Question

A machine part has the shape of a solid uniform sphere of mass 240 g and diameter 2.50 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. Part A Find its angular acceleration. Let the direction the sphere is spinning be the positive sense of rotation.

Answer 1

`-16.6 rad/s^2`

Step-by-step explanation:

The torque exerted on a rigid body is related to the angular acceleration by the equation

`\tau = I \alpha` (1)

where

`\tau` is the torque

I is the moment of inertia of the body

`\alpha` is the angular acceleration

Here we have a solid sphere: the moment of inertia of a sphere rotating about is centre is

`I=(2)/(5)MR^2`

where

M = 240 g = 0.240 kg is the mass of the sphere

`R=(2.50)/(2)=1.25 cm = 0.0125 m` is the radius of the sphere

Substituting,

`I=(2)/(5)(0.240)(0.0125)^2=1.5\cdot 10^(-5) kg m^2`

The torque exerted on the sphere is

`\tau = Fr`

where

F = -0.0200 N is the force of friction

r = 0.0125 m is the radius of the sphere

So

`\tau=(-0.0200)(0.0125)=-2.5\cdot 10^(-4) Nm`

Substituting into (1), we find the angular acceleration:

`\alpha = (\tau)/(I)=(-2.5\cdot 10^(-4))/(1.5\cdot 10^(-5))=-16.6 rad/s^2`