Question

A certain merry-go-round is accelerated uniformly from rest and attains an angular speed of 1.2 rad/s in the first 18 seconds. If the net applied torque is 1200 N· m, what is the moment of inertia of the merry-go-round?
A) This cannot be determined since the radius is not given.
B) 1400 kg*m2
C) 18,000 kg*m2
D) 500 kg*m2
E) 9000 kg*m2

Answer 1

Answer:

Moment of inertia will be $I=18000kgm^2$

So option (c) will be correct answer

Explanation:

We have given initial angular velocity $\omega _i=0rad/sec$

Final angular velocity $\omega _f=1.2rad/sec$

Time taken to reach final angular velocity t = 18 sec

According to first equation of motion

$\omega _f=\omega _i+\alpha t$

$1.2=0+\alpha \times 18$

$\alpha =0.066rad/sec^2$

Torque is given in question as 1200 N -m

We know that torque is equal to $\tau =I\alpha$, here I is moment of inertia and $\alpha$ is angular acceleration

So $1200=I\times 0.0666$

$I=18000kgm^2$

So option (c) will be correct answer