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Let $\alpha_1$ and $\alpha_2$ be the angular acceleration of top pulley and bottom pulley respectively while $a$ is the acceleration of centre of mass of bottom pulley.

Then, $$\alpha_2r+a=\alpha_1r$$ $$\alpha_2r-a=0$$

I understand the maths but I am not able to make sense of these equations logically.

Also, while writing these equations what exactly are we assuming as our system? A point on the pulley? A part of the rope?


2 Answers 2


The speed of the connecting belt must be twice the downward speed of the center of the lower pulley. This means that $α_1 = 2 α_2$. Then: $mg – T – t = ma = mr α_2$ where, $T$, is the connecting tension and, $t$, is the tension on the right side of the lower pulley. Also $Tr = I α_1$ and $tr – Tr = Iα_2$ where , $I = (1/2)mr^2$, (assuming solid cylinders). Solve for, $T$, and, $a$. Then $v^2 = 2ay$ for the final speed.


I believe the first equation should be $\alpha_2r + 2a = \alpha_1r$

In the diagram you have shown, the top pulley (pulley 1) is fixed and can only rotate, while the bottom one (pulley 2) can move. Now, the length of the belt should remain constant, and it being a rough belt indicates that we can neglect slipping at the points of contact.

$\alpha_1r$ is the acceleration of the belt downwards at the point of contact with pulley 1.
When pulley 2 moves down by some distance x, it will cause the belt to move down by 2x. This is because the motion will require x length of the belt to be added to either side of the pulley, which is achieved via the whole belt moving by 2x. The same logic applies to velocities and accelerations.

So, acceleration of belt downwards at point of contact with pulley 2 is $2a$.
But this is only due to the translational motion. If the pulley is only rotating, it would cause the belt to accelerate by $\alpha_2r$ to avoid slipping.
Now, we can superpose both motions. So the acceleration of belt = $$\alpha_2r + 2a = \alpha_1r$$

The second equation comes from the no-slipping condition at pulley 2. To be more specific, look at the rightmost point of contact. The belt there is stationary, so pulley 2 is undergoing pure rolling.

  • $\begingroup$ a is defined as the linear acceleration of bottom pulley. How can it be 2a? $\endgroup$
    – Satya
    May 15 at 14:40
  • $\begingroup$ @Satya a is the acceleration of the bottom pulley, but that doesn't mean the belt also needs to have the same acceleration. As I pointed out in my answer, for the length of belt to be constant, it has to move twice the distance as the pulley. Differentiating, we get that the belt should have twice the velocity and acceleration as the pulley. $\endgroup$
    – Sankalp
    May 15 at 14:53
  • $\begingroup$ @Sankap the length of the belt is not constant. The belt is unwinding continually. AFAIK, the point of contact would have the same acceleration as the belt as there is no slipping. $\endgroup$
    – Satya
    May 15 at 15:04
  • $\begingroup$ I guess I phrased it improperly. What I meant by the length of the belt being constant is that there is no "generation" or "destruction" of the belt in and close to the 2nd pulley. I think the belt being generated at the top of the 1st pulley and unwinding, wouldn't affect the answer. $\endgroup$
    – Sankalp
    May 15 at 15:18
  • $\begingroup$ And as for the acceleration of the belt and point of contact being the same, that is true. In my answer, I divided the motion of the pulley into translational and rotational and recombined them at the end. Considering only translation of the pulley, I agree slipping will occur.. but in the end after recombining I believe the acceleration should be 2a + alpha2_r (i.e. 3a), with no slipping present. $\endgroup$
    – Sankalp
    May 15 at 15:21

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