A softball player swings a bat, accelerating it from rest to 2.2 rev/srev/s in a time of 0.20 ss . Approximate the bat as a 2.3-kgkg uniform rod of length 0.96 mm , and compute the torque the player applies to one end of it.

83.168 Nm

Explanation:

I=mass of bat* ( length )^2 /3

Moment of inertia from one end = 2.2 * (0.96)^2 / 3.

Ang. acc. = 2.2×2 pi / 0.2 = 22pi

Torque =  2.3 * (0.96)^2 / 3 * 22 pi = 83.168 Nm SI units

the answer is b i think

3.2x10^7