A merry-go-round on a playground consists of a horizontal solid disk with a weight of 810 N and a radius of 1.56 m. A child applies a force 49.0 N tangentially to the edge of the disk to start it from rest. What is the kinetic energy of the merry-go-round disk (in J) after 2.95 s

Kinetic Energy of the disk = 252 J

Explanation:

weight of disk = 810 N

radius = 1.56 m

applied force = 49 N

time = 2.95 s

kinetic energy of disk = ?

first, we find the mass of the disk

mass of disk = weight/acceleration due to gravity(9.81 m/s^2) = 810/9.81 m/s^2

mass of disk = 82.57 kg

torque on the disk = force x radius = 49 x 1.56 = 76.44 N-m

moment of inertia I = m = 82.57 x = 200.9 kg-

recall that

Torque T = Iα

where α = angular acceleration

76.44 = 200.9α

α = 76.44/200.9 = 0.38 m/s^2

from the equation of angular motion,

ω = ω' + αt

where ω =  final angular speed

ω' = initial angular speed = 0 rad/s since disk starts from rest

t = time = 2.95 s

imputing values into the equation, we have

ω = 0 + (0.38 x 2.95)

ω = 1.12 rad/s

kinetic energy of the disk = I

KE = 200.9 x

Kinetic Energy of the disk = 252 J