Answer to Question #177272 in Mechanics | Relativity for Abbas

Question

A golf ball with mass 45.90 g and diameter 42.60 mm is struck such that it moves with a speed of 51.85 ⁄ and rotates with a frequency of 2857rpm. What is the kinetic energy of the golf ball?

Answer 1

The kinetic energy of the golf ball can be found as follows:

$$KE=KE_{translational}+KE_{rotational}.$$

We can find the translational kinetic energy of the golf ball as follows:

$$KE_{translational}=\dfrac{1}{2}mv^2,$$ $$KE_{translational}=\dfrac{1}{2}\cdot0.0459\ kg\cdot(51.85\ \dfrac{m}{s})^2=61.7\ J.$$

We can find the rotational kinetic energy as follows:

$$KE_{rotational}=\dfrac{1}{2}I\omega^2=\dfrac{1}{2}\cdot\dfrac{2}{5}\cdot mr^2\omega^2,$$

$KE_{rotational}=\dfrac{1}{5}\cdot0.0459\ kg\cdot(0.0213\ m)^2\cdot(2857\ \dfrac{rev}{min}\cdot\dfrac{1\ min}{60\ s}\cdot2\pi\ rad)^2,$

$$KE_{rotational}=0.372\ J.$$

Then, we can calculate the kinetic energy of the golf ball:

$$KE=KE_{translational}+KE_{rotational}=61.7\ J+0.372\ J=62.1\ J.$$