JEE Physics System Of Particles and Rotational Motion MCQs Set 34

Answer: (d) Downwards at a point D such that BD = 1 m

Explanation:
1. The 6 N force must balance the 6 N upward total.
2. Torque about A: \( 4\times3=6\times x\Rightarrow x=2\ \text{m} \) from A, so BD = 1 m.

Teacher's Note:
1. Take moments about one end; the equilibrant must lie nearer the larger force.

Answer: (b) 4 m/s squared

Explanation:
1. In the cart frame the pseudo force \( ma \) acts at the centre (height 5 cm); the weight acts 2 cm from the tipping edge.
2. \( ma\cdot\frac h2=mg\cdot\frac d2\Rightarrow a=g\frac dh=9.8\times0.4\approx3.9\ \text{m/s}^2 \).
3. Friction needed is \( ma=3.9m \) which is less than the limit \( \mu mg=4.9m \), so it tips before sliding.

Teacher's Note:
1. Compare the tipping condition with the sliding condition and use whichever occurs first.

Answer: (b) B

Explanation:
1. \( I_A=\frac12mR^2=0.5mR^2,\quad I_B=\frac{mR^2}{6}=0.17mR^2,\quad I_C=\frac25mR^2=0.4mR^2 \).
2. B has the smallest.

Teacher's Note:
1. Compare the numerical coefficients of mR squared.

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

Explanation:
1. About the right angle vertex C: \( I_C=\frac{Ma^2}6+\frac{Ma^2}6=\frac{Ma^2}3 \).
2. Centroid G is at \( d^2=\frac{2a^2}{9} \) from C: \( I_G=\frac{Ma^2}3-\frac{2Ma^2}9=\frac{Ma^2}9 \).
3. Midpoint of AB is \( \frac{a^2}{18} \) (squared distance) from G: \( I=\frac{Ma^2}9+\frac{Ma^2}{18}=\frac{Ma^2}6 \)

Teacher's Note:
1. Shift to the centroid with the parallel axis theorem and then out to the required point.

Answer: (a) Less than 1

Explanation:
1. \( I_1=\frac{ML^2}{12} \); the arc radius is \( R=\frac L\pi \) so \( I_2=MR^2=\frac{ML^2}{\pi^2} \).
2. \( \frac{I_1}{I_2}=\frac{\pi^2}{12}=0.82<1 \)

Teacher's Note:
1. All the mass of an arc lies at the same distance from the centre.

Answer: (a) \( \frac12MR^2 \)

Explanation:
1. Four such sectors make the whole disc, each with the same moment of inertia about the centre: \( 4I=\frac12(4M)R^2 \), so \( I=\frac12MR^2 \).

Teacher's Note:
1. A fraction of a disc with the same density has the same I per unit mass about the central axis.

Answer: (a) I

Explanation:
1. For a square plate the moments about all axes in its plane through the centre are equal (the inertia tensor is isotropic in the plane): \( I_{CD}=I\cos^2\theta+I\sin^2\theta=I \).

Teacher's Note:
1. Use the principal axes and the symmetry of the square.

Answer: (b) Exceeds g at some point

Explanation:
1. Treat the pencil as a rod of length L pivoted at the bottom: \( \alpha=\frac{3g\sin\theta}{2L} \).
2. Tangential acceleration of the tip \( =\alpha L=\frac32g\sin\theta \), which exceeds g once \( \sin\theta>\frac23 \).

Teacher's Note:
1. The tip of a falling rod accelerates faster than free fall.

Answer: (c) \( \frac{5mg}{2} \)

Explanation:
1. Energy: \( mg\frac L2=\frac12\cdot\frac{mL^2}3\omega^2\Rightarrow\omega^2=\frac{3g}L \).
2. At the vertical position, centripetal requirement: \( N-mg=m\omega^2\frac L2=\frac{3mg}2 \), so \( N=\frac{5mg}2 \)

Teacher's Note:
1. At the bottom the hinge supports the weight plus the centripetal force of the centre of mass.

Answer: (d) \( \frac32g\cos\theta \)

Explanation:
1. \( \tau=Mg\frac L2\cos\theta \), so \( \alpha=\frac{\tau}{ML^2/3}=\frac{3g\cos\theta}{2L} \).
2. Initially the free end has only tangential acceleration \( a=\alpha L=\frac32g\cos\theta \)

Teacher's Note:
1. The initial acceleration of the end of a hinged rod is alpha times L.

Answer: (b) 2 s

Explanation:
1. \( \frac12I\omega^2=1500\Rightarrow\omega^2=2500\Rightarrow\omega=50\ \text{rad/s} \).
2. \( t=\frac\omega\alpha=\frac{50}{25}=2\ \text{s} \)

Teacher's Note:
1. Find the final angular speed from the kinetic energy, then divide by the angular acceleration.

