Find acceleration in the following system :

enter image description here

This was a solved example in the book (*) which had steps as follows:

$$ W_1 - T_1 = M_1 a $$ $$ W_2 - T_2 = M_2 a (**)$$ $$ T_1 R - T_2 R = I \alpha $$ $$ T_1 - T_2 = I \frac{ \alpha}{R}$$

Now, combining equations,

$$ W_1 - W_2 - \frac{I\alpha}{R^2} = (M_1 +M_2) a$$

Using the fact that $ I = \frac{M_p R^2}{2}$ since pulley is a disc and plugging in weights,

$$ a= \frac{(M_1 - M_2) g}{ M_1 +M_2 + \frac{M_p}{2} }$$

What makes this solution hard for me to think about is that the rotation of pulley is dependent on it having mass, that is, if we took that mass of the disc tends to zero, then we would find that there is no rotation. So, do you need mass to rotate? I'm pretty sure you don't.. which I find weird or is this saying something about the physical world, that is, do you need mass to feel torques and forces unlike where particles can just move for the sake of it in mathematics?


*: Kleppner and Kolenkow's , introduction to mechanics, page-254

**: due to constraint acceleration of both blocks are equal.

Notes for answers/future answers:

  • Why is infinite angular acceleration unrealistic? I'm pretty sure you can make a light object rotate fast if you apply a small torque.

  • Is mass the most fundamental property through which object can interact with surroundings?

  • As we reduce the mass of pulley, it is necessary that it should become unable to rotate but why?

  • As a counterpoint to the previous one, $ \sum \tau = I \alpha$ and rearranging: $ \frac{ \sum \tau}{I} = \alpha$ so if we made the mass small then moment of inertia would go down to zero. This would make the angular acceleration blow up, however some say that it should still be zero.. which I can't understand.

  • $\begingroup$ If you plug $M_p=0$ into the equation you would still have an acceleration. $M_p$ just reduces the acceleration. $\endgroup$
    – Dr Xorile
    Sep 15, 2020 at 14:16
  • $\begingroup$ I'm not talking about the total behaviour, I'm referring to behaviour of the disc $\endgroup$ Sep 15, 2020 at 14:21
  • $\begingroup$ I'm pretty sure that factor of $R$ in your final answer should be a $g$ instead. $\endgroup$ Sep 15, 2020 at 14:23
  • $\begingroup$ ahh nice catch! $\endgroup$ Sep 15, 2020 at 14:26
  • $\begingroup$ Okay Why do you think there would be no rotation if $M_p=0$ $\endgroup$
    – Dr Xorile
    Sep 15, 2020 at 14:27

2 Answers 2


The phrase "a massless rope" or "a massless pulley" in introductory mechanics classes should really be understood to mean that the mass of the object is small, but non-zero. We can do this by taking the limit as $M_p \to 0$, which is effectively a sequence of cases where $M_p$ gets smaller and smaller and smaller but is never quite zero, and seeing what the behavior of the system is.

In the limit $M_p \to 0$, everything is perfectly well-defined. In particular, the angular acceleration of the wheel is, for general $M_p$, $$ \alpha = \frac{(M_1 - M_2)g}{R(M_1 + M_2 + M_p/2)} $$ and in the limit as $M_p \to 0$, this is a finite number. You will also find that in this limit, $T_1 = T_2$, so there is no net torque on the wheel; and since the moment of inertia of the wheel is also undefined, the equation $\sum \tau = I \alpha$ becomes $0 = 0$, which is automatically satisfied.

In practice, saying that an object is "massless" in a mechanics problem is really just saying that its mass is so small that it can be ignored. Functionally, this is equivalent to saying that the object has no net force and no net torque on it. This is because $\sum F = ma$, and if $m \approx 0$, then the net force on the object must be (basically) zero as well. So a "massless rope" automatically exerts the same tension at both of its ends; but if you're given a problem where the mass of the rope is not negligible, this is no longer true.

Follow-up: it's important to note that Newton's Second Law is not terribly useful for objects whose mass is "truly" zero. It predicts that such objects must experience no net forces and no net torques, and doesn't tell us anything about their accelerations (since $m a = 0$ automatically, $a$ can be any number and still satisfy Newton's Second Law.) While you can predict the motion & rotation of very light objects using Newton's Laws, you can't say anything about objects whose mass is truly $m = 0$.

The only way to make meaningful predictions about the behavior of "massless" objects in Newtonian mechanics is to assume that their mass is non-zero, and then look at the behavior of the solution in the $m \to 0$ limit. Depending on the system, this limit might be physically reasonable behavior, as it is in this case; or it might not be physically reasonable (e.g., a fixed force is applied to an object of mass $m \to 0$.)

  • $\begingroup$ Even if mass drops to zero, couldn't the angular acceleration rise so as to cancel the effect out $\endgroup$ Sep 15, 2020 at 14:34
  • $\begingroup$ Secondly , I'm pretty sure that you can make light objects rotate $\endgroup$ Sep 15, 2020 at 14:39
  • $\begingroup$ +1. But another way to understand it is an unrealistic simplification that allows you to focus on the physics in this chapter of the textbook. You can see how much more complex the math gets when you add mass to the rope and pulley. It would add confusion to the first chapters. In this chapter, students are ready for the complexity. All massive objects respond to forces. If a truly massless pulley was possible, you might expect it to act like a photon and move only at the speed of light. This is not the intent of the problem. $\endgroup$
    – mmesser314
    Sep 15, 2020 at 14:40
  • $\begingroup$ Like the idea of taking the limit $\to 9$. +1 from me. $\endgroup$
    – Gert
    Sep 15, 2020 at 14:44
  • $\begingroup$ @Buraian: See my follow-up in the answer. $\endgroup$ Sep 15, 2020 at 15:13

If the pulley is massless then $I=0$ and so $T_1=T_2$ - in other words the tension is the same on each side of the pulley. This makes sense because there should be no net torque on a massless object (otherwise we would get an infinite angular acceleration, which is unrealistic).

So if $I=0$ we have

$W_1-W_2 = (M_1+M_2)a \\ \displaystyle \Rightarrow a = \frac {W_1-W_2}{M_1+M_2} = \frac {M_1-M_2}{M_1+M_2}g$

  • $\begingroup$ I'm not sure why infinite / large angular acceleration is deemed unrealisitc. Imagine applying some torque on a light rod, it'd do some really quick rotations $\endgroup$ Sep 15, 2020 at 14:42
  • $\begingroup$ @Buraian "infinite" does not mean the same as "very large". Very large angular accelerations are realistic (ignoring relativistic effects if they are really large). an "Infinite acceleration" is meaningless, because "infinity" is not a number and therefore doesn't measure the size of anything. $\endgroup$
    – alephzero
    Sep 15, 2020 at 22:43
  • $\begingroup$ Right, I agree with you but that wasn't the point I wanted to convey. What I really wanted to say was about the possibilty of the object still rotating at high speeds. $\endgroup$ Sep 15, 2020 at 22:50
  • $\begingroup$ @Buraian The linear acceleration of the string and hence of the pulley is limited by the inertia of the two masses. The angular acceleration of the pulley is inversely proportional to its radius. So if you had a pulley with a negligible mass and small radius then it would have a very high (but still finite) angular acceleration. $\endgroup$
    – gandalf61
    Sep 16, 2020 at 6:08

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