Physics: Angular velocity (multiple choice)

Question

At a dilapidated playground, people have stolen the support bar to a merry-go-round leaving a flat disk that can rotate freely. While pondering conservation of angular momentum, a curious physics student wonders if he left his physics book on the ground and jumped onto the merry-go-round, and started walking, will he ever move away from the physics book, assume he is less than half its mass?

a. No, the merry-go-round will have angular momentum in one direction and he in the other, they will cancel and he’ll stay right above the book.

b. No, because the initial angular momentum is zero, so the final must be zero too, causing his angular velocity to be zero, so he’ll stay right above the book.

c. Yes, angular momentum is not conserved so he’ll move around away from the book.

d. Yes, because his moment of inertia is smaller than the merry-go-round’s, so his angular velocity will be bigger, so he’ll move around away from the book.

Answer 1

The answer is d - he will move away from the book.  What makes this problem hard to think about is that the mechanics of walking is complicated.   To get around this difficulty replace ordinary walking with a series of "bunny hops" in which the student leaps forward, stays airborne for some time  τ_air, then lands and sticks.  

The analysis of the first bunny hop (where the initial condition is rest) is as follows:

On leaping, the student exerts a force, F for a very short time Δt on the disk and (by Newton's 3rd) the disk exerts an equal an opposite force on the student for the same short time.    According to the principle that impulse = change in momentum, we have, for the student:   F Δt =  m v    , where m is mass of the student and 

   the angular form of this for the disk:    τ Δt =  I ω  

           where I is the rotational inertia (moment of inertia) of the disk and ω is the angular speed of the disk

  ( we don't have to write   Δv and Δω because we start from rest)

  However   the torque, τ, is   F R   where R is the radius of the disk.

 

     The two equations are  thus

        F  Δt =  m  v  

       RF Δt  =  I  ω

 

    Dividing the first one by the second leads, after some rearrangement, to  R ω =  [ m /(I/R²) ] v

 

   Since  I/R²   is probably much greater than m,    ω is fairly small.  

 

   At this point our student is still airborne.   If he stays airborne for  a time τ_air  , he will move a distance 

   d =  v τ_air  away from the book on the ground before he lands and sticks.  The spot on the disk from whence he jumped will move a (probably shorter) distance   R ω τ_air =  v [ m/(I/R²)] τ_air   in the other direction.  

 

What happens when the student lands and sticks?   A similar analysis shows that the both v and ω return to zero. 

Thus the bunny hop does not impart any net rotational motion at all.   But the student will have traveled a distance away from the book.    A series of bunny hops will move the student further and further away from the book. 

A more involved treatment of walking will allow the disk to gradually gain rotational speed as the student accelerates with respect to the disk.