Answer: (b) \( \sqrt{\frac{6g}{5l}} \)

Explanation:
1. Each arm has mass m/2. About P: horizontal arm \( \frac13\cdot\frac m2l^2=\frac{ml^2}6 \); vertical arm \( \frac m2\left(\frac{l^2}{12}+l^2+\frac{l^2}4\right)=\frac{2ml^2}3 \). Total \( I=\frac{5ml^2}6 \).
2. After a 90 degree swing both centres of mass drop by l/2, so the loss of potential energy is \( mg\frac l2 \).
3. \( \frac12\cdot\frac{5ml^2}6\omega^2=\frac{mgl}2\Rightarrow\omega=\sqrt{\frac{6g}{5l}} \)

Teacher's Note:
1. Add the moments of inertia of the two arms about P and use energy conservation.

Answer: (a) Its angular velocity increases, but the force on the hinge remains constant

Explanation:
1. Constant torque \( FR \) gives constant angular acceleration, so the angular velocity increases.
2. The centre does not accelerate, so the hinge supplies a constant force equal and opposite to the thread force F (plus weight).

Teacher's Note:
1. A constant torque gives uniform angular acceleration; the hinge force just balances F.

Answer: (a) Angular momentum remains the same while kinetic energy increases

Explanation:
1. No external torque: \( L=I\omega \) is constant.
2. \( K=\frac{L^2}{2I} \) increases when I decreases (the skater does work pulling the arms in).

Teacher's Note:
1. Kinetic energy at fixed angular momentum is inversely proportional to I.

Answer: (b) \( \frac{2v}{3d} \)

Explanation:
1. Angular momentum about the axle: \( mv\frac d2=I\omega \) with \( I=\frac{6md^2}{12}+m\frac{d^2}4=\frac34md^2 \).
2. \( \omega=\frac{mvd/2}{3md^2/4}=\frac{2v}{3d} \)

Teacher's Note:
1. Conserve angular momentum about the pivot and include the particle in the final moment of inertia.

Answer: (a) 3M

Explanation:
1. Angular momentum: \( \frac{Ml^2}3\omega=mv\frac l3 \), so \( v=\frac{Ml\omega}{m} \).
2. Energy: \( \frac12\cdot\frac{Ml^2}3\omega^2=\frac12mv^2=\frac{M^2l^2\omega^2}{2m} \).
3. Equating: \( \frac13=\frac Mm\Rightarrow m=3M \)

Teacher's Note:
1. An elastic collision that stops the rod conserves both angular momentum and kinetic energy.

Answer: (d) North

Explanation:
1. For rolling towards the east the top of the wheel moves east.
2. Taking east = x, north = y, up = z: \( \vec v=\vec\omega\times\vec r \) with \( \vec r=R\hat z \) requires \( \vec\omega\parallel+\hat y \): north.

Teacher's Note:
1. Use the right hand rule with the top of the wheel moving forward.

Answer: (c) \( \frac{24}{289} \) rad/s

Explanation:
1. After 5 s the position is (15, 8), so \( r^2=15^2+8^2=289 \).
2. \( \omega=\frac{|xv_y-yv_x|}{r^2}=\frac{8\times3}{289}=\frac{24}{289}\ \text{rad/s} \)

Teacher's Note:
1. Angular velocity about a point is v times the perpendicular distance divided by r squared.

Answer: (b) Downward, much less than g

Explanation:
1. \( a=\frac{g}{1+I/(mr^2)} \): for a yo-yo the axle radius r is small compared with the radius of gyration, so \( I/mr^2\gg1 \) and the acceleration is downward and much less than g.

Teacher's Note:
1. The string tension supplies the torque, and the large rotational inertia slows the fall.

Answer: (b) Up the incline while ascending as well as descending

Explanation:
1. The acceleration of the centre is down the incline in both cases, so the angular acceleration must be in the sense that slows the rotation up the slope and speeds it down: friction acts up the incline in both cases (\( f=\frac{mg\sin\theta}{3} \) for a solid cylinder).

Teacher's Note:
1. Friction is the torque source that keeps rolling consistent with the downhill acceleration.

Answer: (d) None of the above

Explanation:
1. Friction depends on the sign of \( v-\omega R \): it can be forward, backward, or zero (if \( v=\omega R \) the sphere rolls).
2. So none of the first three statements is always true.

Teacher's Note:
1. The direction of kinetic friction follows the sense of slipping at the contact point.

Answer: (c) 1 : 2

Explanation:
1. \( \frac{K_{rot}}{K_{trans}}=\frac{\frac12\cdot\frac12MR^2\omega^2}{\frac12Mv^2}=\frac12 \)

Teacher's Note:
1. With v = omega R, the ratio is the coefficient of MR squared in I, divided by 1.