Answer:
The force acting on the rope is 211.95 N.
Explanation:
Given that,
Mass of merry-go-round, m = 180 kg
Radius of merry-go-round, r = 1.5 m
Initial angular speed is 0 it was at rest
Final angular speed, [tex]\omega_f=0.4\ rev/s=3.14\ rad/s[/tex]
Time, t = 2 s
When it moves in circular path, the torque acting on it is given by :
[tex]\tau=F\times r=I\alpha \\\\F=\dfrac{I\alpha}{r}\\\\F=\dfrac{mr^2\omega_f}{2rt}\\\\F=\dfrac{180\times (1.5)^2\times 3.14}{2\times 1.5\times 2}\\\\F=211.95\ N[/tex]
So, the force acting on the rope is 211.95 N.
