A solid sphere of mass ‘M’ and radius ‘R’ is initially at rest. Solid sphere is gradually lowered onto a truck moving with constant velocity $V_{0}$ .
The final speed of sphere’s centre of mass in ground frame when eventually pure rolling sets is (in multiples of $\frac{V_{0}}{7})$ ?

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

answer is 2.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

$v = v e l o c i t y o f c e n t r e o f s p h e r e w r t t r u c k$

Applying conservation of angular momentum about point P on the truck:

$M V_{0} R = M v R + I ω = M v R + \frac{2}{5} M R^{2} ω$ .....(1)

for pure rolling, $v = ω R$ .... (2)

solving (1) and (2),

$v = \frac{5}{7} V_{0}$

velocity wrt ground:

$v' = V_{0} - v = \frac{2}{7} V_{0}$

Watch 3-min video & get full concept clarity

JEE

NEET

Foundation JEE

Foundation NEET

CBSE