Question 734: Centre of Mass and Rotational Motion

Question

A solid sphere of radius $R$ has moment of inertia $I$ about its geometrical axis. If it is melted into a disc of radius $r$ and thickness $t$. If its moment of inertia about the tangential axis (which is perpendicular to plane of the disc), is also equal to $I$, then the value of $r$ is equal to

Accepted Answer

To solve this problem, we need to compare the moments of inertia of the sphere and the disc. For a solid sphere of radius $R$, the moment of inertia about its geometrical axis is given by:
$$I_{	ext{sphere}} = \frac{2}{5} M R^2$$
where $M$ is the mass of the sphere.
When the sphere is melted into a disc of radius $r$ and thickness $t$, the volume (and hence mass) remains the same. So, the volume of the sphere is:
$$\frac{4}{3} \pi R^3$$
And the volume of the disc is:
$$\pi r^2 t$$
Equating the volumes, we get:
$$\frac{4}{3} \pi R^3 = \pi r^2 t$$
This simplifies to:
$$t = \frac{4 R^3}{3 r^2}$$
Next, we need to find the moment of inertia of the disc about the tangential axis (which is perpendicular to the plane of the disc). The moment of inertia of a disc of mass $M$ and radius $r$ about an axis through its center and perpendicular to the plane of the disc is:
$$I_{	ext{disc, center}} = \frac{1}{2} M r^2$$
Using the parallel axis theorem, the moment of inertia about a tangential axis is:
$$I_{	ext{disc, tangential}} = I_{	ext{disc, center}} + M r^2 = \frac{1}{2} M r^2 + M r^2 = \frac{3}{2} M r^2$$
Since we are given that the moment of inertia of the sphere and the disc about their respective axes are equal, we set:
$$\frac{2}{5} M R^2 = \frac{3}{2} M r^2$$
Canceling the mass $M$ from both sides and solving for $r$, we get:
$$\frac{2}{5} R^2 = \frac{3}{2} r^2$$
Solving for $r$, we have:
$$r^2 = \frac{4}{15} R^2$$
Taking the square root of both sides:
$$r = \frac{2}{\sqrt{15}} R$$
So, the correct answer is:
Option A: $$\frac{2}{\sqrt{15}} R$$

Answer

$\frac{2}{\sqrt{15}} R$

Answer

$\frac{2}{\sqrt{5}} R$

Answer

$\frac{3}{\sqrt{15}} R$

Answer

$\frac{\sqrt{3}}{\sqrt{15}} R$

Centre of Mass and Rotational Motion

A solid sphere of radius R has moment of inertia I about its geometrical axis. If it is melted into a disc of radius r and thickness t. If its moment of inertia about the tangential axis (which is perpendicular to plane of the disc), is also equal to I, then the value of r is equal to

Choose the correct answer: