0
$\begingroup$

I'm having some troubles in understanding why my reasoning doesn't work in the following problem:

Problem Two pulleys of mass $m_1$,$m_2$ and radius $r_1,r_2$ are connected by a belt (like chain rings in a bycicle) of negligible mass. Torques $\Gamma _1$ and $\Gamma_2$ are applied to $m_1$ and $m_1$. The belt is not slipping over the two pulleys. What is the angular acceleration of 1?

I've already solved the problem writing down the rotational equations for the two pulleys with the constraint $\omega _1 r_1 =\omega _2 r_2$ given by the non-slipping condition. What I got is:$$\alpha _1=2\dfrac{\frac{\Gamma _1}{r_1}+\frac{\Gamma _2}{r_2}}{r_1(m_1+m_2)},$$ which is correct, according to my book.

Then I've tried to attack the problem from a different point of view. Since the tensions in the belt are internal forces, the only torques producing a change in total angular momentum are the two torques applied to the two axis of the pulleys. At first my idea was to write:$$\dot L = \Gamma _1+ \Gamma _2,$$since the angular momentum here is independent from the pole, the solve for $\dot \omega _1$ ($\dot L=k\dot \omega_ 1$ for some constant). However, the two values $\Gamma _1$ and $\Gamma _2$ are the torques referred to the center of the respective pulleys and this approach doesn't work.

In fact the derivative of $L$ is really:$$\dot L=\dot L_1+\dot L_2 = \Gamma _1+(T_1-T_2)r_1+\Gamma_2+(T_2-T_1) r_2,$$where $T_1,T_2$ are the two tensions in the belt and the angular momentums are respect the center of mass of 1 and 2. So the sentence that I wrote above "Since the tensions... internal forces..." seems to be wrong.

Could you please point out the wrong parts of my reasonings?

Also, how could I write the two torques $\Gamma _1$ and $\Gamma _2$ respect to a different pole?

Finally, is there a different way from the (working) one I wrote above to solve this problem?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
1
$\begingroup$

Torques of the tension forces will cancel out internally when you write the equation of angular momentum about one single axis. What you are doing is cancelling out torque of T1 about axis 1 with T2 about axis 2.

Now if the two pulleys are connected by a chain, they will rotate about different axes, and so the torque of tension about one axis wont cancel out with torque of tension about the other.

Practically, if one pulley rotates clockwise, the other will also rotate clockwise and torques of both tensions about their simultaneous axes will add up and not cancel out.

P.S. I'm not too sure about how to write the torques or angular momentum about a single axis, so your first method is good enough.

Improve this answer
$\endgroup$
3
  • $\begingroup$ Thank you, I see the problem here. I was trying to cancel out the torques of the tensions about two different axes, that in general won't cancel out if they are referred to different axes. However, I wonder if is there a simpler solution that does not involve tensions, since they are only constraints. $\endgroup$
    – pppqqq
    Jun 29, 2013 at 19:37
  • $\begingroup$ constraints are only useful to correlate kinematic quantities, i.e. velocity, acceleration or for your case angular velocities and accelerations. But they do not involve forces or torques. So they might not be too useful. Anyway, I think your first method by which you solved the problem is simple enough. And yeah, please accept my answer if you're satisfied by it. $\endgroup$
    – udiboy1209
    Jun 30, 2013 at 17:50
  • $\begingroup$ Hi @udiboy, your answer is good since it points out where was the confusion in my reasoning. However I did not accept it because it doesn't answer my last two questions in my OP, that are interesting as well in my opinion. $\endgroup$
    – pppqqq
    Jun 30, 2013 at 18:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.