Question

A merry-go-round on a playground consists of a horizontal solid disk with a weight of 805 N and a radius of 1.47 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?

Answer 1

Answer:

255.34 J

Explanation:

Given,

Weight of disk = 805 N

radius = 1.47 m

Force applied by the child = 49 N

time = 2.95 s

KE = ?

mass of the disk

$M = \dfrac{W}{g}= \dfrac{805}{9.81} = 82.059\ Kg$

Moment of inertia of the disk

$I = \dfrac{1}{2}Mr^2$

$I = \dfrac{1}{2}\times 82.059\times 1.47^2 =88.66\ kgm^2$

Torque on the child

$\tau = F \times r = 49 \times 1.47 = 72.03 Nm$

Angular acceleration

$\alpha = \dfrac{\tau}{I}=\dfrac{72.03}{88.66} = 0.812\ rad/s^2$

So, angular speed at t = 2.95 s

$\omega = \alpha t = 0.812 \times 2.95 = 2.4\ rad/s$

Now, KE of the merry go round

$KE = \dfrac{1}{2} I \omega^2 = \dfrac{1}{2}\times 88.66\times 2.4^2 = 255.34 J$

Hence, the Kinetic energy of the merry go round = 255.34 J