Question

A hollow sphere is released from the top of a wedge as shown in the fig. There is no friction between the wedge and the ground. There is sufficient friction to provide pure rolling . The velocity of the sphere just before it leaves the wedge will be:
[Assume:Masses of the wedge and sphere are equal and $h>>R$ to be the radius of sphere]

• A. $\sqrt{\frac{3gh}{7}}$
• B. $\sqrt{\frac{3}{4}gh}$
• C. $\sqrt{\frac{5}{7}gh}$
• D. $\sqrt{\frac{6}{5}gh}$

Answer 1

The correct option is A $\sqrt{\frac{3gh}{7}}$

A hollow sphere is released from the top of a wedge as shown. There is no friction between the wedge and ground.

The velocity of sphere just before it leaves the wedge will be $=V$

Given

Masses of the wedge and sphere are equal

$h>>R$ to be the radius of sphere

Given,

mass$=m$

Initial height$=h$

gravity$=g$

Final velocity$=v$

radius $=r$

angular velocity $\omega =\frac{v}{r}$

Moment of inertia $I=Km{v}^2$

(where K is a measure how close to the edge the mass is on average)

Then by conservation of energy

$mgh=\frac{1}{2}m{v}^2+\frac{1}{3}I{\omega }^{2}2gh=v^2+K{v}^{2}v=\sqrt{\frac{3gh}{1+K}}=\sqrt{\frac{3gh}{7}}$