Question

A hollow sphere is moving with linear velocity and angular velocity as shown in figure. The time after which pure rolling starts is :

• A. $\frac{2v}{3\mu g}$
• B. $\frac{4v}{7\mu g}$
• C. $\frac{4v}{5\mu g}$
• D. $\frac{5v}{4\mu g}$

Answer 1

Answer: (C)

4v/5\mu g

Answer 2

Explanation of the solution:

1. Identify Initial State: The hollow sphere has initial linear velocity v (right) and angular velocity ω = 3v/R (clockwise).

2. Determine Point of Contact Velocity: Let right be positive for linear velocity and counter-clockwise be positive for angular velocity. So, v_cm = v and ω_0 = -3v/R. The velocity of the point of contact v_P = v_cm - ωR = v - (-3v/R)R = 4v. Since v_P > 0, the point of contact slides to the right.

3. Apply Friction: Friction f_k = μmg acts to the left on the sphere, opposing the sliding motion.

4. Linear Motion Equations:

   • Acceleration a_cm = -f_k/m = -μg.
   • Linear velocity v_cm(t) = v - μgt.

5. Rotational Motion Equations:

   • Moment of inertia for a hollow sphere I = (2/3)MR^2.
   • Torque τ = f_k R = μmgR. This torque is counter-clockwise (positive).
   • Angular acceleration α = τ/I = (μmgR) / ((2/3)MR^2) = (3/2)(μg/R).
   • Angular velocity ω(t) = ω_0 + αt = -3v/R + (3/2)(μg/R)t.

6. Pure Rolling Condition: Pure rolling occurs when v_cm(t) = ω(t)R.

   • v - μgt = R(-3v/R + (3/2)(μg/R)t)
   • v - μgt = -3v + (3/2)μgt
   • 4v = (1 + 3/2)μgt = (5/2)μgt
   • t = (8v)/(5μg).

7. Reconciliation with Options: The calculated result (8v)/(5μg) does not match any option. However, if the initial angular velocity ω=3v/R was interpreted as counter-clockwise (i.e., in the direction of forward rolling), then:

   • v_cm = v, ω_0 = 3v/R.
   • v_P = v_cm - ωR = v - (3v/R)R = -2v. The point of contact slides to the left.
   • Friction f_k = μmg acts to the right.
   • a_cm = μg. v_cm(t) = v + μgt.
   • Torque τ = -f_k R = -μmgR (clockwise, negative).
   • α = τ/I = (-μmgR) / ((2/3)MR^2) = -(3/2)(μg/R).
   • ω(t) = ω_0 + αt = 3v/R - (3/2)(μg/R)t.
   • Pure rolling: v_cm(t) = ω(t)R.
   • v + μgt = R(3v/R - (3/2)(μg/R)t)
   • v + μgt = 3v - (3/2)μgt
   • (1 + 3/2)μgt = 3v - v
   • (5/2)μgt = 2v
   • t = (4v)/(5μg). This matches option (C).

Given the options, it is highly probable that the intended angular velocity direction was counter-clockwise, despite the diagram showing it as clockwise.