Question

A 0.68 kg tennis ball is given an angular momentum of 2.72 X 10-3 kg.m2/s when struck by a racquet. If its radius of gyration is 2 cm what is its angular velocity?

50 POINT QUESTION!! Any help is appreciated, thank you!

Answer 1

The angular velocity of the tennis ball is 10 radians per second.

Explanation:

The tennis ball can be represented as a particle, the angular momentum (
`L`), in kilogram-square meter per second, of the tennis ball is described by the following formula:

`L = m\cdot r^(2)\cdot \omega` (1)

Where:

`m` - Mass of the tennis ball, in kilograms.

`r` - Radius of gyration, in meters.

`\omega` - Angular velocity, in radians per second.

If we know that
`m = 0.68\,kg`,
`r = 0.02\,m` and
`L = 2.72* 10^(-3)\,(kg \cdot m^(2))/(s)`, then the angular velocity of the tennis ball is:

`\omega = (L)/(m\cdot r^(2))`

`\omega = (2.72* 10^(-3)\,(kg\cdot m^(2))/(s) )/((0.68\,kg)\cdot (0.02\,m)^(2))`

`\omega = 10\,(rad)/(s)`

The angular velocity of the tennis ball is 10 radians per second.