Posted by Sarah on .
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?
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.