Question

Inside a hollow uniform sphere of mass $M$, a uniform rod of length $R_2$ is released from the state of rest. The mass of the rod is same as that of the sphere. If the inner radius of the hollow sphere is $R$ then find out horizontal displacement of sphere with respect to earth in the time in which the rod becomes horizontal.

• A. $\frac{R}{2\sqrt{2}}$
• B. $\frac{R}{2}$
• C. $\frac{R}{4}$
• D. None

Answer 1

The correct option is C $\frac{R}{4}$

Draw a rough diagram of sphere and rod.
Given: $M_{rod}=M_{sphere}=M$

Left figure shown initial position of system and right figure shown the position when rod becomes horizontal.

Find Displacement of sphere.
Let $x=$ displacement of centre of mass of rod with respect to sphere along $x$ axis, then

$x=\frac{\sqrt{2}R}{2}\cos {45}^{\circ }=\frac{R\sqrt{2}}{2\sqrt{2}}=\frac{R}{2}$

And let $y=$displacement of sphere with respect to ground

$My=M(x-y)$

$y=\frac{x}{2}=\frac{R}{4}$