301. Arrangement A contains a point mass \(m\) at radius \(a\) and a point mass \(2m\) at radius \(2a\). Arrangement B contains a point mass \(3m\) at radius \(a\) and another point mass \(m\) at radius \(x\). Equal net torques applied to the two arrangements produce equal angular accelerations. The value of \(x\) is
ⓐ. \(\sqrt3\,a\)
ⓑ. \(\sqrt5\,a\)
ⓒ. \(\sqrt6\,a\)
ⓓ. \(3a\)
Correct Answer: \(\sqrt6\,a\)
Explanation: Equal torque and equal angular acceleration imply equal moments of inertia because
\[
\tau=I\alpha
\]
\(\textbf{Arrangement A:}\)
\[
I_A=m(a)^2+2m(2a)^2
\]
Therefore,
\[
I_A=ma^2+8ma^2=9ma^2
\]
\(\textbf{Arrangement B:}\)
\[
I_B=3m(a)^2+mx^2
\]
Thus,
\[
I_B=3ma^2+mx^2
\]
The experimental condition requires
\[
I_A=I_B
\]
Therefore,
\[
9ma^2=3ma^2+mx^2
\]
Cancelling \(m\),
\[
x^2=6a^2
\]
Hence,
\[
x=\sqrt6\,a
\]
Although the two systems have different total masses and different detailed distributions, they possess the same rotational inertia about their respective axes.
\(\textbf{Final answer:}\)
\[
\boxed{x=\sqrt6\,a}
\]
302. A rigid assembly contains a central component of moment of inertia \(2ma^2\) and two movable point masses, each of mass \(m\). Initially both point masses are at radius \(a\). The same constant torque is used before and after a rearrangement. After one movable mass is placed on the axis, the measured angular acceleration becomes exactly half its initial value. The final radius \(x\) of the other movable mass is
ⓐ. \(2a\)
ⓑ. \(\sqrt6\,a\)
ⓒ. \(3a\)
ⓓ. \(2\sqrt2\,a\)
Correct Answer: \(\sqrt6\,a\)
Explanation: \(\textbf{Initial moment of inertia:}\)
\[
I_i=2ma^2+ma^2+ma^2
\]
Thus,
\[
I_i=4ma^2
\]
The same torque is applied in both configurations.
Since
\[
\alpha=\frac{\tau}{I},
\]
the condition
\[
\alpha_f=\frac12\alpha_i
\]
requires
\[
I_f=2I_i
\]
Therefore,
\[
I_f=8ma^2
\]
\(\textbf{Final mass distribution:}\)
One point mass lies on the axis and contributes zero.
The other is at radius \(x\), so
\[
I_f=2ma^2+mx^2
\]
Hence,
\[
8ma^2=2ma^2+mx^2
\]
Therefore,
\[
x^2=6a^2
\]
Thus,
\[
x=\sqrt6\,a
\]
The reduced angular acceleration is therefore evidence that the final mass distribution has a larger moment of inertia despite one mass being moved inward to the axis.
\(\textbf{Final answer:}\)
\[
\boxed{x=\sqrt6\,a}
\]
303. A rigid body has total mass \(4m\) and moment of inertia \(20ma^2\) about a specified axis. Its radius of gyration about that axis is
ⓐ. \(5a\)
ⓑ. \(\sqrt5\,a\)
ⓒ. \(2\sqrt5\,a\)
ⓓ. \(\frac{\sqrt5}{2}a\)
Correct Answer: \(\sqrt5\,a\)
Explanation: Radius of gyration \(k\) is defined through
\[
I=Mk^2
\]
Here the total mass is
\[
M=4m
\]
and the moment of inertia is
\[
I=20ma^2
\]
Therefore,
\[
20ma^2=(4m)k^2
\]
Cancelling \(4m\),
\[
k^2=5a^2
\]
Hence,
\[
k=\sqrt5\,a
\]
Physically, this means the body's entire mass \(4m\) could be imagined concentrated at perpendicular distance \(\sqrt5\,a\) from the specified axis and would then have the same moment of inertia.
It does not mean that the real body must possess material precisely at radius \(\sqrt5\,a\).
