Question

A hollow spherical shell has mass 8.35 kg and radius 0.225 m. It is initially at rest and then rotates about a stationary axis that lies along a diameter with a constant acceleration of 0.885 rad/s^2.

Required:
What is the kinetic energy of the shell after it has turned through 6.00 rev?

Answer 1

K = 9.41 J

Step-by-step explanation:

The kinetic energy of a spherical shell is given as:

`k = (1)/(2) I \omega^2`

where I = moment of inertia and

ω = angular velocity

Let us find I:

For a hollow sphere:

`I = (2)/(3) MR^2`

where M = mass = 8.35 kg

R = radius = 0.225 m

`=> I = (2)/(3) * 8.35 * 0.225^2 = 0.282 kgm^2`

Let us find ω:

Since angular acceleration is constant:

`\omega^2 = \omega_0^2 + 2\alpha(\theta - \theta_0)\\\\\omega_0 = 0\\\\=>\omega^2 = 2\alpha(\theta - \theta_0)\\\\\theta - \theta_0 = 6 * (2 * \pi) = 37.70 rad\\\\\omega^2 = 2 * 0.885 * 37.70 = 66.729`

Therefore, its kinetic energy is:

`K = (1)/(2) * 0.282 * 66.729\\\\K = 9.41 J`