JEE Physics System Of Particles and Rotational Motion MCQs Set 25

Answer: (b) \( L=\sqrt3\,R \)

Explanation:
1. About its own axis: \( I_1=\frac12MR^2 \).
2. About the transverse axis through the centre: \( I_2=M\left(\frac{R^2}{4}+\frac{L^2}{12}\right) \).
3. Equate: \( \frac{R^2}{2}=\frac{R^2}{4}+\frac{L^2}{12}\Rightarrow L^2=3R^2 \), so \( L=\sqrt3R \).

Teacher's Note:
1. The transverse moment of inertia of a cylinder has two parts, one from the radius and one from the length.
2. Memorise both: MR squared over 2 (axis) and M(R squared over 4 + L squared over 12) (transverse).

Answer: (b) \( \frac12MR^2+4mR^2 \)

Explanation:
1. Disc: \( \frac12MR^2 \).
2. Each point mass is at distance R from the axis: \( 4\times mR^2 \).
3. Total \( I=\frac12MR^2+4mR^2 \).

Teacher's Note:
1. Moments of inertia of parts about the same axis simply add.
2. A point mass at distance r contributes m r squared.

Answer: (c) 20 kg m\( ^2 \)

Explanation:
1. By the perpendicular axes theorem, the moment of inertia about the axis perpendicular to the plane equals the sum of those about two perpendicular in-plane axes: \( I_Y=I_X+I_Z \).
2. \( I_Z=100-80=20 \) kg m\( ^2 \).

Teacher's Note:
1. Identify first which axis is perpendicular to the plane; that one carries the sum.
2. The theorem holds only for plane (two-dimensional) bodies.

Answer: (d) \( 6ma^2 \)

Explanation:
1. Distance from the Y-axis equals the absolute value of the x-coordinate: 0, 2a, a, a.
2. \( I=m(0+4a^2+a^2+a^2)=6ma^2 \).

Teacher's Note:
1. For the moment of inertia about the Y-axis use the x-distances, not the y-distances.
2. The mass at the origin lies on the axis and contributes nothing.

Answer: (a) 0.4 kg m\( ^2 \)

Explanation:
1. \( I=MK^2 \).
2. \( I=10\times(0.20)^2=0.4 \) kg m\( ^2 \).

Teacher's Note:
1. Radius of gyration is the distance at which the whole mass could be concentrated to give the same I.
2. The mass or shape of the body is already contained in K.

Answer: (b) \( \frac{m\ell^2}{6} \)

Explanation:
1. Each rod about a perpendicular axis through its centre: \( \frac{m\ell^2}{12} \).
2. Both rods give the same value about this axis, so \( I=2\times\frac{m\ell^2}{12}=\frac{m\ell^2}{6} \).

Teacher's Note:
1. The axis is perpendicular to both rods and passes through the centre of each.
2. Rotating the second rod by 90 degrees does not change its I about this axis.

Answer: (b) \( \frac43M\ell^2 \)

Explanation:
1. For one rod, by the parallel axes theorem: \( I=\frac{M\ell^2}{12}+M\left(\frac\ell2\right)^2=\frac{M\ell^2}{3} \).
2. For four rods: \( 4\times\frac{M\ell^2}{3}=\frac43M\ell^2 \).

Teacher's Note:
1. The centre of each side is at a distance l/2 from the centre of the square.
2. Use the parallel axes theorem for each rod and then add.

Answer: (c) \( \sqrt5:\sqrt3 \)

Explanation:
1. Solid: \( \frac25MR_1^2 \). Hollow (thin shell): \( \frac23MR_2^2 \).
2. Equate: \( \frac{R_1^2}{R_2^2}=\frac{2/3}{2/5}=\frac53 \).
3. \( R_1:R_2=\sqrt5:\sqrt3 \).

Teacher's Note:
1. The solid sphere needs a bigger radius, because its mass is closer to the axis.
2. Always take the hollow sphere as a thin spherical shell unless told otherwise.

Answer: (c) \( \frac32MR^2 \)

Explanation:
1. About a diameter: \( I_{CM}=\frac12MR^2 \).
2. Parallel axes theorem with \( d=R \): \( I=\frac12MR^2+MR^2=\frac32MR^2 \).

Teacher's Note:
1. Start from the moment of inertia about the parallel axis through the centre of mass.
2. The distance d is the separation between the two parallel axes.

Answer: (d) \( \frac L4 \)

Explanation:
1. \( K=\frac12I\omega^2 \). Doubling \( \omega \) multiplies \( \omega^2 \) by 4, and K is halved, so \( I'=\frac I8 \).
2. \( L'=I'\omega'=\frac I8\cdot2\omega=\frac{I\omega}{4}=\frac L4 \).

Teacher's Note:
1. Both I and omega change here, so track each separately.
2. Use K = L squared over 2I to cross-check: L' squared = 2 I' K' .

Answer: (d) \( \vec r\cdot\vec\tau=0 \) and \( \vec F\cdot\vec\tau=0 \)

Explanation:
1. \( \vec\tau=\vec r\times\vec F \) is perpendicular to both \( \vec r \) and \( \vec F \).
2. So both dot products vanish: \( \vec r\cdot\vec\tau=0 \) and \( \vec F\cdot\vec\tau=0 \).

Teacher's Note:
1. A cross product is always perpendicular to the two vectors that form it.
2. So torque has no component along either the position vector or the force.

