Question

A merry-go-round rotates at the rate of 0.17 rev/s with an 79 kg man standing at a point 1.6 m from the axis of rotation.

What is the new angular speed when the man walks to a point 0 m from the center? Consider the merry-go-round is a solid 34 kg cylinder of radius of 1.6 m.
Answer in units of rad/s.

Answer 1

Hi there!

We can use the conservation of angular momentum to solve.

`L_i = L_f\\\\I\omega_i = I\omega_f`

I = moment of inertia (kgm²)

ω = angular velocity (rad/sec)

Recall the following equations for the moment of inertia.

`\text{Solid cylinder:} I = (1)/(2)MR^2\\\\\text{Object around center:} = MR^2`

Begin by converting rev/sec to rad sec:

`(0.17rev)/(s) * (2\pi rad)/(1 rev) = 1.068 (rad)/(s)`

According to the above and the given information, we can write an equation and solve for ωf.

`1.068((1)/(2)(34)(1.6)^2 + (79)(1.6)^2) = \omega_f((1)/(2)(34)(1.6^2) + 79(0^2))\\\\\omega_f = \boxed{6.03 (rad)/(sec)}`