The value of \(k\) is tied to both the mass distribution and the chosen axis.
\(\textbf{Final answer:}\)
\[
\boxed{k=\sqrt5\,a}
\]
304. The same rigid body of mass \(M\) has moments of inertia \(12MR^2\) and \(27MR^2\) about two different axes A and B. If equal torque magnitudes are applied about these axes in separate trials, then the ratios \(k_B/k_A\) and \(\alpha_A/\alpha_B\) are
ⓐ. \(\frac32\) and \(\frac94\)
ⓑ. \(\frac94\) and \(\frac32\)
ⓒ. \(\frac32\) and \(\frac49\)
ⓓ. \(\sqrt{\frac32}\) and \(\frac94\)
Correct Answer: \(\frac32\) and \(\frac94\)
Explanation: For the same body,
\[
I=Mk^2
\]
\(\textbf{Compare radii of gyration:}\)
\[
\frac{k_B}{k_A}
=\sqrt{\frac{I_B}{I_A}}
\]
Therefore,
\[
\frac{k_B}{k_A}
=\sqrt{\frac{27MR^2}{12MR^2}}
\]
Thus,
\[
\frac{k_B}{k_A}
=\sqrt{\frac94}
=\frac32
\]
\(\textbf{Compare angular accelerations under equal torque:}\)
Since
\[
\alpha=\frac{\tau}{I},
\]
we have
\[
\frac{\alpha_A}{\alpha_B}
=\frac{I_B}{I_A}
\]
Hence,
\[
\frac{\alpha_A}{\alpha_B}
=\frac{27}{12}
=\frac94
\]
A larger radius of gyration corresponds to a larger moment of inertia for the same total mass and therefore to a smaller angular acceleration under the same torque.
\(\textbf{Final answer:}\)
\[
\boxed{\frac{k_B}{k_A}=\frac32,\qquad \frac{\alpha_A}{\alpha_B}=\frac94}
\]
305. Three point masses \(m,\ 2m,\) and \(3m\) are situated at perpendicular distances \(a,\ 2a,\) and \(x\), respectively, from a fixed axis. The radius of gyration of the complete \(6m\) system about this axis is \(2a\). The value of \(x\) and the fraction of the total moment of inertia contributed by the \(3m\) particle are
ⓐ. \(\sqrt3\,a\) and \(\frac38\)
ⓑ. \(\sqrt5\,a\) and \(\frac58\)
ⓒ. \(2a\) and \(\frac12\)
ⓓ. \(\sqrt6\,a\) and \(\frac34\)
Correct Answer: \(\sqrt5\,a\) and \(\frac58\)
Explanation: The total mass is
\[
M=6m
\]
The given radius of gyration is
\[
k=2a
\]
\(\textbf{Total moment of inertia from the radius of gyration:}\)
\[
I=Mk^2
\]
Therefore,
\[
I=(6m)(2a)^2=24ma^2
\]
\(\textbf{Use the actual mass distribution:}\)
\[
I=m(a)^2+2m(2a)^2+3mx^2
\]
Thus,
\[
24ma^2=ma^2+8ma^2+3mx^2
\]
Hence,
\[
15ma^2=3mx^2,
\]
so
\[
x=\sqrt5\,a
\]
The \(3m\) particle contributes
\[
I_3=3m(\sqrt5\,a)^2=15ma^2
\]
Therefore its fractional contribution is
\[
\frac{I_3}{I}
=\frac{15}{24}
=\frac58
\]
\(\textbf{Final answer:}\)
\[
\boxed{x=\sqrt5\,a,\qquad \frac{I_3}{I}=\frac58}
\]
306. A mass distribution is enlarged geometrically so that every mass element remains unchanged in mass but its perpendicular distance from the chosen axis becomes twice its original value. If the original radius of gyration is \(k\), which statement is correct for the enlarged configuration?
