A baseball pitcher loosens up his pitching arm.He tosses a 0.15-kg ball using only the rotation of his forearm,0.32 m in length,to accelerate the ball.If the ball starts at rest and is released with a speed of 24 m/s in a time of 0.40 s,what torque is applied to the ball while being held by the pitcher's hand to produce the angular acceleration?
A) 1.1 N·m
B) 11 N·m
C) 2.9 N·m
D) 1.4 N·m

Question

Quizplus · Accepted Answer

We can use the formula for torque:

τ = Iα

Where τ is the torque, I is the moment of inertia, and α is the angular acceleration.

We can use the formula for moment of inertia for a point mass:

I = mr^2

Where m is the mass of the ball and r is the length of the pitcher's forearm.

Plugging in the given values:

I = (0.15 kg)(0.32 m)^2 = 0.01536 kg·m^2

We can rearrange the formula for torque to solve for it:

τ = Iα

τ = (0.01536 kg·m^2)(24 m/s / 0.40 s)

τ = 0.9216 N·m

Therefore, the torque applied to the ball is 0.9216 N·m. The closest answer choice to this value is C) 2.9 N·m.

A Baseball Pitcher Loosens Up His Pitching Arm