Atwood Machine

arildno

3. Motion about pulley 2:
If you are uncertain about accelerated frames of reference, note that the acceleration of block 2 in the inertial frame fulfills:
[tex]a_{2,abs}=a_{1}+a_{2}[/tex]
whereas the acceleration of block 3 in the inertial frame is:
[tex]a_{3,abs}=a_{1}-a_{2}[/tex]

Hence, we get:
[tex]T_{2,L}-m_{2}g=m_{2}a_{2,abs} (4)[/tex]
[tex]T_{2,R}-m_{3}g=m_{3}a_{3,abs} (5)[/tex]

The frictional torque about pulley 2 must fulfill:
[tex]R_{2}(T_{2,R}-T_{1,L})=I_{2}\frac{a_{2}}{R_{2}}(6)[/tex]

Equations (1)-(6) is the system to be solved.

e(ho0n3

That is exactly what I needed! My physics book makes no mention of this, so I was left "in the air". It's funny how my book assigns such problems without explaining anything about Newton's laws in non-inertial frames. I feel cheated here.

I think you meant

[tex]R_{2}(T_{2,R}-T_{2,L})=I_{2}\frac{a_{2}}{R_{2}}[/tex]

Thanks alot. I really appreciate it.

arildno

Yes, that's what I meant..

e(ho0n3

Going back to my first post, I found in my book a way of solving the problem using the relationship [itex]\tau = dL/dt[/itex]. In the derivation of the solution, there is a part that says and I quote:

The external torque on the system is

[tex]\tau = m_2gr - m_1gr[/tex]

I was wondering how they came to this conclusion. Isn't the torque supposed to be [itex](T_2 - T_1)r[/itex]. What am I missing here?

arildno

If the author uses the system consisting of pulley+two objects, they are correct in this being the torque of external forces about the pulley axle (the tensile forces becomes internal forces in the system, along with, most importantly, the friction between cord and pulley).
However, L then becomes the angular momentum of a non-rigid body about the pulley axle.

I would be interested to see how they proceed.

arildno

Note that [tex](T_{2}-T_{1})r[/tex] is the external torque acted upon the pulley (and only the pulley) with respect to the pulley axle.
[tex](m_{2}-m_{1})gr[/tex] is the external torque acted upon the non-rigid system pulley+2 objects with respect to the pulley axle.

I honestly hadn't thought of the author's approach here, I hope you don't feel cheated!
IMO, I think it's great that solutions of physics problems can be found in a variety of ways, all sort of adjust itself to each other, so that everything adds up.

Now that I think of it, I regard the author's approach to be far "cleverer" than my own, since it is shorter, and hence, more elegant.
I still think though, that my approach is rather instructive in showing how the internal forces in the author's system actually works in order to generate the motion predicted by his analysis.

arildno

The author's approach suggests that mechanical energy is conserved, since only gravity seems to be important here (i.e doing work)
In order to prove this, let's set up the energy integral, assume it is constant
in time and show that the implication is the result we're seeking:
[tex]\frac{1}{2}m_{1}v_{1}^{2}+\frac{1}{2}m_{2}v_{2}^{2}+\frac{1}{2}I\omega^ {2}+m_{1}gz_{1}(t)+m_{2}gz_{2}(t)=C[/tex]

Now, clearly, we have: [tex]v_{2}=-v_{1},\omega=\frac{v_{1}}{r}[/tex]

Differentiating with respect to time yields:
[tex]m_{1}v_{1}a+m_{2}(-v_{1})(-a)+\frac{I}{r}v_{1}\frac{a}{r}-(m_{2}-m_{1})gv_{1}=0[/tex]

Recognizing [tex]v_{1}[/tex] as a common factor and rearranging, yields the expression for a.

Now, what about the existence of friction?
Wouldn't that violate conservation of mechanical energy?
It doesn't because it is static friction, and static friction can be shown to do no net work on the system consisting of the two objects that are concerned!

Edited an ambiguity in the last line

e(ho0n3

Yes, yes. I forgot about the external torque issue. This wasn't stressed in the explanation of the solution so it lead me astray. I understand now. Thanks again.