ⓐ. \(k'=2k,\quad I'=4I,\quad \alpha'=\alpha/4\)
ⓑ. \(k'=4k\), \(I'=4I\), and \(\alpha'=\alpha/4\)
ⓒ. \(k'=2k\), \(I'=2I\), and \(\alpha'=\alpha/2\)
ⓓ. \(k'=k\), \(I'=4I\), and \(\alpha'=4\alpha\)
Correct Answer: \(k'=2k,\quad I'=4I,\quad \alpha'=\alpha/4\)
Explanation: Initially,
\[
I=Mk^2
\]
If every perpendicular distance is doubled, each \(r^2\) contribution increases by a factor of \(4\).
Therefore,
\[
I'=4I
\]
The total mass is unchanged, so
\[
I'=M(k')^2
\]
Hence,
\[
M(k')^2=4Mk^2
\]
Thus,
\[
k'=2k
\]
For the same applied torque,
\[
\alpha=\frac{\tau}{I}
\]
Therefore,
\[
\alpha'
=\frac{\tau}{4I}
=\frac{\alpha}{4}
\]
Radius of gyration scales like an effective distance, whereas moment of inertia scales with the square of that distance.
\(\textbf{Final answer:}\)
\[
k'=2k,\qquad I'=4I,\qquad \alpha'=\frac{\alpha}{4}
\]
307. The radius of gyration of a rigid body of mass \(M\) is \(R\) about axis A and \(\sqrt3R\) about axis B. The body is rotated with the same angular velocity about each axis in separate trials, and then the same torque magnitude is applied in separate experiments about those axes. Which pair is correct?
ⓐ. \(\displaystyle \frac{L_B}{L_A}=3,\qquad \frac{\alpha_B}{\alpha_A}=3\)
ⓑ. \(\displaystyle \frac{L_B}{L_A}=\sqrt3,\qquad \frac{\alpha_B}{\alpha_A}=\frac1{\sqrt3}\)
ⓒ. \(\displaystyle \frac{L_B}{L_A}=\frac13,\qquad \frac{\alpha_B}{\alpha_A}=3\)
ⓓ. \(\displaystyle \frac{L_B}{L_A}=3,\qquad \frac{\alpha_B}{\alpha_A}=\frac13\)
Correct Answer: \(\displaystyle \frac{L_B}{L_A}=3,\qquad \frac{\alpha_B}{\alpha_A}=\frac13\)
Explanation: For a body of mass \(M\),
\[
I=Mk^2
\]
Therefore,
\[
I_A=MR^2
\]
For axis B,
\[
I_B=M(\sqrt3R)^2=3MR^2
\]
Thus,
\[
I_B=3I_A
\]
\(\textbf{Same angular velocity comparison:}\)
For fixed-axis rotation,
\[
L=I\omega
\]
Since the angular velocities are equal,
\[
\frac{L_B}{L_A}
=\frac{I_B}{I_A}
=3
\]
\(\textbf{Same torque comparison:}\)
Using
\[
\alpha=\frac{\tau}{I},
\]
equal torques give
\[
\frac{\alpha_B}{\alpha_A}
=\frac{I_A}{I_B}
=\frac13
\]
Thus a larger radius of gyration increases angular momentum at fixed \(\omega\) but decreases angular acceleration under a fixed torque.
\(\textbf{Final answer:}\)
\[
\boxed{\frac{L_B}{L_A}=3,\qquad \frac{\alpha_B}{\alpha_A}=\frac13}
\]
308. Consider the following statements about radius of gyration.
Statement I: It is defined by \(I=Mk^2\).
Statement II: It has the dimensions of length.
Statement III: It must always be equal to the outer physical radius of the body.
Statement IV: The same rigid body may have different radii of gyration about different axes.
Statement V: If two bodies have equal mass and equal radius of gyration about specified axes, their moments of inertia about those axes are equal.
Which combination is correct?
ⓐ. I, II and III only
ⓑ. I, II, IV and V only
ⓒ. II, III, IV and V only
ⓓ. I, III and V only
Correct Answer: I, II, IV and V only
Explanation: Radius of gyration is introduced through
\[
I=Mk^2,
\]
so Statement I is correct.
From this relation,
\[
[k]=\sqrt{\frac{[I]}{[M]}}
=\sqrt{\frac{ML^2}{M}}
=L,
\]
so Statement II is correct.
Statement III is false.
The radius of gyration is an effective radius representing the mass distribution with respect to a specified axis; it need not coincide with an actual outer radius of the body.
