Answer to Question #112422 in Mechanics | Relativity for Niya

Question

A golf ball with mass 45.90 g and diameter 42.60 mm is struck such that it moves with a speed of 54.15 m/s while rotating. The golf ball has a kinetic energy of 67.67 J. What is the rotational frequency of the golf ball (in rpm)?

Answer 1

The total kinetic energy of the ball is the sum of the linear kinetic energy and angular kinetic energy:

$K_{tot} = K_{linear}+K_{angular}$.

By definition:

$K_{linear} = \dfrac12mv^2; \space K_{angular} = \dfrac12I\omega^2$ ,

where $m$ is body mass, $v$ - linear velocity, $I$ - moment of inertia around the axis of rotation, $\omega$ - angular rotational frequency.

The moment of inertia of the ball is $I = \dfrac{1}{10}md^2$ , where $d$ is the diameter of the ball.

Putting it all together in the first expression, obtain:

$K_{tot} =\dfrac12mv^2+ \dfrac12\dfrac{1}{10}md^2\omega^2$ .

Express $\omega$ and substitute numerical values:

$\omega^2 =\dfrac{ 20(K_{tot}- \dfrac12mv^2)}{md^2}=\dfrac{ 20(67.67- 0.5\cdot 0.0459\cdot 54.15^2)}{0.0459\cdot 0.0426^2} = 90157$ .

Thus, $\omega = \sqrt{90157} \approx300$ rad/s

The rotational frequency in rpm is:

$f = 60\cdot\dfrac{\omega}{2\pi} = 2864.8$ rpm.

Answer. 2864.8 rpm.