Classical Physics - Pulley Problem

[danny271828]

A massless string is placed over a massless pulley, and each end is wound around and fastened to a vertical hoop. The hoops have masses M1 and M2 and radii R1 and R2. The apparatus is placed in a uniform gravitation field g and released with each end of the string aligned along the field.

I have to show that the tension is T = gM1M2/(M1+M2)

I can sort of solve the problem by just saying that (M1M2)/(M1+M2) is the reduced mass. Then all we have to do is say the tension is balanced with the weight, so that T = Mg = gM1M2/(M1+M2). But then doesn't this imply that the hoop isn't moving? I think I'm missing something here... It is also important to realize that both hoops are rolling down the string so the acceleration of one hoop downward is not necessarily the acceleration of the other one upward... Any help / hints would be appreciated... thanks

 

[BlackWyvern]

<statement retracted>

 

[BlackWyvern]

<statement retracted>

 

[Dick]

Try this. Each hoop has a torque on it of T times the appropriate radius. This determines the angular acceleration alpha of each hoop. Let s be the amount of string as a function of time. Then I claim s''=a1+a2=R1*alpha1+R2*alpha2 where the a's are the linear acceleration of each hoop. (The sum of the center of mass accelerations determines the second derivative of the length of string but the amount is also determined by the angular accelerations). Put this together with the usual force balance diagram for the vertical forces on each hoop and you will get what you want.

 

[Dick]

I just checked.

Your formula is supposed to be:

$$T = \frac{2gmM}{(M + m)}$$

That's the result if the hoops don't spin.

 

[BlackWyvern]

I see.
<edits>

 

[danny271828]

ok I'm here... about to pull my hair out

 

[Dick]

Can you work out a1 and a2 from linear force balance? Can you work out alpha1 and alpha2 from moment of inertia and torque? Let's get started... Do you accept the equality of the two expressions for s''? Does it seem right?

 

[danny271828]

yes to all

 

[danny271828]

so just using T - M1g = M1A1
and T - M2g = M2A2

 

[danny271828]

but these accelerations are center of mass acceleration for each hoop

 

[danny271828]

and the equations for s'' make perfect sense

 

[Dick]

My convention was to label positive acceleration down. Suggest you do the same if you want s''=a1+a2. Now the rotational part.

 

[Dick]

but these accelerations are center of mass acceleration for each hoop

Center of mass acceleration is also related to the rates of string being played out. Think about it.

 

[danny271828]

and Torque1 = Inertia_hoop1*alpha1 = Tension*R1
Torque2 = Inertia_hoop2*alpha2 = Tension*R2

 

[Dick]

Doing great! Put in the I's, solve for the alphas and put into the s'' equation along with the a's.

 

[danny271828]

hmm having trouble relating alphas to Acom

 

[danny271828]

oh just A = alpha*r right?

 

[danny271828]

hmm that's tangential though not center of mass accel

 

[danny271828]

and Inertia1 = (1/2)MR1^2
Inertia2 = (1/2)MR2^2

 

[Dick]

Individually the tangential accelerations DON'T have to equal the linear accelerations. I only claimed the sums do (due to the seldom used physics law "conservation of string"). :)

 

[Dick]

and Inertia1 = (1/2)MR1^2
Inertia2 = (1/2)MR2^2

Dump the (1/2). It's a hoop, not a disk. I made that mistake.

 

[danny271828]

so I am getting these equations

M1R1alpha1 = T = M2R2alpha2

 

[Dick]

I believe that.

 

[danny271828]

so T = ((M1 + M2)s'')/2

 

[danny271828]

oops nevermind

 

[Dick]

You can forget s'' now. Just use a1+a2=R1*alpha1+R2*alpha2.

 

[danny271828]

hmm but that doesn't make sense does it? then M1a1 = T = M2a2 which means the tension is the only force present

 

[Dick]

hmm but that doesn't make sense does it? then M1a1 = T = M2a2 which means the tension is the only force present

It means tension is the only force that produces a torque. Gravity doesn't, it acts at the center of mass=center of rotation.

 

[danny271828]

ok now I am just going in circles

 

[danny271828]

3 unknowns and only 2 eqs

 

[Dick]

You have expressions for the a's in terms of T and you can get expressions for the alpha's in terms of T. Put them both into a1+a2=r1*alpha1+r2*alpha2. All that's left is T. Solve for it. You aren't going in circles. You are just giving up. Don't be a weakling! It's ONE equation in ONE unknown.

 

[danny271828]

you still there? i don't really feel like I am getting anywhere... in fact i think I've been here before heh..

 

[Dick]

Yes, I'm still here. But I'm getting impatient. Solve this thing, ok?

 

[danny271828]

T = ((m1 + m2)G)/2