301. A body has a large mass, but most of its mass lies close to the axis of rotation. Another body has smaller mass, but its mass lies far from the same kind of axis. The second body can have a larger moment of inertia because
ⓐ. distance from the axis is ignored in \(I\)
ⓑ. smaller masses always rotate more slowly
ⓒ. moment of inertia depends only on total mass
ⓓ. mass farther from the axis contributes more
Correct Answer: mass farther from the axis contributes more
Explanation: Moment of inertia depends on both mass and distribution of mass about the axis. The formula \(I=\sum m_ir_i^2\) shows that the distance from the axis is squared. Because of this squared distance, mass placed far from the axis can contribute very strongly. A body with smaller total mass can therefore have a larger moment of inertia if that mass is distributed much farther from the axis. Total mass alone does not decide resistance to rotation.
302. For the same total mass \(M\) and radius \(R\), a thin ring and a uniform disc rotate about their common central axis perpendicular to their planes. The ring has greater moment of inertia because
ⓐ. its angular velocity must be larger
ⓑ. its centre of mass is outside the ring
ⓒ. its mass is zero at the centre
ⓓ. all its mass is farther from the axis
Correct Answer: all its mass is farther from the axis
Explanation: A thin ring has nearly all its mass at distance \(R\) from the central axis. A uniform disc has mass spread from the centre up to the rim, so much of its mass is at distances less than \(R\). Since moment of inertia depends on \(r^2\), mass farther from the axis contributes more. For the same \(M\) and \(R\), \(I_{\text{ring}}=MR^2\), while \(I_{\text{disc}}=\frac{1}{2}MR^2\). The difference comes from mass distribution, not from the total mass alone.
303. Match each body with its standard moment of inertia about the stated symmetry axis.
| Column I | Column II |
|---|
| P. Thin ring about central axis perpendicular to plane | 1. \(MR^2\) |
| Q. Uniform disc about central axis perpendicular to plane | 2. \(\frac{1}{2}MR^2\) |
| R. Uniform solid sphere about diameter | 3. \(\frac{2}{5}MR^2\) |
| S. Thin spherical shell about diameter | 4. \(\frac{2}{3}MR^2\) |
ⓐ. P-1, Q-3, R-2, S-4
ⓑ. P-4, Q-2, R-3, S-1
ⓒ. P-1, Q-2, R-3, S-4
ⓓ. P-2, Q-1, R-3, S-4
Correct Answer: P-1, Q-2, R-3, S-4
Explanation: A thin ring about its central perpendicular axis has moment of inertia \(MR^2\). A uniform disc about the same type of central axis has \(I=\frac{1}{2}MR^2\). A uniform solid sphere about a diameter has \(I=\frac{2}{5}MR^2\). A thin spherical shell about a diameter has \(I=\frac{2}{3}MR^2\). The formulas differ because mass is distributed differently with respect to the chosen axis.
304. A thin ring and a uniform disc have the same mass \(2\,\text{kg}\) and radius \(0.50\,\text{m}\). Their moments of inertia about central axes perpendicular to their planes are compared. The difference \(I_{\text{ring}}-I_{\text{disc}}\) is
ⓐ. \(0.250\,\text{kg m}^2\)
ⓑ. \(1.000\,\text{kg m}^2\)
ⓒ. \(0.500\,\text{kg m}^2\)
ⓓ. \(0.125\,\text{kg m}^2\)
Correct Answer: \(0.250\,\text{kg m}^2\)
Explanation: \( \textbf{Given data:} \) \(M=2\,\text{kg}\), and \(R=0.50\,\text{m}\).
\( \textbf{Ring formula:} \)
\[
I_{\text{ring}}=MR^2
\]
\( \textbf{Disc formula:} \)
\[
I_{\text{disc}}=\frac{1}{2}MR^2
\]
\( \textbf{Difference:} \)
\[
I_{\text{ring}}-I_{\text{disc}}=MR^2-\frac{1}{2}MR^2
\]
\[
I_{\text{ring}}-I_{\text{disc}}=\frac{1}{2}MR^2
\]
\( \textbf{Substitution:} \)
\[
I_{\text{ring}}-I_{\text{disc}}=\frac{1}{2}(2)(0.50)^2
\]
\( \textbf{Square radius:} \)
\[
(0.50)^2=0.25
\]
\( \textbf{Calculation:} \)
\[
I_{\text{ring}}-I_{\text{disc}}=0.250\,\text{kg m}^2
\]
\( \textbf{Final answer:} \) The difference is \(0.250\,\text{kg m}^2\).