Since moment of inertia changes when the chosen axis changes, the corresponding value
\[
k=\sqrt{\frac IM}
\]
can also change for the same rigid body.
Therefore Statement IV is correct.
Finally, if two bodies have equal \(M\) and equal \(k\), then
\[
I=Mk^2
\]
gives equal moments of inertia about the stated axes.
Hence Statement V is also correct.
Only Statements I, II, IV and V are correct.
309. The same rigid body of mass \(M\) has radii of gyration \(k_A\) and \(k_B\) about two different fixed axes A and B. The same torque magnitude is applied about the two axes in separate experiments, and the measured angular accelerations satisfy
\[
\frac{\alpha_A}{\alpha_B}=\frac94.
\]
If the body is instead rotated with the same angular speed about each axis, which pair is correct?
ⓐ. \(\displaystyle \frac{k_B}{k_A}=\frac23,\quad \frac{L_B}{L_A}=\frac49\)
ⓑ. \(\displaystyle \frac{k_B}{k_A}=\frac94,\quad \frac{L_B}{L_A}=\frac32\)
ⓒ. \(\displaystyle \frac{k_B}{k_A}=\frac32,\quad \frac{L_B}{L_A}=\frac94\)
ⓓ. \(\displaystyle \frac{k_B}{k_A}=\frac32,\quad \frac{L_B}{L_A}=\frac32\)
Correct Answer: \(\displaystyle \frac{k_B}{k_A}=\frac32,\quad \frac{L_B}{L_A}=\frac94\)
Explanation: For a body of mass \(M\),
\[
I=Mk^2
\]
\(\textbf{Use the equal-torque experiment:}\)
Since
\[
\alpha=\frac{\tau}{I},
\]
equal torque magnitudes give
\[
\frac{\alpha_A}{\alpha_B}
=\frac{I_B}{I_A}
\]
Therefore,
\[
\frac{I_B}{I_A}=\frac94
\]
Using \(I=Mk^2\) for the same body,
\[
\frac{k_B^2}{k_A^2}=\frac94
\]
Hence,
\[
\frac{k_B}{k_A}=\frac32
\]
\(\textbf{Now compare equal-angular-speed states:}\)
For fixed-axis rotation,
\[
L=I\omega
\]
Since the angular speeds are equal,
\[
\frac{L_B}{L_A}
=\frac{I_B}{I_A}
=\frac94
\]
Thus the radius-of-gyration ratio involves the square root of the inertia ratio, whereas the angular-momentum ratio at equal \(\omega\) equals the inertia ratio directly.
\(\textbf{Final answer:}\)
\[
\boxed{\frac{k_B}{k_A}=\frac32,\qquad \frac{L_B}{L_A}=\frac94}
\]
310. A rigid body of mass \(3m\) has radius of gyration \(a\) about a fixed axis. A point mass \(m\) is attached rigidly at perpendicular distance \(2a\) from the same axis. For the combined system, the radius of gyration and the ratio of its angular acceleration to that of the original body under the same net torque are
ⓐ. \(\displaystyle \frac{\sqrt7}{2}a\) and \(\displaystyle \frac37\)
ⓑ. \(\sqrt7\,a\) and \(\displaystyle \frac73\)
ⓒ. \(\displaystyle \frac{\sqrt7}{2}a\) and \(\displaystyle \frac73\)
ⓓ. \(\sqrt{\frac73}\,a\) and \(\displaystyle \frac47\)
Correct Answer: \(\displaystyle \frac{\sqrt7}{2}a\) and \(\displaystyle \frac37\)
Explanation: The original body has
\[
I_0=Mk^2
\]
Therefore,
\[
I_0=(3m)a^2=3ma^2
\]
\(\textbf{Add the point-mass contribution:}\)
The attached mass contributes
\[
I_m=m(2a)^2=4ma^2
\]
Hence,
\[
I_{\rm total}=3ma^2+4ma^2=7ma^2
\]
The new total mass is
\[
M_{\rm total}=4m
\]
Using
\[
I_{\rm total}=M_{\rm total}k'^2,
\]
we obtain
\[
7ma^2=(4m)k'^2
\]
Thus,
\[
k'^2=\frac74a^2,
\]
so
\[
k'=\frac{\sqrt7}{2}a
\]
\(\textbf{Compare torque responses:}\)
For the same net torque,
\[
\alpha\propto\frac1I
\]
Therefore,
\[
\frac{\alpha'}{\alpha_0}
=\frac{I_0}{I_{\rm total}}
=\frac{3}{7}
\]
\(\textbf{Final answer:}\)
\[
\boxed{k'=\frac{\sqrt7}{2}a,\qquad \frac{\alpha'}{\alpha_0}=\frac37}
\]
311. Every mass element of a rigid body lies at a perpendicular distance between \(R\) and \(2R\) from a specified axis. A measurement claims that the radius of gyration about this axis is \(\sqrt5\,R\). Which conclusion is correct?
