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
[tex]M = \dfrac{W}{g}= \dfrac{805}{9.81} = 82.059\ Kg[/tex]
Moment of inertia of the disk
[tex]I = \dfrac{1}{2}Mr^2[/tex]
[tex]I = \dfrac{1}{2}\times 82.059\times 1.47^2 =88.66\ kgm^2 [/tex]
Torque on the child
[tex]\tau = F \times r = 49 \times 1.47 = 72.03 Nm[/tex]
Angular acceleration
[tex]\alpha = \dfrac{\tau}{I}=\dfrac{72.03}{88.66} = 0.812\ rad/s^2[/tex]
So, angular speed at t = 2.95 s
[tex]\omega = \alpha t = 0.812 \times 2.95 = 2.4\ rad/s[/tex]
Now, KE of the merry go round
[tex]KE = \dfrac{1}{2} I \omega^2 = \dfrac{1}{2}\times 88.66\times 2.4^2 = 255.34 J[/tex]
Hence, the Kinetic energy of the merry go round = 255.34 J
