Question

A baseball pitcher throws the ball in a motion where there is rotation of the forearm about the elbow joint as well as other movements. If the linear velocity of the ball relative to the elbow joint is 17.1 m/s at a distance of 0.470 m from the joint and the moment of inertia of the forearm is 0.500 kg·m^2. What is the rotational kinetic energy of the forearm?

Answer 1

Answer: 330.88 J

Step-by-step explanation:

Given

Linear velocity of the ball, v = 17.1 m/s

Distance from the joint, d = 0.47 m

Moment of inertia, I = 0.5 kgm²

The rotational kinetic energy, KE(rot) of an object is given by

KE(rot) = 1/2Iw²

Also, the angular velocity is given

w = v/r

Firstly, we calculate the angular velocity. Since it's needed in calculating the Kinetic Energy

w = v/r

w = 17.1 / 0.47

w = 36.38 rad/s

Now, substituting the value of w, with the already given value of I in the equation, we have

KE(rot) = 1/2Iw²

KE(rot) = 1/2 * 0.5 * 36.38²

KE(rot) = 0.25 * 1323.5

KE(rot) = 330.88 J