Question 6

Question

A 24.0 kg uniform solid cylinder has a radius of 0.259 m. If the cylinder accelerates at 3.90 × 10−2 rad/s2 as it rotates about an axis through its center, how large is the torque acting on the cylinder?

Answer 1

torque = moment of inertia x angular acceleration. Or,

 

τ = Iα

The moment of inertia of a cylinder about its axis is

1/2mr²

T = 1/2*24kg*(.259m)²*10^-2rad/s²

T = .0804 Nm