A solid cylinder of mass M and radius R rolls without slipping down an inclined plane of length L and height h. What is the speed of its centre of mass when the cylinder reaches its bottom?

Question

A solid cylinder of mass M and radius R rolls without slipping down an inclined plane of length L and height h. What is the speed of its centre of mass when the cylinder reaches its bottom? (A) √2 g h (B) √(3/4) g h (C) √(4/3) g h (D) √4 g h

Tardigrade Admin · Accepted Answer

m g h=(1/2) M v2+(1/2) (M R2/2) × (v2/R2) ⇒ g h=(2 v2+v2/4) ⇒ (3 v2/4)=g h ⇒ v=√(4/3) g h

Q. A solid cylinder of mass $M$ and radius $R$ rolls without slipping down an inclined plane of length $L$ and height $h$ . What is the speed of its centre of mass when the cylinder reaches its bottom?

$2 g h$

$\frac{3}{4} g h$

$\frac{4}{3} g h$

$4 g h$

$m g h = \frac{1}{2} M v^{2} + \frac{1}{2} \frac{M R ^{2}}{2} \times \frac{v ^{2}}{R ^{2}}$
$\Rightarrow g h = \frac{2 v ^{2} + v ^{2}}{4}$
$\Rightarrow \frac{3 v ^{2}}{4} = g h$
$\Rightarrow v = \frac{4}{3} g h$

Q. A solid cylinder of mass M and radius R rolls without slipping down an inclined plane of length L and height h. What is the speed of its centre of mass when the cylinder reaches its bottom?

2gh​

43​gh​

34​gh​

4gh​