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

Answer:

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

Step-by-step 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 $L = 2.72\times 10^{-3}\,\frac{kg \cdot m^{2}}{s}$, then the angular velocity of the tennis ball is:

$\omega = \frac{L}{m\cdot r^{2}}$

$\omega = \frac{2.72\times 10^{-3}\,\frac{kg\cdot m^{2}}{s} }{(0.68\,kg)\cdot (0.02\,m)^{2}}$

$\omega = 10\,\frac{rad}{s}$

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