ROTATIONAL MOTION MULTIPLE CHOICE QUESTIONS 1. Athinspnerizal shell of mass m ard radius R rolls down a parahchc path PQR from a beight H without slipping. Part PQ 1s sutficizntly rcugh while nart QR is smooth If the neignt reachad by the shell on the part QR be h. then hiH 15 given by (assume R << H) (A) 1 (B 075 (C) 06 (0) 05

1. Rolling Without Slipping on a Rough Track (PQ)

When an object rolls without slipping, the point of contact is instantaneously at rest. The condition is
$v = \omega R$ where $v$ is its centre-of-mass speed and $\omega$ its angular speed.

For a thin spherical shell (a hollow sphere), the moment of inertia about its centre is
$I = \frac{2}{3} m R^2$

Total kinetic energy while rolling:

Using energy conservation between the top (height H) and the bottom of PQ (reference height 0):

2. Motion on the Smooth Track (QR)

On a smooth segment there is zero friction, so no external torque acts about the centre; the rotational kinetic energy stays constant. Only the translational part can convert into gravitational potential.

Let the ball climb to height h before momentarily stopping its translation (so $v = 0$) but it continues spinning with its bottom-of-PQ angular speed.

Energy conservation from the bottom of PQ to the highest point on QR:

(We omit rotational energy on both sides because it is the same and cancels.)

Substitute $v_b^2$:

Hence $\frac{h}{H} = 0.6$

Why Each Step?

1. Identify type of object: Moment of inertia depends on geometry—thin spherical shell ➔ $I = \frac{2}{3}mR^2$.
2. Pick energy conservation because no non-conservative work (rolling without slipping is pure rolling on a rough but stationary track).
3. Separate energies on the smooth part: spin energy stays; translational part changes.
4. Solve algebraically for $h$ in terms of $H$.

Option (C) 0.6 is correct.

What’s Happening?

Imagine you have a hollow cricket ball (a thin spherical shell) that starts at the top of a hill at height H.

1. First part (PQ) of the track is rough, so the ball rolls without sliding—meaning it both spins and moves forward.
2. Second part (QR) is perfectly smooth, so no extra force can change its spin; the spin just keeps going.
3. We want to know how high (call it h) the ball will climb on the smooth part before it stops moving upward.

Key Ideas

• When the ball rolls on the rough part, gravity’s energy (at height H) changes into both forward motion energy and spin energy.
• On the smooth part, only the forward motion energy can turn back into height; the spin energy stays locked in (no friction to change it).
• Because some energy is ‘trapped’ in spin, the ball cannot reach the same height H again; it only gets up to 0.6 H.

So the answer is 0.6 or 60 % of the original height.

Step-by-Step Solution

1. Moment of inertia of a thin spherical shell
   $I = \frac{2}{3} m R^2$

2. Energy at the top of PQ (height H)
   $\text{Potential Energy} = mgH$

3. At the bottom of PQ
   Using $v = \omega R$ and energy conservation,

   $$mgH = \frac{1}{2}mv_b^2 + \frac{1}{2}I\omega_b^2 = \frac{1}{2}mv_b^2 + \frac{1}{2}\left(\frac{2}{3}mR^2\right)\left(\frac{v_b}{R}\right)^2 = \frac{1}{2}mv_b^2 + \frac{1}{3} m v_b^2 = \frac{5}{6} m v_b^2$$

   Solving,
   $v_b^2 = \frac{6}{5} g H$

4. Climb on the smooth part (QR)
   Rotational kinetic energy remains constant (no friction), so only translational kinetic converts to potential:
   $\frac{1}{2} m v_b^2 = mgh$ Substituting $v_b^2$,

   $$\frac{1}{2} \left(\frac{6}{5} g H\right) = g h \Rightarrow h = \frac{3}{5} H$$

5. Ratio asked
   $\boxed{\dfrac{h}{H} = 0.6}$

Hence, the correct option is (C) 0.6.