A horizontal `9 kg` merry-go-round is a solid disk of radius `1.50 m` and is started from rest by a constant horizontal force of `50.0 N` applied tangentially to the edge of the disk. The kinetic energy of the disk after `3.00 s` is.

Options

A. 125 J
B. 500 J
C. 250 J
D. 150 J

Answer

C

Explanation

( c) From the rigid object under a net torque model,
`sum tau = I prop rarr prop = (sum tau)/(l) = (FR)/(1/(2) MR^2) = (2 F)/(MR)`
From the definition of rotational kinetic energy and the rigid object under constant angular acceleration model,
`K = (1)/(2) I omega^2 = (1)/(2) I (omega_i + prop t)^2`
=`(1)/(2) I prop^2 t^2 = ((1)/(2) MR^2)((2F)/(MR))^2 t^2 = (F^2 t^2)/(M)`
Substituting, `K = ((50.0 N)^2 (3.00 s)^2)/((9.00 kg)) =250 J`.