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(sorry for the shabby figure!)

The problem-a bob of mass $m$ is hanging from a cart of mass $M$. The system is released from rest from the position shown. Find the maximum speed of the cart relative to ground. String length is $l$. The answer - $v= \sqrt{\frac{m^2gl}{M(M+m)}}$

My try- the speed of the cart must be maximum when the bob is at the lowest point relative the the cart. As there is no external force along X-axis, the speed of COM along X axis must be constant. Now, wrt cart, the speed of bob at that point must be $-\sqrt{gl}$ towards left. So, wrt ground, it is $(v-\sqrt{gl})$ towards right, where $v$ is the velocity of cart at that instant relative to ground, which must be the maximum speed, as said earlier. Hence,

$v_{com}=\frac{Mv+m(v-\sqrt{gl})}{M+m}=0$(initially, it was at rest)

Which gives the answer as $v=\frac{m\sqrt{gl}}{M+m}$.

I don't understand where I have gone wrong. Any help would be appreciated. Thanks in advance!!


As there is no external force along X-axis, the speed of COM along X axis must be constant.

There was no linear momentum at the start and so the centre of mass of the system does not move.

One way of doing the problem is to treat it as as a rotation about the centre of mass with the loss in gravitational potential energy of mass $m$ equal to the gain in rotational kinetic energy of the two masses $M$ and $m$.

  • $\begingroup$ but I am getting the same answer as before! $\endgroup$ – GRrocks Jun 9 '16 at 9:53
  • $\begingroup$ Find the position of the centre of mass in terms of $l, m$ and $M$. Then find the moment of inertia of the two masses about the centre of mass. Initially use $\omega$ in your kinetic energy formula which later you can convert into the linear speed of mass $M$. $\endgroup$ – Farcher Jun 9 '16 at 10:52
  • $\begingroup$ I am still not getting the right answer...are you sure this works?? $\endgroup$ – GRrocks Jun 9 '16 at 11:51
  • $\begingroup$ Okay got it thanks!!!! Can you point out though what is wrong with my approach??? $\endgroup$ – GRrocks Jun 9 '16 at 12:26
  • 2
    $\begingroup$ I think that your error is saying that the bob moves at a speed of $\sqrt{gl}$ because in doing that you have given all the gravitational potential energy to kinetic energy of the mass $m$ whereas some of that energy is taken up as kinetic energy by the mass $M$. $\endgroup$ – Farcher Jun 9 '16 at 13:15

Your error is the assumption that the speed of the bob is $\sqrt{gl}$. You do not provide any explanation for this value, which is not obvious.

I agree with your other assumptions :

(1) The COM of the system has constant velocity. In fact zero velocity, as Farcher points out, because the system starts from rest. This means that, if the horizontal components of speed of the bob and the cart are v and V relative to the ground then at all times $mv = MV$.

(2) The cart has maximum speed V when the bob swings through its lowest point. The justification for this is (presumably) that the system has minimum PE at this point, hence maximum KE. By symmetry, this is the equilibrium point for both masses.

The loss of PE at this point is $mgl(1-cos60)=\frac12mgl$. Applying conservation of energy (2) and momentum (1) gives the correct value of V.


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