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

Answer:

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

$\frac{2F}{mR} \delta t$

Explanation:

Given that

the mass of the child = m

The radius of the disc = R

moment of inertia I = $\frac{1}{2} mR^2$

change in time = $\delta \ t$

By using the torque around the inertia ; we have:

T = I×∝

where

R×F = I × ∝

R×F = $\frac{1}{2} mR^2$∝

F = $\frac{1}{2} mR$∝

∝ = $\frac{2F}{mR}$           ( expression for angular  angular acceleration)

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

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

where ;

∝ = $\frac{2F}{mR}$    

Then;

$\omega = 0+ \frac{2F}{mR} \delta t$

$\omega = \frac{2F}{mR} \delta t$

 

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

$\frac{2F}{mR} \delta t$