Question

a baseball pitcher throws the ball. If the linear velocity of the ball relative to the elbow joint 23.0 (m)/(s) at a distance of 0.340m from the joint and the moment of inertia of the forearm is 0.5 kg*m^(2), what is the rotational kinetic energy?

Answer 1

By definition, the rotational kinetic energy is:

$$K = \dfrac{I\omega^2}{2}$$

where $I = 0.5 kg\cdot m^2$ is the moment of inertia, and $\omega$ is the angular frequency. The angular frequency relates to the linear velocity as follows:

$$\omega = \dfrac{v}{R}$$

where $v = 23m/s$, and $R = 0.34m$ is the distance from the joint. Substituting $\omega$ into $K$, obtain:

$$K = \dfrac{Iv^2}{2R^2}\\ K = \dfrac{0.5\cdot 23^2}{2\cdot 0.34^2} \approx 1144J$$

Answer. 1144 J.