ⓐ. The measurement is possible only if the body is rotating sufficiently fast
ⓑ. The measurement is impossible under the stated mass-distribution condition
ⓒ. The measurement is possible because radius of gyration need not be related to actual mass distances
ⓓ. The measurement proves that all the mass lies at distance \(\sqrt5\,R\)
Correct Answer: The measurement is impossible under the stated mass-distribution condition
Explanation: Radius of gyration satisfies
\[
k^2=\frac{I}{M}
=\frac{\int r^2\,dm}{M}.
\]
Thus \(k^2\) is the mass-weighted mean of the values \(r^2\).
Every mass element satisfies
\[
R\le r\le2R.
\]
Therefore,
\[
R^2\le r^2\le4R^2.
\]
A weighted mean of quantities all lying between \(R^2\) and \(4R^2\) must itself lie in the same interval.
Hence,
\[
R^2\le k^2\le4R^2.
\]
Therefore,
\[
R\le k\le2R.
\]
But the claimed value is
\[
k=\sqrt5\,R\gt 2R.
\]
That contradicts the stated geometry.
Angular speed does not affect the moment of inertia or radius of gyration of a given rigid mass distribution.
Thus the claimed measurement is impossible under the stated condition.
312. A body of total mass \(M\) has half of its mass at perpendicular distance \(R\) from an axis and the remaining half at distance \(2R\). The outer half is moved onto the axis while the inner half remains at \(R\). The initial and final radii of gyration are
ⓐ. \(\displaystyle \sqrt{\frac52}R\) and \(\displaystyle \frac{R}{\sqrt2}\)
ⓑ. \(\displaystyle \frac32R\) and \(\displaystyle \frac R2\)
ⓒ. \(\sqrt3R\) and \(R\)
ⓓ. \(\displaystyle \sqrt{\frac52}R\) and \(R\)
Correct Answer: \(\displaystyle \sqrt{\frac52}R\) and \(\displaystyle \frac{R}{\sqrt2}\)
Explanation: Radius of gyration is defined by
\[
I=Mk^2
\]
\(\textbf{Initial mass distribution:}\)
The moment of inertia is
\[
I_i=\frac M2R^2+\frac M2(2R)^2
\]
Thus,
\[
I_i=\frac12MR^2+2MR^2
=\frac52MR^2
\]
Therefore,
\[
Mk_i^2=\frac52MR^2
\]
Hence,
\[
k_i=\sqrt{\frac52}R
\]
\(\textbf{After moving the outer half onto the axis:}\)
The mass on the axis contributes zero.
Thus,
\[
I_f=\frac M2R^2
\]
Hence,
\[
Mk_f^2=\frac12MR^2
\]
Therefore,
\[
k_f=\frac{R}{\sqrt2}
\]
The total mass has not changed; only its distribution relative to the axis has changed.
\(\textbf{Final answer:}\)
\[
\boxed{k_i=\sqrt{\frac52}R,\qquad k_f=\frac{R}{\sqrt2}}
\]
313. Two rigid bodies P and Q have the same total mass \(M\). Their radii of gyration about their respective fixed axes are \(R\) and \(2R\). Starting from rest, the same constant torque acts on each for the same time \(t\). Which statement is correct at the end of the interval?
