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 17.117.1 m/s at a distance of 0.4700.470 m from the joint and the moment of inertia of the forearm is 0.5500.550 kg·m2, what is the rotational kinetic energy of the forearm?

Answer 1

Answer:

363.96 J

Explanation:

v = 17.1 m/s

r = 0.47 m

I = 0.55 kgm^2

Let ω be the angular velocity

ω = v / r = 17.1 / 0.47 = 36.38 rad/s

The kinetic energy of rotation is

K = 1/2 I ω^2 = 0.5 x 0.55 x 36.38 x 36.38 = 363.96 J

Answer 2

Answer:

364.4 J

Explanation:

I = Moment of inertia of the forearm = 0.550 kgm²

v = linear velocity of the ball relative to elbow joint = 17.1 m/s

r = distance from the joint = 0.470 m

w = angular velocity

Using the equation

v = r w

17.1 = (0.470) w

w = 36.4 rad/s

Rotational kinetic energy of the forearm is given as

RKE = (0.5) I w²

RKE = (0.5) (0.550) (36.4)²

RKE = 364.4 J