Question

A machine part has the shape of a solid uniform sphere of mass 240 g and diameter 4.50 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. Find its angular acceleration. Let the direction the sphere is spinning be the positive sense of rotation. How long will it take to decrease its rotational speed by 28.0 rad/s ?

Answer 1

1) $-9.3rad/s^2$

2) 3.0 s

Explanation:

1)

The angular acceleration of a rigid body in rotation can be  found by using the equivalent of Newton's second law for rotational motions:

$\tau = I\alpha$ (1)

where

$\tau$ is the torque on the object

I is the moment of inertia of the object

$\alpha$ is the angular acceleration

Here we have:

$I=\frac{2}{5}MR^2$ is the moment of inertia of a solid sphere about its central axis, where

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

R = 4.50 cm /2= 2.25 cm = 0.0225 m is the radius of the sphere

$\tau = F r$ is the torque exerted by the frictional force, where

$F=-0.0200 N$ is the force of friction (negative because the direction is opposite to the motion)

r = R = 0.0225 m is the distance of the point of application of the force from the centre

Substituting into eq(1) we find

$FR=\frac{2}{5}MR^2 \alpha$

And solving for $\alpha$, we find the angular acceleration:

$\alpha = \frac{5F}{2MR}=\frac{5(-0.0200)}{2(0.240)(0.0225)}=-9.3rad/s^2$

2)

Here we know that the motion of the sphere is an angular accelerated motion.

Therefore, we can use the equivalent of suvat equations for rotational motion:

$\omega_f = \omega_i +\alpha t$

where

$\omega_f$ is the final angular velocity

$\omega_i$ is the initial angular velocity

$\alpha$ is the angular acceleration

t is the time

In this problem, we know that

$\omega_f-\omega_i = -28.0 rad/s$, since we are told that the rotational speed decreases by 28.0 rad/s

$\alpha=-9.3 rad/s^2$ is the angular acceleration of the sphere

Solving for t, we find how long it takes for the sphere to decelerate by 28.0 rad/s:

$t=\frac{\omega_f - \omega_i}{\alpha}=\frac{-28.0}{-9.3}=3.0 s$