# Answer to Question #112422 in Mechanics | Relativity for Niya

Question #112422
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)?
1
2020-04-28T09:46:44-0400

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.

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