Question

A merry-go-round rotates at the rate of 0.14 rev/s with an 84 kg man standing at a point 2.4 m from the axis of rotation. What is the new angular speed when the man walks to a point 0 m from the center

Answer 1

The given question is incomplete. The complete question is as follows.

A merry-go-round rotates at the rate of 0.14 rev/s with an 84 kg man standing at a point 2.4 m from the axis of rotation. What is the new angular speed when the man walks to a point 0.6 m from the center? Assume that the merry-go-round is a solid 25-kg cylinder of radius 2.0 m.

Step-by-step explanation:

The given data is as follows.

Initial angular velocity (
`\omega_(1)`) = 0.14 rev/s

`\omega_(1) = (0.14 * 2 \pi) rad/s`

= 0.87 rad/s

Mass of man (M) = 84 kg, Mass of cylinder (m) = 25 kg

Radius of the cylinder (r) = 2.4 m

Let
`r_(1)` be the initial distance of man from the axis of rotation = 2.0 m

Final distance of man from the axis of rotation (
`r_(2)`) = 0.6 m

Formula to calculate the moment of inertia of cylinder is as follows.

I =
`(Mr^(2))/(2)`

And, moment of inertia of man =
`mr^(2)`

Therefore, initial moment of inertia of the system is as follows.

`I_(1) = (Mr^(2))/(2) + mr^(2)_(1)`

=
`(84 * (2.4 m)^(2))/(2) + 25 * (2)^(2)`

= 241.92 + 100

= 341.92
`kg m^(2)`

Now, final moment of inertia of the system will be calculated as follows.

`I_(2) = (Mr^(2))/(2) + mr^(2)_(2)`

=
`(84 * (2.4)^(2))/(2) + 25 * (0.6)^(2)`

= 241.92 + 9

= 250.92

Now, according to the law of conservation of angular momentum is as follows.

Initial angular momentum = Final angular momentum

`I_(1) \omega_(1) = I_(2) \omega_(2)`

or,
`\omega_(2) = (I_(1)\omega_(1))/(I_(2))`

=
`(341.92 * 0.87)/(250.92)`

= 1.18 rad/s

Thus, we can conclude that the new angular speed is 1.18 rad/s.