ⓐ. \(L_P=2L_Q\) and \(\omega_P=2\omega_Q\)
ⓑ. \(L_P=4L_Q\), but \(\omega_P=\omega_Q\)
ⓒ. \(L_P=L_Q\), but \(\omega_Q=4\omega_P\)
ⓓ. \(L_P=L_Q\), but \(\omega_P=4\omega_Q\)
Correct Answer: \(L_P=L_Q\), but \(\omega_P=4\omega_Q\)
Explanation: Their moments of inertia are
\[
I_P=MR^2
\]
and
\[
I_Q=M(2R)^2=4MR^2
\]
\(\textbf{Angular momentum after equal angular impulses:}\)
Both start from rest and experience the same torque for the same time.
Therefore,
\[
\Delta L=\tau t
\]
is identical for both.
Hence,
\[
L_P=L_Q=\tau t
\]
\(\textbf{Compare angular velocities:}\)
Using
\[
L=I\omega,
\]
we have
\[
\omega_P=\frac{L}{MR^2},
\]
whereas
\[
\omega_Q=\frac{L}{4MR^2}
\]
Therefore,
\[
\omega_P=4\omega_Q
\]
Equal torque acting for equal time gives equal angular-momentum changes, not equal angular velocities when the moments of inertia differ.
\(\textbf{Final answer:}\)
\[
L_P=L_Q,\qquad \omega_P=4\omega_Q
\]
314. A rigid body has radius of gyration \(k\) about a fixed axis. Its mass is unchanged, but its mass distribution is altered so that its new radius of gyration is \(3k/2\). The same constant net torque acts before and after the redistribution. If the original angular acceleration was \(\alpha\), the new angular acceleration is
ⓐ. \(\frac23\alpha\)
ⓑ. \(\frac49\alpha\)
ⓒ. \(\frac94\alpha\)
ⓓ. \(\frac32\alpha\)
Correct Answer: \(\frac49\alpha\)
Explanation: For fixed total mass,
\[
I=Mk^2
\]
Initially,
\[
I_i=Mk^2
\]
After redistribution,
\[
I_f=M\left(\frac{3k}{2}\right)^2
\]
Thus,
\[
I_f=\frac94Mk^2=\frac94I_i
\]
For the same net torque,
\[
\tau=I\alpha
\]
Therefore,
\[
\frac{\alpha_f}{\alpha_i}
=\frac{I_i}{I_f}
\]
Hence,
\[
\frac{\alpha_f}{\alpha}
=\frac{1}{9/4}
=\frac49
\]
The radius of gyration increases only by a factor \(3/2\), but the moment of inertia increases by its square, \(9/4\).
\(\textbf{Final answer:}\)
\[
\boxed{\alpha_f=\frac49\alpha}
\]
315. Match the body-axis combinations in List I with the correct moments of inertia in List II.
| List I | Body and axis | List II | Moment of inertia |
|---|
| P | Thin ring, perpendicular to plane through centre | 1 | \(\frac25MR^2\) |
| Q | Circular disc, perpendicular to plane through centre | 2 | \(MR^2\) |
| R | Solid sphere, about a diameter | 3 | \(\frac12MR^2\) |
| S | Solid cylinder, about its cylinder axis | 4 | \(\frac14MR^2\) |
ⓐ. P-2, Q-4, R-3, S-1
ⓑ. P-3, Q-2, R-1, S-4
ⓒ. P-2, Q-3, R-1, S-3
ⓓ. P-1, Q-3, R-2, S-4
Correct Answer: P-2, Q-3, R-1, S-3
Explanation: For a thin circular ring about the central axis perpendicular to its plane,
\[
I_{\rm ring}=MR^2.
\]
Therefore P-2.
For a circular disc about the corresponding central perpendicular axis,
\[
I_{\rm disc}=\frac12MR^2.
\]
Therefore Q-3.
For a solid sphere about a diameter,
\[
I_{\rm sphere}=\frac25MR^2.
\]
Therefore R-1.
For a solid cylinder about its own symmetry axis,
\[
I_{\rm cyl}=\frac12MR^2.
\]
Therefore S-3.
Two different bodies can have the same coefficient for particular specified axes.