305. A uniform rod of mass \(M\) and length \(L\) has moment of inertia \(\frac{1}{12}ML^2\) about an axis through its centre and perpendicular to its length. About a parallel axis through one end, its moment of inertia is
ⓐ. \(\frac{1}{6}ML^2\)
ⓑ. \(\frac{1}{3}ML^2\)
ⓒ. \(\frac{1}{12}ML^2\)
ⓓ. \(ML^2\)
Correct Answer: \(\frac{1}{3}ML^2\)
Explanation: \( \textbf{Known central-axis value:} \)
\[
I_{\text{CM}}=\frac{1}{12}ML^2
\]
\( \textbf{Parallel-axis theorem:} \)
\[
I=I_{\text{CM}}+Md^2
\]
\( \textbf{Distance between axes:} \) The centre of the rod is \(\frac{L}{2}\) from either end, so \(d=\frac{L}{2}\).
\( \textbf{Substitution:} \)
\[
I=\frac{1}{12}ML^2+M\left(\frac{L}{2}\right)^2
\]
\( \textbf{Square the distance:} \)
\[
\left(\frac{L}{2}\right)^2=\frac{L^2}{4}
\]
\( \textbf{Add terms:} \)
\[
I=\frac{1}{12}ML^2+\frac{1}{4}ML^2
\]
\[
I=\left(\frac{1}{12}+\frac{3}{12}\right)ML^2
\]
\[
I=\frac{1}{3}ML^2
\]
\( \textbf{Final answer:} \) The moment of inertia about the end is \(\frac{1}{3}ML^2\).
306. The parallel-axis theorem states that if an axis is shifted parallel to a centre-of-mass axis by distance \(d\), then
ⓐ. \(I=Md\)
ⓑ. \(I=I_{\text{CM}}-Md^2\)
ⓒ. \(I=I_{\text{CM}}+Md^2\)
ⓓ. \(I=\frac{I_{\text{CM}}}{Md^2}\)
Correct Answer: \(I=I_{\text{CM}}+Md^2\)
Explanation: The parallel-axis theorem connects moments of inertia about two parallel axes. One of the axes must pass through the centre of mass of the body. If the other axis is at a perpendicular distance \(d\), then \(I=I_{\text{CM}}+Md^2\). The added term \(Md^2\) is always non-negative, so the shifted parallel-axis moment of inertia is larger than or equal to \(I_{\text{CM}}\). The distance \(d\) is the separation between the axes, not the length of the body in any direction.
307. A student writes \(I=I_{\text{CM}}-Md^2\) for a parallel axis away from the centre of mass. The best correction is that
ⓐ. the correct form is \(I=I_{\text{CM}}+Md^2\)
ⓑ. the formula is correct only when \(d\) is large
ⓒ. moment of inertia is independent of axis position
ⓓ. \(M\) should be replaced by angular velocity
Correct Answer: the correct form is \(I=I_{\text{CM}}+Md^2\)
Explanation: The correct parallel-axis theorem is \(I=I_{\text{CM}}+Md^2\). Moving to a parallel axis away from the centre-of-mass axis increases the effective distance of the mass distribution from the new axis. Therefore, the additional term \(Md^2\) is added, not subtracted. A subtraction could even give an impossible negative moment of inertia for some values. Moment of inertia cannot be negative because it is built from terms of the form \(m_ir_i^2\).
308. A uniform disc of mass \(M\) and radius \(R\) has \(I_{\text{CM}}=\frac{1}{2}MR^2\) about its central axis perpendicular to its plane. About a parallel tangent axis in its plane's perpendicular direction, the moment of inertia is
ⓐ. \(2MR^2\)
ⓑ. \(MR^2\)
ⓒ. \(\frac{1}{2}MR^2\)
ⓓ. \(\frac{3}{2}MR^2\)
Correct Answer: \(\frac{3}{2}MR^2\)
Explanation: \( \textbf{Central-axis value:} \)
\[
I_{\text{CM}}=\frac{1}{2}MR^2
\]
\( \textbf{Parallel-axis theorem:} \)
\[
I=I_{\text{CM}}+Md^2
\]
\( \textbf{Axis separation:} \) A tangent axis parallel to the central axis is at distance \(d=R\) from the centre.
