How Does a Man Walking on a Merry-Go-Round Affect Its Motion?

[asafbuch]

1. A man with mass m walks with a constant velocity v with respect to the ground on the edge of a marry-go-round that has a radius of R and moment of inertia I. The system starts from rest.
a) what is the angular velocity of the marry-go-round?
b) when the man completes a full cycle (returns to its initial position) at what angle did the marry-go-round rotate?
c)what is the friction coefficient between the man and the marry-go-round for the man not to get off the marry-go-round?



2. No relevant equations.



3. I tried to use conservation of angular momentum and got this:
I\omega+mR²v/R=0

 

[tiny-tim]

welcome to pf!

hi asafbuch! welcome to pf!

(have an omega: ω and btw it's "merry-go-round" )

3. I tried to use conservation of angular momentum and got this:

that's right!

so ω = … ? ​

 

[asafbuch]

I thought so... but shouldn't there be any torque?

If there is a conservation of angular momentum than there is no torque acting on the marry-go-round, means there is no friction between the man and the marry-go-round...
but if there is a torque (and thus a friction) than you don't have a constant angular velocity...

whats wrong with the theory?

 

[tiny-tim]

hi asafbuch!

there will be conservation of angular momentum about any point if there is no external torque about that point …

the external friction is at the centre (and so has zero torque about the centre), so angular momentum will be conserved about the centre

(the friction between the man and the merry-go-round is an internal torque, so it doesn't matter )

 

[asafbuch]

oh, thank you tiny-tim!