The formula must therefore always be associated with both the body and the exact axis.
The correct match is P-2, Q-3, R-1, S-3.
316. A thin ring, a uniform disc and a solid sphere have the same mass \(M\) and radius \(R\). They rotate about a central axis perpendicular to the ring/disc and about a diameter of the sphere. The same net torque magnitude is applied separately to each. The correct order of angular accelerations is
ⓐ. \(\alpha_{\rm ring}\gt \alpha_{\rm disc}\gt \alpha_{\rm sphere}\)
ⓑ. \(\alpha_{\rm sphere}\gt \alpha_{\rm disc}\gt \alpha_{\rm ring}\)
ⓒ. \(\alpha_{\rm disc}\gt \alpha_{\rm sphere}\gt \alpha_{\rm ring}\)
ⓓ. \(\alpha_{\rm sphere}\gt \alpha_{\rm ring}\gt \alpha_{\rm disc}\)
Correct Answer: \(\alpha_{\rm sphere}\gt \alpha_{\rm disc}\gt \alpha_{\rm ring}\)
Explanation: For the specified axes,
\[
I_{\rm ring}=MR^2,
\]
\[
I_{\rm disc}=\frac12MR^2,
\]
and
\[
I_{\rm sphere}=\frac25MR^2.
\]
Thus the moments of inertia satisfy
\[
I_{\rm ring}\gt I_{\rm disc}\gt I_{\rm sphere}.
\]
For the same torque,
\[
\alpha=\frac{\tau}{I}.
\]
Hence the angular-acceleration order is the reverse:
\[
\alpha_{\rm sphere}\gt \alpha_{\rm disc}\gt \alpha_{\rm ring}.
\]
The solid sphere responds most strongly because its mass distribution produces the smallest moment of inertia among the three stated cases.
The ring responds least strongly because all of its mass is concentrated at radius \(R\).
\(\textbf{Final answer:}\)
\[
\alpha_{\rm sphere}\gt \alpha_{\rm disc}\gt \alpha_{\rm ring}.
\]
317. A thin ring, a solid cylinder and a solid sphere have equal mass \(M\) and radius \(R\). Each starts from rest about its standard central symmetry axis, and the same constant torque \(\tau\) acts for the same time \(t\). Which statement is correct at the end of the interval?
ⓐ. \(L_{\rm ring}=L_{\rm cylinder}=L_{\rm sphere}\), and \(\omega_{\rm sphere}\gt \omega_{\rm cylinder}\gt \omega_{\rm ring}\)
ⓑ. \(\omega_{\rm ring}=\omega_{\rm cylinder}=\omega_{\rm sphere}\), and \(L_{\rm ring}\gt L_{\rm cylinder}\gt L_{\rm sphere}\)
ⓒ. \(L_{\rm ring}\gt L_{\rm cylinder}\gt L_{\rm sphere}\), and \(\omega_{\rm sphere}\gt \omega_{\rm cylinder}\gt \omega_{\rm ring}\)
ⓓ. \(L_{\rm ring}=L_{\rm cylinder}=L_{\rm sphere}\), and \(\omega_{\rm ring}=\omega_{\rm cylinder}=\omega_{\rm sphere}\)
Correct Answer: \(L_{\rm ring}=L_{\rm cylinder}=L_{\rm sphere}\), and \(\omega_{\rm sphere}\gt \omega_{\rm cylinder}\gt \omega_{\rm ring}\)
Explanation: The same torque acts for the same time on each body.
Since all start from rest,
\[
L=\tau t
\]
for each body.
Therefore,
\[
L_{\rm ring}=L_{\rm cylinder}=L_{\rm sphere}.
\]
Their standard moments of inertia are
\[
I_{\rm ring}=MR^2,
\]
\[
I_{\rm cylinder}=\frac12MR^2,
\]
and
\[
I_{\rm sphere}=\frac25MR^2.
\]
Using
\[
L=I\omega,
\]
equal angular momentum gives
\[
\omega\propto\frac1I.
\]
Therefore,
\[
\omega_{\rm sphere}\gt \omega_{\rm cylinder}\gt \omega_{\rm ring}.