\( \textbf{Substitution:} \)
\[
I=\frac{1}{2}MR^2+M(R)^2
\]
\( \textbf{Combine terms:} \)
\[
I=\frac{1}{2}MR^2+MR^2
\]
\[
I=\frac{3}{2}MR^2
\]
\( \textbf{Final answer:} \) The moment of inertia about the tangent parallel axis is \(\frac{3}{2}MR^2\).
309. The perpendicular-axis theorem for a plane lamina lying in the \(xy\)-plane is written as
ⓐ. \(I_z=I_x-I_y\)
ⓑ. \(I_z=I_x+I_y\)
ⓒ. \(I_x=I_y+I_z\)
ⓓ. \(I_xI_y=I_z\)
Correct Answer: \(I_z=I_x+I_y\)
Explanation: The perpendicular-axis theorem applies to a plane lamina. If the lamina lies in the \(xy\)-plane, and the \(z\)-axis is perpendicular to the lamina through the same point, then \(I_z=I_x+I_y\). All three axes must pass through the same point, with \(x\)- and \(y\)-axes lying in the plane of the lamina. The theorem is not a general three-dimensional solid-body theorem. It is especially useful for finding the moment of inertia of flat bodies.
310. A uniform circular disc has moment of inertia \(I_z=\frac{1}{2}MR^2\) about the central axis perpendicular to its plane. By symmetry, \(I_x=I_y\) for two perpendicular diameters in its plane. The moment of inertia about a diameter is
ⓐ. \(\frac{1}{4}MR^2\)
ⓑ. \(\frac{1}{8}MR^2\)
ⓒ. \(MR^2\)
ⓓ. \(\frac{1}{2}MR^2\)
Correct Answer: \(\frac{1}{4}MR^2\)
Explanation: \( \textbf{Given perpendicular-axis value:} \)
\[
I_z=\frac{1}{2}MR^2
\]
\( \textbf{Perpendicular-axis theorem:} \)
\[
I_z=I_x+I_y
\]
\( \textbf{Symmetry of disc:} \) For two perpendicular diameters in the plane, \(I_x=I_y\).
\( \textbf{Define the diameter moment:} \)
\[
I_z=I_d+I_d=2I_d
\]
\( \textbf{Substitute the perpendicular-axis moment:} \)
\[
\frac{1}{2}MR^2=2I_d
\]
\( \textbf{Solve:} \)
\[
I_d=\frac{1}{4}MR^2
\]
\( \textbf{Final answer:} \) The moment of inertia about a diameter is \(\frac{1}{4}MR^2\).
311. Consider the following statements about the two axis theorems.
I. The parallel-axis theorem contains the term \(Md^2\).
II. The perpendicular-axis theorem applies to plane laminas.
III. The perpendicular-axis theorem can be used directly for any solid sphere.
ⓐ. I and III only
ⓑ. I and II only
ⓒ. I, II, and III
ⓓ. II and III only
Correct Answer: I and II only
Explanation: Statement I is true because the parallel-axis theorem is \(I=I_{\text{CM}}+Md^2\). Statement II is true because the perpendicular-axis theorem is meant for plane laminas. Statement III is not suitable because a solid sphere is a three-dimensional body, not a plane lamina. The perpendicular-axis theorem connects two in-plane axes and one perpendicular axis for a flat body. Using the correct theorem requires checking the geometry of the body and the axes first.
312. A plane lamina has \(I_x=3\,\text{kg m}^2\) and \(I_y=5\,\text{kg m}^2\) about two perpendicular axes in its plane through the same point. Its moment of inertia about the perpendicular axis through that point is
ⓐ. \(8\,\text{kg m}^2\)
ⓑ. \(4\,\text{kg m}^2\)
ⓒ. \(15\,\text{kg m}^2\)
ⓓ. \(2\,\text{kg m}^2\)
Correct Answer: \(8\,\text{kg m}^2\)
Explanation: \( \textbf{Given data:} \) \(I_x=3\,\text{kg m}^2\), and \(I_y=5\,\text{kg m}^2\).
\( \textbf{Body type:} \) The body is a plane lamina, so the perpendicular-axis theorem can be used.
\( \textbf{Theorem:} \)
\[
I_z=I_x+I_y
\]
\( \textbf{Substitution:} \)
\[
I_z=3\,\text{kg m}^2+5\,\text{kg m}^2
\]
\( \textbf{Calculation:} \)
\[
I_z=8\,\text{kg m}^2
\]
\( \textbf{Axis condition:} \) The three axes must pass through the same point for this direct addition to be valid.
