Question

A horizontal 694 N merry-go-round of radius 1.96 m is started from rest by a constant horizontal force of 61.5 N applied tangentially to the merry-go-round. Find the kinetic energy of the merry-go round after 3.89 is. The acceleration of gravity is 9.8 m/s 2 . Assume the merry-go-round is a solid cylinder. Answer in units of J.

Answer 1

808.19J

Step-by-step explanation:

Image attached gives a proper explanation

Answer 2

Answer: 808.24625J

Step-by-step explanation:

Moment of inertia is the physical quantity, that expresses the tendency of a body to resist angular acceleration. It determines the Torque needed for a desired angular acceleration, about a rotational axis

I = 1/2mr²

I = 1/2(W/g)r²

I = 1/2 * 694/9.8 * 1.96²

I = 1/2 * 272.048

I = 136.024kgm²

Also,

Iα = Fr

136.024α = 61.5*1.96

136.024α = 120.54

α = 120.54/136.024

α = 0.8862 rad/s²

Angular Velocity,

ω = αt

ω = 0.8862*3.89

ω = 3.4473 rad/s

K = 1/2Iω²

K = 1/2*136.024*3.4473²

K = 1/2 * 1616.4925

K = 808.24625J