Question

A child of mass m is standing at the edge of a carousel. Both the carousel and the child are initially stationary. The carousel is a large flat, uniform disc of mass M, and radius R, which turns on a frictionless pivot at its center. It has moment of inertia I = MR2 /2. The child is very small compared to the carousel. The child starts to run around the edge of the carousel at speed v. The direction of child is counter-clockwise when the carousel is viewed from above. Taking the z-axis as upwards, what is the angular velocity of the carousel after the child has started running?

Answer 1

the angular velocity of the carousel after the child has started running =

`(2F)/(mR) \delta t`

Step-by-step explanation:

Given that

the mass of the child = m

The radius of the disc = R

moment of inertia I =
`(1)/(2) mR^2`

change in time =
`\delta \ t`

By using the torque around the inertia ; we have:

T = I×∝

where

R×F = I × ∝

R×F =
`(1)/(2) mR^2`∝

F =
`(1)/(2) mR`∝

∝ =
`(2F)/(mR)` ( expression for angular angular acceleration)

The first equation of motion of rotating wheel can be expressed as :

`\omega = \omega_0 + \alpha \delta t`

where ;

∝ =
`(2F)/(mR)`

Then;

`\omega = 0+ (2F)/(mR) \delta t`

`\omega = (2F)/(mR) \delta t`

∴ the angular velocity of the carousel after the child has started running =

`(2F)/(mR) \delta t`