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)?

Expert's answer

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.

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