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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 :

ajeigbeibraheem

Answer:

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

[tex]\frac{2F}{mR} \delta t[/tex]

Explanation:

Given that

the mass of the child = m

The radius of the disc = R

moment of inertia I = [tex]\frac{1}{2} mR^2[/tex]

change in time = [tex]\delta \ t[/tex]

By using the torque around the inertia ; we have:

T = I×∝

where

R×F = I × ∝

R×F = [tex]\frac{1}{2} mR^2[/tex]∝

F = [tex]\frac{1}{2} mR[/tex]∝

∝ = [tex]\frac{2F}{mR}[/tex]           ( expression for angular  angular acceleration)

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

[tex]\omega = \omega_0 + \alpha \delta t[/tex]

where ;

∝ = [tex]\frac{2F}{mR}[/tex]    

Then;

[tex]\omega = 0+ \frac{2F}{mR} \delta t[/tex]

[tex]\omega = \frac{2F}{mR} \delta t[/tex]

 

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

[tex]\frac{2F}{mR} \delta t[/tex]

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