A child exerts a tangential 43.5-N force on the rim of a disk-shaped merry-go-round with a radius of 2.41 m. If the merry-go-round starts at rest and acquires an angular speed of 0.086 rev/s in 3.54 s, what is its mass?

Answer 1

Applied torque = 43.5 * 2.41 = 104.84 Newton-meters

Angular acceleration rate =
0.086*2*pi/3.54 = 0.1526 rad/s^2

The moment of inerti, I , is torque divided by angular acceleration.

I = 687 kg*m^2

For a uniform disc, The moment of inertia is

I = (1/2) M R^2.

Use that equation and the known value of I so solve for M.