Answer: (d) \( (-1.5\hat i-4.0\hat j-\hat k) \) N m

Explanation:
1. Position relative to the point: \( \vec r'=(0-2)\hat i+0.5\hat j+(-2+3)\hat k=-2\hat i+0.5\hat j+\hat k \).
2. \( \vec\tau=\vec r'\times\vec F \) with \( \vec F=(2,0,-3) \):
3. \( \tau_x=0.5(-3)-1(0)=-1.5 \), \( \tau_y=1(2)-(-2)(-3)=-4 \), \( \tau_z=(-2)(0)-0.5(2)=-1 \).
4. \( \vec\tau=(-1.5\hat i-4\hat j-\hat k) \) N m.

Teacher's Note:
1. Torque must be taken about the given point, so first find the position vector from that point.
2. Use the determinant form to avoid sign mistakes.

Answer: (d) 157 N

Explanation:
1. \( \alpha=2\ \text{rev/s}^2=4\pi \) rad/s\( ^2 \).
2. \( I=\frac12MR^2=\frac12(50)(0.25)=6.25 \) kg m\( ^2 \).
3. \( TR=I\alpha\Rightarrow T=\frac{6.25\times4\pi}{0.5}=50\pi\approx157 \) N.

Teacher's Note:
1. Convert revolutions per second squared to radians per second squared first.
2. The tension acts at the rim, so its torque is T times R.

Answer: (b) more angular acceleration is produced in (b)

Explanation:
1. In both cases the torque is \( F\times L \), with the same lever arm, so the torque is equal.
2. Steel is denser, so in (a) the heavy half is far from the pivot and the moment of inertia is larger; in (b) the heavy half is near the pivot and the moment of inertia is smaller.
3. \( \alpha=\frac\tau I \), so the angular acceleration is larger in (b).

Teacher's Note:
1. With equal torque the body with the smaller moment of inertia accelerates more.
2. Mass far from the axis increases the moment of inertia much more than mass near the axis.

Answer: (c) increasing the angle of inclination of the ladder with the ground

Explanation:
1. For a ladder of weight W and angle \( \theta \) with the ground, balance of torques about the foot gives \( N_w=\frac W2\cot\theta \), and the friction needed at the ground equals \( N_w \).
2. As \( \theta \) increases, \( \cot\theta \) decreases, so the required friction decreases and the ladder can be balanced.
3. Changing the length alone does not change the required ratio, as \( \cot\theta \) depends on the angle only.

Teacher's Note:
1. A steeper ladder needs less friction at the ground.
2. Length cancels out because both the torque of the weight and that of the wall reaction scale with it.

Answer: (b) (i) equal, (ii) B

Explanation:
1. The block is in rotational equilibrium, so the torques of the two wires about the centre of mass are equal and opposite.
2. \( T_Ad_A=T_Bd_B \). Since \( d_B<d_A \), \( T_B>T_A \).
3. So (i) torques are equal in magnitude and (ii) wire B exerts more force.

Teacher's Note:
1. The wire closer to the centre of mass carries more of the load.
2. This is the lever principle: force times distance is the same on both sides of the balance point.

Answer: (c) \( \frac{W(d-x)}{d} \)

Explanation:
1. Torque about B: \( N_A\times d=W\times(d-x) \).
2. \( N_A=\frac{W(d-x)}{d} \).

Teacher's Note:
1. Take torques about the other support so that its reaction drops out.
2. The support nearer the centre of mass carries the larger share of the weight.

Answer: (a) 8.0

Explanation:
1. \( \omega=\alpha t=4 \) rad/s.
2. Tangential acceleration \( a_t=\alpha R=2\times0.5=1 \) m/s\( ^2 \); centripetal acceleration \( a_c=\omega^2R=16\times0.5=8 \) m/s\( ^2 \).
3. \( a=\sqrt{a_t^2+a_c^2}=\sqrt{65}\approx8 \) m/s\( ^2 \).

Teacher's Note:
1. The net acceleration of a point on the rim has a tangential part and a radial part at right angles.
2. The radial part dominates once the angular speed has built up.

Answer: (a) \( \frac\pi3 \) rad/s

Explanation:
1. Initially \( \omega=\frac{2\pi}{2}=\pi \) rad/s.
2. Finally \( I'=100+50\times2^2=300 \) kg m\( ^2 \).
3. Angular momentum is conserved: \( 100\pi=300\omega' \Rightarrow\omega'=\frac\pi3 \) rad/s.

Teacher's Note:
1. No external torque acts on the table and person, so angular momentum is conserved.
2. A person at the centre contributes almost nothing to the moment of inertia.

Answer: (d) \( 12\omega \)

Explanation:
1. The pull is central, so angular momentum is conserved: \( mr^2(3\omega)=m\left(\frac r2\right)^2\omega' \).
2. \( \omega'=4\times3\omega=12\omega \).

Teacher's Note:
1. A force directed towards the axis exerts no torque, so L is constant.
2. Halving the radius reduces I to one-quarter, so omega becomes four times.

Answer: (a) \( 2\pi \) s

Explanation:
1. Angular momentum is conserved (initially zero): \( 50\times1\times2=200\,\omega_p \Rightarrow\omega_p=0.5 \) rad/s, in the opposite sense.
2. Angular velocity of the man about the axis: \( \frac vR=0.5 \) rad/s.
3. Relative to the platform: \( \omega_{rel}=0.5+0.5=1 \) rad/s.
4. Time for one revolution \( =\frac{2\pi}{1}=2\pi \) s.

Teacher's Note:
1. The platform turns the other way, so the man's angular velocity relative to it is the sum of the two.
2. Total angular momentum stays zero throughout.