Problem: Two masses $A$ and $B$ connected with an inextensible string of length $l$ lie on a smooth horizontal plane. $A$ is given a velocity of $v$ m/s along the ground perpendicular to line $AB$ as shown in the figure. Find the tension in string just after this instant?

enter image description here

Solution: I solved this problem in the Center of Mass frame of reference by getting the velocities of both the masses in C.M frame and applying $T = Mv^2/r$ for one of the bodies. And I got the answer. (as here)

My Question: I am not satisfied with my solution. I am not convinced why I had to solve this problem in C.M frame.

  1. Because mass B is at rest (just after), why isn't it the instantaneous axis of rotation? and then, if I treat the mass 2m to be rotating about mass m and then apply this: enter image description here Why I am not getting the same answer enter image description here?

  2. How to solve this problem in ground frame of reference?

How do we know what is the system rotating about? Is there any other way to solve the problem? I just feel I didn't get it properly.

  • $\begingroup$ I think that solving it in the CM frame gives the system something to pivot about, allowing circular motion and hence making the problem solvable. In the lab frame, B will not remain fixed due to the criteria in the question, making it difficult to analyse. $\endgroup$ – Physics Jan 5 at 16:48
  • $\begingroup$ In the absence of any further external forces, the linear momentum of the system is constant so the centre of mass travels at a constant velocity $\frac{2v}{3}$ wrt the lab frame. The COM frame is then an inertial frame, so we can apply Newton II to either of the masses. Though since A and B are both accelerating wrt this inertial frame, if we transform into either of them we will need to add some fictitious forces. $\endgroup$ – James Wirth Jan 5 at 17:32
  • $\begingroup$ You are not satisfied with your solution? It looks like someone else's solution which you are not satisfied with. $\endgroup$ – sammy gerbil Jan 5 at 22:22
  • $\begingroup$ No. I am not satisfied with _my_solution. That solution isn’t how I solved it. I just linked it because I used to verify my answer. And I clearly described what is that I am not satisfied also. I am trying to solve it in other frames of references. $\endgroup$ – claws Jan 6 at 4:19
  • 1
    $\begingroup$ doubtnut.com/question-answer-physics/… $\endgroup$ – sammy gerbil Jan 7 at 2:53

To solve it in the ground plane you need to separate out in your mind the rotation about a common centre of mass and the movement of the centre of mass.

You might then need to think about the special case of such motion in which one of the masses stops moving from time to time in a particular reference frame.

Googling 'cycloid' might also shed some light.

Your second approach is wrong because if you fix the smaller mass in place it is not free to yield to the tension in the string, so you get a greater tension for the answer.


An impulse acting at point A gives the 2m mass its velocity, and the center of mass a parallel velocity of (2/3)V. Since the impulse is not acting in line with the center of mass, it also provide an impulsive torque which causes the system to rotate around the center of mass. Working in the center of mass system is valid and gives a tension which is the same at each end of the string.


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