Question

A uniform sphere with mass 28 kg and radius 0.38 m is rotating at constant angular velocity about a stationary axis that lies along a diameter of the sphere. If the kinetic energy of the sphere is 236 J, what is the tangential velocity of a point on the rim of the sphere?

Answer 1

To solve this problem it is necessary to apply the concepts related to the equations of description of the rotational movement, also include there Inertia and Rotational Kinetic Energy.

From the definition for a sphere the moment of inertia can be defined as

$I = \frac{2}{5} mr^2$

Where,

m = mass

r = radius

At the same time the rotational kinetic energy is described as

$KE = \frac{1}{2} I\omega^2$

Where,

I = Moment of Inertia

$\omega =$ Angular velocity

From the description of the angular movement we know that the product of the angular velocity and the radius is equivalent to the linear / tangential velocity:

$V = r \omega$

Our values are given as

m = 28Kg

r = 0.38m

KE = 236J

Let's start by finding the moment of Inertia for which,

$I = \frac{2}{5} mr^2$

$I = \frac{2}{5} (28)(0.38)^2$

$I = 1.6172Kg\cdot m^2$

With the moment of Inertia we can find the angular velocity through energy, that is

$KE = \frac{1}{2} I\omega^2 \rightarrow \omega = \sqrt{\frac{2KE}{I}}$

$\omega = \sqrt{\frac{2KE}{I}}$

$\omega = \sqrt{\frac{2(236)}{1.6172}}$

$\omega = 17.084rad/s$

Finally the linear velocity would be given by

$v = r\omega$

$v = 0.38*17.084$

$v = 6.4919m/s$

Therefore the tangential velocity of a point on the rim of the sphere is 6.5m/s