JEE Physics System Of Particles and Rotational Motion MCQs Set 31

Answer: (b) 5 : 7

Explanation:
1. K_t = ½mv² and K_r = ½(2/5 mR²)(v/R)² = (1/5)mv².
2. K_t + K_r = (7/10)mv².
3. K_t : (K_t + K_r) = (1/2) : (7/10) = 5 : 7.

Teacher's Note:
1. Find K_r in terms of mv².
2. Add them for the total.

Answer: (d) angular momentum

Explanation:
1. No external torque acts on the sphere, so its angular momentum is conserved.
2. I = (2/5)MR² increases, so ω = L/I decreases and K = L²/(2I) decreases.

Teacher's Note:
1. Free rotation means no external torque.
2. I changes with R.

Answer: (a) 3 J

Explanation:
1. Work needed = total kinetic energy of the rolling disc = ¾Mv².
2. W = ¾ × 100 × (0.2)² = 3 J.

Teacher's Note:
1. Work to stop equals the kinetic energy.
2. Convert 20 cm/s to 0.2 m/s.

Answer: (d) 2.4 m

Explanation:
1. Total kinetic energy = ¾mv² = ¾ × 2 × 16 = 24 J.
2. It becomes potential energy: mg s sin30° = 2 × 10 × s × 0.5 = 10s.
3. s = 24/10 = 2.4 m.

Teacher's Note:
1. Use K = (3/4)mv² for a rolling cylinder.
2. Height gained = s sin30°.

Answer: (a) \( I_1 + I_2 \)

Explanation:
1. By the perpendicular axes theorem, the moment of inertia about the axis perpendicular to the plate equals the sum of the moments of inertia about any two mutually perpendicular axes in the plane of the plate through O.
2. Axes 1 and 2 are perpendicular, so I_z = I₁ + I₂ (also I₃ + I₄).
3. I₁ + I₂ + I₃ + I₄ would be 2I_z, and the differences I₁ − I₃ and I₂ − I₄ are zero for a square.

Teacher's Note:
1. Apply the perpendicular axes theorem.
2. Choose the pair of perpendicular in-plane axes.

Answer: (a) \( \dfrac{ML^2}{4} \)

Explanation:
1. Each slant rod has one end on the axis and makes 30° with it: I = (ML²/3) sin²30° = ML²/12. Two rods give ML²/6.
2. The base rod is perpendicular to the axis and its centre lies on the axis: I = ML²/12.
3. Total I = ML²/6 + ML²/12 = ML²/4.

Teacher's Note:
1. Use I = (ML²/3)sin²θ for a rod pivoted at an end and inclined at θ to the axis.
2. The base rod is a rod about a perpendicular axis through its centre.

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

Explanation:
1. The block is in equilibrium, so the net torque about the centre of mass is zero; the weight has no torque about it. The torques of the two tensions must therefore be equal in magnitude.
2. T_A d_A = T_B d_B, with d_B < d_A, so T_B > T_A.
3. Hence wire B exerts the larger force.

Teacher's Note:
1. Take torques about the centre of mass.
2. The wire nearer the centre of mass carries more load.

Answer: (d) 8 N

Explanation:
1. The three lines of action are at the same distance d from O.
2. F₁ and F₂ give torques in one sense and F₃ in the opposite sense.
3. (F₁ + F₂)d = F₃d, so F₃ = 5 + 3 = 8 N.

Teacher's Note:
1. The lever arm is the same for all three forces.
2. Add the two torques that act in the same sense.

Answer: (c) 10 rad/s²

Explanation:
1. τ = FR = 50 × 0.4 = 20 N m.
2. I = ½MR² = ½ × 25 × 0.16 = 2 kg m².
3. α = τ/I = 10 rad/s².

Teacher's Note:
1. Treat the flywheel as a uniform disc.
2. α = τ/I.

Answer: (c) \( \sqrt{\dfrac{24g}{7\ell}} \)

Explanation:
1. The centre of mass is ℓ/2 from the end, so it is ℓ/4 from the hinge and falls through ℓ/4.
2. I about the hinge = mℓ²/12 + m(ℓ/4)² = 7mℓ²/48.
3. Energy conservation: mg(ℓ/4) = ½(7mℓ²/48)ω², so ω² = 24g/(7ℓ).

Teacher's Note:
1. Find I about the hinge by the parallel axes theorem.
2. The fall of the centre of mass is ℓ/4.

Answer: (d) \( \sqrt{4mgh/(2m + M)} \)

Explanation:
1. Energy conservation: mgh = ½mv² + ½Iω² with I = ½MR² and ω = v/R.
2. mgh = ½mv² + ¼Mv² = (2m + M)v²/4.
3. v = √(4mgh/(2m + M)).

Teacher's Note:
1. The disc also gains rotational kinetic energy.
2. v = ωR links the two motions.

