign, the magnitude of the negative angular velocity grows rather than decreases.
Therefore Statement IV is incorrect.
520. Two point masses \(m\) and \(3m\) are connected by a light rigid connector of length \(4a\) and rotate about a fixed axis through their centre of mass, perpendicular to the connector. By an internal mechanism the separation is reduced to \(2a\), with the centre of mass remaining on the rotation axis. The net external torque during the change is zero. If the initial angular speed is \(\omega\), the final angular speed and the ratio of final to initial rotational kinetic energy are
ⓐ. \(2\omega\) and \(2\)
ⓑ. \(4\omega\) and \(4\)
ⓒ. \(4\omega\) and \(1\)
ⓓ. \(2\omega\) and \(4\)
Correct Answer: \(4\omega\) and \(4\)
Explanation: The unequal masses do not lie at equal distances from the centre of mass.
\(\textbf{Initial distances from the centre of mass:}\)
For separation \(4a\), the inverse mass-distance condition gives
\[
m\,r_1=3m\,r_2
\]
and
\[
r_1+r_2=4a
\]
Hence,
\[
r_1=3a,\qquad r_2=a
\]
The initial moment of inertia is
\[
I_i=m(3a)^2+3m(a)^2
\]
Therefore,
\[
I_i=12ma^2
\]
\(\textbf{After reducing the separation:}\)
The same mass ratio gives
\[
r_1'=\frac{3a}{2},
\qquad
r_2'=\frac a2
\]
Thus,
\[
I_f
=m\left(\frac{3a}{2}\right)^2
+3m\left(\frac a2\right)^2
\]
Hence,
\[
I_f=3ma^2=\frac14I_i
\]
\(\textbf{Conserve angular momentum:}\)
With zero external torque,
\[
I_i\omega=I_f\omega_f
\]
Therefore,
\[
\omega_f=4\omega
\]
\(\textbf{Compare rotational kinetic energies:}\)
For conserved angular momentum,
\[
K=\frac{L^2}{2I}
\]
Hence,
\[
\frac{K_f}{K_i}
=\frac{I_i}{I_f}
=4
\]
The increase in kinetic energy is supplied by internal work performed during the redistribution.
\(\textbf{Final answer:}\)
\[
\boxed{\omega_f=4\omega,\qquad
\frac{K_f}{K_i}=4}
\]
