So I have probably got a major conceptual error here, But I was just confused by two problems while going through rotational dynamics.

So the first problem goes like this: A thin uniform rod of mass $m$ and length $l$ is attached to two strings at its ends. Find the tension in one string when the other is cut.

Problem 1

So the way we approach this problem goes like this:

  1. Calculate Torque of $mg$ about the other string and equate it with $I\alpha$
  2. Convert the angular acceleration into linear acceleration
  3. Use the equation $mg$ - $T$ = $ma_c$
    $\tau = \frac{mgl}{2} = \frac{ml^2}{3}\alpha$
    $\alpha = \frac{3g}{2l}$
    $\Rightarrow$ Linear accelration of Centre of mass = $\frac{3g}{4}$ (As the centre of mass is $\frac{l}{2}$ away from AOR)
    The tension can be found to be $\frac{mg}{4}$

    But the point to remember here is that we calculated the torque about an axis and then got linear acceleration of centre of mass from there

    Now consider this problem:
    A thin uniform square plate of mass $m$ and side $a$ is attached to two strings. Find the tension in one string when the other is cut.
    Problem 2

I approached this problem in a similar way. Calculated torque about what I think is the axis of rotation (a line perpendicular to the plane passing through $A$) and then got the angular acceleration, then the acceleration of centre of mass and then the force equation. It goes like this:

$\tau$ = $I\alpha$ = $\frac{mga}{2}$ = $\frac{2ma^2\alpha}{3}$
$\Rightarrow$ $\alpha = \frac{3g}{4a}$
Now linear acceleration $a_l$ = $\frac{3g}{4\sqrt{2}}$ (as the centre is $\frac{a}{\sqrt{2}}$ away from $A$)

$T = mg - m\frac{3g}{4\sqrt{2}}$
This is where everything falls apart. When I use this value in the force equation I get some answer which does not match with the key. The solution proceeds by taking the torque of Tension force about the centre of mass and then calculating $a_l$ of point $A$ from that equation, and then the acceleration of point $A$ in the vertical direction is taken to be zero, thus we can get a value for the angular acceleration, which then gives the value of Tension as $\frac{2mg}{5}$

Solution 1
Solution 2 (has a few typos)

But why does a similar approach of calculating torque about the axis of rotation not work here? I cannot understand why in one case it works and in the other it doesn't.
This was a little long, thanks to anyone who read this till the end and an extra thanks to all suggestions and answers.

Edit [05.01.2021]: I made a setup and filmed the motion of point $A$. Can't make solid conclusions. Could be helpful, I guess. The experiment conditions aren't all perfect but if anyone can study the footage closely then that would be helpful too. And from what I think, as the string is in full extension after it is cut, the length is same but now it makes an angle with the original position, which means point $A$ must have moved up. That could be a fair contradiction to the solutions.
Link to video

  • $\begingroup$ > "torque of Tension force about the centre of mass and then calculating al of point A from that equation" This makes no sense, acceleration of point A can't be determined from torque equation for center of mass alone. Please post the solution you are referring to. $\endgroup$ Jan 4, 2021 at 20:41
  • $\begingroup$ @archmundada Thank you for the updates .I agree with your solution . I couldn't spot the mistake . I am also waiting for an answer. $\endgroup$ Jan 4, 2021 at 20:59
  • $\begingroup$ @Ján Lalinský I have added two solutions; I originally referred to the second one, which was quite unclear. I just discovered the one from quora and that has some quality to it, kinda made things clearer, but still can't understand why my approach fails here $\endgroup$ Jan 4, 2021 at 22:36
  • $\begingroup$ Hi archmundadaE. Linking to private clouds, dropbox, etc, is for various reasons not acceptable on SE, cf. this meta post. $\endgroup$
    – Qmechanic
    Feb 6, 2023 at 9:59

2 Answers 2


This problem is not nearly as simple as it is presented. I'm pretty sure the solution you've linked to is either wrong or hiding some major assumptions. The key seems to be "But the net acc. of A in the vertical direction should be zero," but no justification for that claim is given, and I don't think it's even right.

The question is the relationship between the angular acceleration $\alpha$ and the linear acceleration of the center of mass $\vec{a}$. In many simple cases in homework problems you can use $|a|=r\alpha$ and the direction of the acceleration is obvious. This usually happens when you have a solid object that is rotating around some fixed point of rotation. But that's not the case here! The pivot point is free to swing with the string so $|\vec{a}_CM| = r\alpha$" is not necessarily true.

Start with what we do know. When in doubt, go to Newton's Laws.

First, use $m a_{CM} = \vec{F}_{net}$ and that see the object must accelerate vertically. If $T$ is vertical then the net force is vertical and therefore acceleration of the center of mass must be vertical or zero.

Second, use $I_{CM}\alpha = \sum \tau_{CM}$, ie the rate of change of rotation around the center of mass is determined by the torques around the center of mass. This is always true, and it implies that the object must begin to rotate around its center of mass. Given that the tension is non-zero there is definitely a non-zero torque and so rotation must happen.

But beyond this, I'm pretty sure the problem doesn't have a unique answer without some additional assumption. And the assumption the solution makes is probably wrong. Imagine holding the string in your hand and winding it up a bit so there's extra tension in it (you'll have to hold one end down while doing this so it stays horizontal). Then when you let the object go you should see it snap up with more acceleration than it would have otherwise even though it started in the same position. The point is that there are many valid answers for the tension in the string. This is true both for your simpler problem and the square one.

I would say that the only really valid answer to "what is the tension in the string the instant the other string is cut" is "the same as the tension in the string the instant right before the other string was cut, $mg/2$." We usually ignore the time it takes for the tension in a string to change, but this is a situation where we can't do that.

  • 1
    $\begingroup$ It actually seems plausible that A can move up after the string is cut. I did a quick experiment by making a setup of the situation and filming it in slo-mo. (drive.google.com/file/d/11Uk1EfKOw2iKhYcmtmH5HePOzvcQfGTQ/…) Any new thoughts from this? The motion of A seems quite horizontal here from my view, which is weird, again. A hint of upward motion, maybe? $\endgroup$ Jan 5, 2021 at 15:26
  • $\begingroup$ I tried some similar experiments myself when I was working on this. The only thing that's clear is that the pivot point definitely moves. That proves my point about the relationship between the angular motion and the linear acceleration not being simple. Tracking the motion of the center of mass is obviously harder to do, though it might make an interesting experiment for some of my students in the future. One suspicion I have is that the motion will depend on the length of the string. If that's the case it would prove that the solutions you linked to are too simple. $\endgroup$ Jan 5, 2021 at 15:57

In the second case, immediately after the second string is cut, the first sting is still vertical and there is no horizontal force acting on the object. The acceleration of the center of mass is vertical and would be the same as the point mid-way between, A, and, B; rather than tangential.

  • $\begingroup$ Why there is a horizontal force in the first case ? $\endgroup$ Jan 4, 2021 at 21:42
  • $\begingroup$ In the first case, with the thin rod, the initial forces are vertical, but the center of mass is at the mid-point between, A, and, B, so that the tangential acceleration is also vertical. $\endgroup$
    – R.W. Bird
    Jan 5, 2021 at 16:00

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