\]
The same result follows from their different angular accelerations under the common torque.
Equal angular impulse fixes \(L\), while the mass distribution determines the resulting \(\omega\).
318. A thin ring, a circular disc and a solid sphere all have the same mass \(M\) and radius \(R\). Their radii of gyration about the ring/disc central perpendicular axis and a diameter of the sphere are \(k_r,\ k_d,\ k_s\), respectively. The correct order is
ⓐ. \(k_r\gt k_d\gt k_s\)
ⓑ. \(k_s\gt k_d\gt k_r\)
ⓒ. \(k_d\gt k_r\gt k_s\)
ⓓ. \(k_r\gt k_s\gt k_d\)
Correct Answer: \(k_r\gt k_d\gt k_s\)
Explanation: Radius of gyration is related to moment of inertia through
\[
I=Mk^2.
\]
For the thin ring,
\[
MR^2=Mk_r^2,
\]
so
\[
k_r=R.
\]
For the circular disc,
\[
\frac12MR^2=Mk_d^2,
\]
giving
\[
k_d=\frac{R}{\sqrt2}.
\]
For the solid sphere,
\[
\frac25MR^2=Mk_s^2,
\]
so
\[
k_s=\sqrt{\frac25}R.
\]
Numerically,
\[
1\gt \frac1{\sqrt2}\gt \sqrt{\frac25}.
\]
Therefore,
\[
k_r\gt k_d\gt k_s.
\]
The ordering expresses how far from the chosen axis the mass is effectively distributed.
319. A thin circular ring and a solid cylinder have the same mass \(M\) and radius \(R\). The ring rotates about one of its diameters, while the cylinder rotates about its own symmetry axis. The same torque magnitude acts about the stated axis in each case. Which statement is correct?
ⓐ. The ring has twice the moment of inertia and half the angular acceleration
ⓑ. The cylinder has twice the moment of inertia and half the angular acceleration
ⓒ. Both have the same moment of inertia and therefore the same angular acceleration
ⓓ. Their moments of inertia cannot be compared because the axes are of different types
Correct Answer: Both have the same moment of inertia and therefore the same angular acceleration
Explanation: For a thin ring about a diameter in its plane,
\[
I_{\rm ring,diameter}
=\frac12MR^2
\]
For a solid cylinder about its own symmetry axis,
\[
I_{\rm cylinder}
=\frac12MR^2
\]
Thus,
\[
I_{\rm ring}=I_{\rm cylinder}
\]
Under equal net torque,
\[
\alpha=\frac{\tau}{I}
\]
Therefore,
\[
\alpha_{\rm ring}=\alpha_{\rm cylinder}
\]
The equality does not mean the bodies have identical mass distributions.
It means only that these two specific body-axis combinations happen to have the same rotational inertia for equal \(M\) and \(R\).
Correct use of standard values requires careful attention to the stated axis.
320. A uniform thin rod of mass \(M\) and length \(L\) rotates about an axis through its midpoint and perpendicular to the rod. A thin circular ring of the same mass \(M\) and radius \(R\) rotates about its central axis perpendicular to its plane. For their moments of inertia to be equal, the rod length must be
ⓐ. \(\sqrt6\,R\)
ⓑ. \(2\sqrt3\,R\)
ⓒ. \(3R\)
ⓓ. \(4R\)
Correct Answer: \(2\sqrt3\,R\)
Explanation: For the uniform thin rod about a perpendicular axis through its midpoint,
\[
I_{\rm rod}=\frac1{12}ML^2
\]
For the thin ring about its central perpendicular axis,
\[
I_{\rm ring}=MR^2
\]
The condition of equal moments of inertia is
\[
\frac1{12}ML^2=MR^2
\]
Cancelling \(M\),
\[
\frac{L^2}{12}=R^2
\]
Therefore,
\[
L^2=12R^2
\]
Taking the positive physical length,
\[
L=\sqrt{12}\,R
\]
Hence,
\[
L=2\sqrt3\,R
\]
If equal torques were then applied about the stated axes, the two bodies would also acquire equal angular accelerations.
\(\textbf{Final answer:}\)
\[
\boxed{L=2\sqrt3\,R}
\]
