Question

A horizontal 808-N merry-go-round is a solid disk of radius 1.54 m, started from rest by a constant horizontal force of 50.9 N applied tangentially to the edge of the disk. Find the kinetic energy of the disk after 3.08 s.

Answer 1

In order to find the angular velocity (w), we first need to find the angular acceleration (a). We know that the force is acting tangentially to the disk, creating a torque on the disk. In this case, the torque is equal to the force times the radius (r) of the disk. Torque is also equal to the moment of inertia times the radial acceleration (a). Set the two Torque values equal and solve for the radial acceleration (a):

F x r = I x a

a = (F x r) / I

I = 1/2m x r^2 for a disk, so:

a = (F x r) / (.5 x m x r^2)

a = (50.9N x 1.54m) / (.5 x 82.45kg x (1.54m)^2 ) = .802 rad/s^2

Solve for the radial velocity (w) using this equation:

w = wo + a x t

wo = zero because the disk starts from rest

w = a x t

w = .802 rad/s^2 x 3.08s = 2.469rad/s

Now solve for rotational KE:

KE = 1/2 x I x w^2

I for a disk = 1/2m x r^2

KE = 1/2 x (1/2m x r^2) x w^2

KE = 1/4 x (82.45kg) x (1.54m)^2 x (2.469rad/s)^2 = 297.99 J