Answer: (c) 12.8 J

Explanation:
1. τ = FR = 80 × 0.1 = 8 N m, so α = τ/I = 8/40 = 0.2 rad/s².
2. θ = ½αt² = ½ × 0.2 × 16 = 1.6 rad.
3. W = τθ = 8 × 1.6 = 12.8 J.

Teacher's Note:
1. Work done by a torque is τθ.
2. Find θ from rotational kinematics.

Answer: (d) \( \dfrac{mu^3\sin\theta\sin 2\theta}{4g} \)

Explanation:
1. At the highest point the velocity is horizontal: v = u cosθ, and the height is H = u² sin²θ/(2g).
2. L = m v H = m u cosθ · u² sin²θ/(2g) = mu³ sin²θ cosθ/(2g).
3. Since sin2θ = 2 sinθ cosθ, this equals mu³ sinθ sin2θ/(4g).

Teacher's Note:
1. Angular momentum = m × speed × perpendicular distance.
2. The perpendicular distance from the point of projection is the maximum height.

Answer: (b) \( \dfrac{M\omega_0}{M + 6m} \)

Explanation:
1. Initially I₀ = ML²/12 (beads on the axis).
2. Finally I = ML²/12 + 2m(L/2)² = (L²/12)(M + 6m).
3. Angular momentum is conserved: I₀ω₀ = Iω, so ω = Mω₀/(M + 6m).

Teacher's Note:
1. Conserve angular momentum.
2. Beads at the ends are at distance L/2 from the axis.

Answer: (c) 3 m/s

Explanation:
1. In pure rolling the contact point of the disc has the same velocity as the surface: 2 m/s to the right.
2. For clockwise rotation the bottom point moves to the left relative to the centre with speed ωR = 5 × 0.2 = 1 m/s.
3. v_cm − 1 = 2, so v_cm = 3 m/s to the right.

Teacher's Note:
1. The contact point moves with the surface.
2. Velocity of bottom point = v_cm − ωR for clockwise rotation.

Answer: (b) 2L

Explanation:
1. The string leaves the cylinder from its top point. In rolling without slipping the top point moves with speed 2v, where v is the speed of the centre.
2. The hand end of the string moves with the top point, so the length of string pulled is twice the displacement of the centre.
3. Centre moves L, so the string pulled is 2L.

Teacher's Note:
1. The top point moves twice as fast as the centre.
2. The string follows the top point.

Answer: (c) \( \dfrac{5}{7} \)v₀

Explanation:
1. Take angular momentum about the point of contact (friction acts there, so it has no torque): mv₀R = mvR + Iω.
2. For a solid sphere I = (2/5)mR² and ω = v/R at the end of slipping: mv₀R = mvR + (2/5)mvR = (7/5)mvR.
3. v = (5/7)v₀.

Teacher's Note:
1. Angular momentum about the line of contact is conserved.
2. Pure rolling means v = ωR.

Answer: (d) \( \text{velocity is zero, acceleration is } \dfrac{v^2}{R} \)

Explanation:
1. In pure rolling the contact point is instantaneously at rest: its velocity is zero.
2. The centre has no acceleration, so the acceleration of the point is only the centripetal acceleration ω²R = v²/R about the centre (directed upwards).
3. So the velocity is zero but the acceleration is v²/R.

Teacher's Note:
1. The point of contact is at rest instantaneously, but not unaccelerated.
2. Use a = ω²R about the centre.

Answer: (c) \( \dfrac{3}{2}MR^2\omega \)

Explanation:
1. Angular momentum about O = spin part about the centre + orbital part of the centre of mass.
2. Spin part: Iω = ½MR²ω. Orbital part: M v_cm × (height R) = M(ωR)R = MR²ω; both have the same sense.
3. L = ½MR²ω + MR²ω = (3/2)MR²ω.

Teacher's Note:
1. Split L into spin and orbital parts.
2. The centre of mass moves along a line at distance R from O.

Answer: (a) \( \dfrac{17}{7} \)mg

Explanation:
1. The centre of the small sphere falls through (R − r). Energy conservation: mg(R − r) = ½mv² + ½(2/5 mr²)(v/r)² = (7/10)mv², so v² = (10/7)g(R − r).
2. At the bottom the centre moves on a circle of radius (R − r): N − mg = mv²/(R − r) = (10/7)mg.
3. N = mg + (10/7)mg = (17/7)mg.

Teacher's Note:
1. Rolling: use the total kinetic energy (7/10)mv².
2. The centre of the small sphere moves on a circle of radius R − r.

Answer: (b) \( \dfrac{7v^2}{10g} \)

Explanation:
1. Total kinetic energy of the rolling sphere = ½Mv² + ½(2/5 MR²)(v/R)² = (7/10)Mv².
2. It is converted into potential energy: Mgh = (7/10)Mv².
3. h = 7v²/(10g).

Teacher's Note:
1. Use the total kinetic energy of rolling.
2. Equate to Mgh.