Need help- Force of friction in a circle.

AI Thread Summary
To determine the force of friction necessary to keep a 45-kg worker from falling off a merry-go-round while moving at 4.1 m/s, the relevant equations include centripetal acceleration and frictional force. The radius (r) of the circular path is 6.3 m, which is essential for calculating the centripetal acceleration. The derived equation for friction, μ = v²/(rg), indicates that both the speed and radius are crucial in finding the coefficient of friction needed. Clarification on the role of the radius in the equations is sought to properly approach the problem. Understanding these relationships is key to solving the friction force requirement.
Socom
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Homework Statement


A 45-kg merry-go-round worker stands on the ride's platform 6.3m from the center. If her speed as she goes around the circle is 4.1 m/s, what is the force of friction necessary to keep her from falling off the platform?


Homework Equations


f=ma
Ac=v^2/r
mu=v^2/rg (derived from [mu]g=v^2/r


The Attempt at a Solution


I don't believe the 6.3m is relevant to the problem.

I believe Fnet=sigmaFx=theta=m*a

I am not really sure how to attack this problem and some help would be very nice.
 
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Socom said:

Homework Statement


A 45-kg merry-go-round worker stands on the ride's platform 6.3m from the center. If her speed as she goes around the circle is 4.1 m/s, what is the force of friction necessary to keep her from falling off the platform?

Homework Equations


f=ma
Ac=v^2/r
mu=v^2/rg (derived from [mu]g=v^2/r

The Attempt at a Solution


I don't believe the 6.3m is relevant to the problem.

I believe Fnet=sigmaFx=theta=m*a

I am not really sure how to attack this problem and some help would be very nice.

You got the right equation:

\mu = \frac{v^2}{rg}

now tell me... what does the r mean in:

A_c = \frac{v^2}{r}
 
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