Question

A machine part has the shape of a solid uniform sphere of mass 245 g and diameter 4.30 cm . It is spinning about a frictionless axle through its center, but at one point on its equator it is scraping against metal, resulting in a friction force of 0.0200 N at that point. Find its angular acceleration. Let the direction the sphere is spinning be the positive sense of rotation.

Answer 1

Answer:

Angular acceleration, $\alpha =9.49\ rad/s^2$

Explanation:

It is given that,

Mass of the solid sphere, m = 245 g = 0.245 kg

Diameter of the sphere, d = 4.3 cm = 0.043 m

Radius, r = 0.0215 m

Force acting at a point, F = 0.02 N

Let $\alpha$ is its angular acceleration. The relation between the angular acceleration and the torque is given by :

$\tau=I\times \alpha$

I is the moment of inertia of the solid sphere

For a solid sphere, $I=\dfrac{2}{5}mr^2$

$\alpha =\dfrac{\tau}{I}$

$\alpha =\dfrac{F.r}{(2/5)mr^2}$

$\alpha =\dfrac{5F}{2mr}$

$\alpha =\dfrac{5\times 0.02}{2\times 0.245\times 0.0215}$

$\alpha =9.49\ rad/s^2$

So, its angular acceleration is $9.49\ rad/s^2$. Hence, this is the required solution.