A rotating disk with two attached masses that slide without friction

1. Mar 13, 2013

cacofolius

1. The problem statement, all variables and given/known data

A disk rotates with angular velocity w. Two masses, Ma and Mb, slide without friction in a groove passing through the cnter of the disk. They are connected by a light string of length L, and are initially held in position by a catch, with mass Ma at distance Ra from the center. Neglect gravity. At t=0 the catch is removed and the masses are free to slide. Find r''_a immediately after the catch is removed in terms of Ma, Mb, L, Ra, and w.

2. Relevant equations

a=(r''-rw^2)êr + (rθ''+2r'we) êθ

3. The attempt at a solution

I figured that immediately after the catch is removed Ma only experiences a centripetal tension that must be equal to the one Mb is experiencing. So we'll have

Mb*r''_b=Ma*r''_a

Mb*r''_b=Mb*(w^2*Rb) = Mb*w^2*(L-Ra)

Ma*r''_a = Mb*w^2*(L-Ra) → r''_a= w^2*(L-Ra)*(Mb/Ma)

I would like to know if this is correct, or if I'm missing something (or everything!).

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2. Mar 13, 2013

TSny

That sounds right.
What's the reasoning behind these two equations? I don't think they're correct.

[EDIT: By the way, Welcome to PF! ]

3. Mar 13, 2013

cacofolius

Hi TSny! Thanks for your repply.

I've been thinking again, and saw that I was equating the centripetal acceleration of Mb to the radial acceleration of Ma.

Now I've reasoned that (centripetal force on of Mb)-(centripetal force on Ma) should result in the radial force on Ma. In this case

Mb(-Rb*w^2) - Ma(-Ra*w^2)= Ma*r''_a

Mb(-(L-Ra)*w^2) + Ma(Ra*w^2)= Ma*r''_a

Mb(Ra-L)w^2) + Ma(Ra*w^2) = Ma*r''_a

(w^2) [Mb(Ra-L)+Ma(Ra)] (1/Ma) = r''_a

the radial acceleration of Mb would be the same, but in opposite direction.

4. Mar 13, 2013

haruspex

That sounds suspiciously like thinking in terms of centrifugal force.
I suggest it will keep things clearer if you introduce T as the tension and treat each mass separately.

5. Mar 13, 2013

cacofolius

Thanks, I thought a little bit more:

T=Ma*Aa; Aa= (r_a''-Ra*w^2)êr
T=Mb*Ab; Ab= (r_b''-Rb*w^2)êr

(I eliminated the êθ part of the vector, since it hasn't started yet to move -"immediately after the catch is removed"- therefore r'=0, and of course θ''=0).

Now, both centripetal accelerations point to the origin, therefore have the same sign, but, no matter how the whole system moves (whether Ma moves inwards or outwards), the radial accelerations will have opposite sign, and equal magnitud (since I assume also that the string is inextensible). Then r''_b= -r''_a
Equating the T's:

Mb(r''_b'-Rb*w^2)=Ma(r''_a-Ra*w^2)

Replacing r''_b=-r''_a', and Rb=(L-Ra):

Mb(-r''_a-(L-Ra)*w^2)=Ma(r''_a-Ra*w^2)
Mb(-r''_a+(Ra-L)*w^2)=Ma(r''_a-Ra*w^2)

Distributing Mb inside the left parenthesis, and then dividing all through Ma:

-r''_a(Mb/Ma)+(Mb/Ma)*(Ra-L)*w^2=(r''_a-Ra*w^2)

-r''_a(Mb/Ma)+(Mb/Ma)*(Ra-L)*w^2+Ra*w^2=r''_a

(Mb/Ma)*(Ra-L)*w^2)+Ra*w^2= r''_a + r''_a(Mb/Ma)

(Mb/Ma)*(Ra-L)*w^2)+Ra*w^2= r''_a(1+(Mb/Ma))

And finally

[(Mb/Ma)*(Ra-L)*w^2)+Ra*w^2] / (1+(Mb/Ma)= r''_a

I hope is clear (and ok)

6. Mar 13, 2013

TSny

Looks good!

7. Mar 14, 2013

cacofolius

Great! Thanks TSny, haruspex, for your help.