\( \textbf{Final answer:} \) The moment of inertia about the perpendicular axis is \(8\,\text{kg m}^2\).
313. A plane lamina lies in the \(xy\)-plane. If \(I_x=I_y\) about two perpendicular axes in its plane through the same point, and \(I_z=12\,\text{kg m}^2\) about the perpendicular axis through that point, then \(I_x\) is
ⓐ. \(6\,\text{kg m}^2\)
ⓑ. \(24\,\text{kg m}^2\)
ⓒ. \(3\,\text{kg m}^2\)
ⓓ. \(12\,\text{kg m}^2\)
Correct Answer: \(6\,\text{kg m}^2\)
Explanation: \( \textbf{Given data:} \) \(I_z=12\,\text{kg m}^2\), and \(I_x=I_y\).
\( \textbf{Applicable theorem:} \) Since the body is a plane lamina, the perpendicular-axis theorem may be used.
\[
I_z=I_x+I_y
\]
\( \textbf{Use symmetry condition:} \)
\[
I_x=I_y
\]
\( \textbf{Substitute into the theorem:} \)
\[
I_z=I_x+I_x=2I_x
\]
\( \textbf{Solve for the in-plane moment:} \)
\[
I_x=\frac{I_z}{2}
\]
\[
I_x=\frac{12}{2}\,\text{kg m}^2=6\,\text{kg m}^2
\]
\( \textbf{Final answer:} \) The moment of inertia about each in-plane symmetry axis is \(6\,\text{kg m}^2\).
314. A student tries to use the perpendicular-axis theorem for a solid cylinder about three mutually perpendicular axes through its centre. The main problem with this method is that
ⓐ. the theorem works only when mass is zero
ⓑ. the theorem is used only for centre-of-mass position
ⓒ. the theorem requires all axes to be parallel
ⓓ. the theorem applies directly only to plane laminas
Correct Answer: the theorem applies directly only to plane laminas
Explanation: The perpendicular-axis theorem is a special result for plane laminas. It connects the moment of inertia about an axis perpendicular to the plane with the moments about two perpendicular axes lying in the plane. A solid cylinder is a three-dimensional body, so this theorem cannot be applied directly in the same way. The three axes may still be mutually perpendicular, but that alone is not enough. The shape and mass distribution must satisfy the plane-lamina condition.
315. A uniform rod has moment of inertia \(I_{\text{CM}}\) about an axis through its centre and perpendicular to its length. About a parallel axis through a point at distance \(d\) from its centre, the graph of \(I\) versus \(d^2\) is
ⓐ. a straight line with slope \(M\) and intercept \(I_{\text{CM}}\)
ⓑ. a straight line with slope \(I_{\text{CM}}\) and intercept \(M\)
ⓒ. a curve with zero intercept and slope \(d\)
ⓓ. a horizontal line independent of \(d\)
Correct Answer: a straight line with slope \(M\) and intercept \(I_{\text{CM}}\)
Explanation: The parallel-axis theorem gives \(I=I_{\text{CM}}+Md^2\). If \(I\) is plotted on the vertical axis and \(d^2\) on the horizontal axis, the equation has the form \(Y=c+mX\). The intercept is \(I_{\text{CM}}\), because that is the value of \(I\) when \(d=0\). The slope is \(M\), because \(M\) multiplies \(d^2\). This graph also shows why shifting the axis farther away increases the moment of inertia.
316. A uniform rod of mass \(2\,\text{kg}\) and length \(1.2\,\text{m}\) is rotated about an axis through one end and perpendicular to its length. Using \(I=\frac{1}{3}ML^2\), its moment of inertia is
ⓐ. \(1.44\,\text{kg m}^2\)
ⓑ. \(0.48\,\text{kg m}^2\)
ⓒ. \(0.72\,\text{kg m}^2\)
ⓓ. \(0.96\,\text{kg m}^2\)
Correct Answer: \(0.96\,\text{kg m}^2\)
Explanation: \( \textbf{Given data:} \) \(M=2\,\text{kg}\), and \(L=1.2\,\text{m}\).
\( \textbf{Required quantity:} \) Moment of inertia about an end perpendicular to the rod.
