A horizontal 810-N merry-go-round of radius 1.40 m is started from rest by a constant horizontal force of 55 N applied tangentially to the merry-go-round. Find the kinetic energy of the merry-go-round after 2.0 s. (Assume it is a solid cylinder. Also assume the force is applied at the outside edge.)

52.218 J

Explanation:

Mass of the merry go round m = weight / g

= 810 / 9.8

= 82.65 kg

radius r = 1.40 m

torque applied τ = r * F

= 1.40 * 55

= 77 N - m

Moment of inertia (of cylinder) I = (1/2) * m * r²

= 0.5 * 82.65 * 1.96

= 80.997 kg-m²

angular acceleration α = τ / I

= 77 / 119.43

= 0.6447 rad/s²

According to first equation in angular motion final angular velocity ω = ω0 + α * t

ω = 0 + 0.6447 * 2.0

= 1.2894 rad/s

Hence k. e. after 2 seconds K = (1/2) * I * ω²

= 0.5 * 80.997 * 1.2894

= 52.218 J