A softball player swings a bat, accelerating it from rest to 8 rev/s in a time of 0.3 s. Find the torque the player applies to one end of the bat. Approximate the bat as a 2.2 kg uniform rod of length 0.95 m.

Answer 1

alpha = ( 8 * 2 pi radians/s )/.3s
= 168 radians/s^2

torque = I alpha

I = (1/3) m L^2 = (1/3)(2.2)(.95)^2 = .662 kg m^2

so
.662 * 168 kg m^2/s^2 = 111 N m