\( \textbf{Formula:} \)
\[
I=\frac{1}{3}ML^2
\]
\( \textbf{Substitution:} \)
\[
I=\frac{1}{3}(2)(1.2)^2
\]
\( \textbf{Square the length:} \)
\[
(1.2)^2=1.44
\]
\( \textbf{Multiply:} \)
\[
I=\frac{1}{3}(2)(1.44)
\]
\[
I=\frac{2.88}{3}=0.96\,\text{kg m}^2
\]
\( \textbf{Final answer:} \) The moment of inertia is \(0.96\,\text{kg m}^2\).
317. The moment of inertia of a body about a given axis is doubled while the same torque acts on it. Its angular acceleration becomes
ⓐ. twice the earlier value
ⓑ. four times the earlier value
ⓒ. unchanged
ⓓ. half the earlier value
Correct Answer: half the earlier value
Explanation: For rotation about a fixed axis, the rotational form of Newton's second law is \(\tau=I\alpha\). For the same torque \(\tau\), angular acceleration is \(\alpha=\frac{\tau}{I}\). If \(I\) is doubled, the denominator in this expression doubles. Therefore, \(\alpha\) becomes half of its previous value. Moment of inertia plays the role of rotational resistance, so larger \(I\) means smaller angular acceleration for the same torque.
318. The rotational form of Newton's second law for a rigid body rotating about a fixed axis is
ⓐ. \(L=I\alpha\)
ⓑ. \(\tau=I\alpha\)
ⓒ. \(\tau=I\omega\)
ⓓ. \(F=I\alpha\)
Correct Answer: \(\tau=I\alpha\)
Explanation: For rotation about a fixed axis, the net external torque is related to angular acceleration by \(\tau=I\alpha\). Here, \(I\) is the moment of inertia about the same axis, and \(\alpha\) is the angular acceleration about that axis. This equation is analogous to \(F=ma\) in translational motion. It applies cleanly when the axis is fixed and \(I\) about that axis is constant. The equation does not say torque equals angular momentum; torque is related to the rate of change of angular momentum.
319. A wheel has moment of inertia \(4\,\text{kg m}^2\). A net torque of \(12\,\text{N m}\) acts on it about its fixed axle. The angular acceleration is
ⓐ. \(0.33\,\text{rad s}^{-2}\)
ⓑ. \(3\,\text{rad s}^{-2}\)
ⓒ. \(8\,\text{rad s}^{-2}\)
ⓓ. \(48\,\text{rad s}^{-2}\)
Correct Answer: \(3\,\text{rad s}^{-2}\)
Explanation: \( \textbf{Given data:} \) \(I=4\,\text{kg m}^2\), and \(\tau=12\,\text{N m}\).
\( \textbf{Required quantity:} \) Angular acceleration \(\alpha\).
\( \textbf{Rotational dynamics relation:} \)
\[
\tau=I\alpha
\]
\( \textbf{Rearrange:} \)
\[
\alpha=\frac{\tau}{I}
\]
\( \textbf{Substitution:} \)
\[
\alpha=\frac{12\,\text{N m}}{4\,\text{kg m}^2}
\]
\( \textbf{Unit check:} \) Since \(1\,\text{N}=1\,\text{kg m s}^{-2}\), \(\frac{\text{N m}}{\text{kg m}^2}=\text{s}^{-2}\), written as \(\text{rad s}^{-2}\) for angular acceleration.
\( \textbf{Calculation:} \)
\[
\alpha=3\,\text{rad s}^{-2}
\]
\( \textbf{Final answer:} \) The angular acceleration is \(3\,\text{rad s}^{-2}\).
320. A light string is wound around a solid cylinder of radius \(R\) and moment of inertia \(I\). If the string is pulled with tension \(T\) tangentially without slipping on the rim, the angular acceleration of the cylinder about its fixed axis is
ⓐ. \(\frac{I}{TR}\)
ⓑ. \(ITR\)
ⓒ. \(\frac{TR}{I}\)
ⓓ. \(\frac{T}{IR}\)
Correct Answer: \(\frac{TR}{I}\)
Explanation: The tension acts tangentially at the rim, so the torque magnitude about the cylinder's axis is \(\tau=TR\). The rotational equation for a fixed axis is \(\tau=I\alpha\). Substituting the torque gives \(TR=I\alpha\). Solving for angular acceleration gives \(\alpha=\frac{TR}{I}\). The radius appears in the numerator because a larger radius gives a larger moment arm for the same tension.
