The problem is presented in the following diagram

enter image description here

I'm refreshing things I've already learnt and I know I have some major gaps but I've searched and haven't managed to find a similar problem. The axis of rotation in most of them is at the center of the pulley but here I think it goes through point A in the diagram. Point A seems to be an instantaneous center of rotation. I assume this means that the axis of rotation goes through it. I could be wrong.

This is a problem I've found in some notes and the solution on this problem confuses me a lot. Let's get into the given solution.

We move the object a distance $x$ which translates to a $x/2$ for the spring. The author seems to take the sum of torques with respect to point A. The given moment of inertia was $I=1/2MR^2$. Going back to high school I think this stands if the center of rotation is at the center of the pulley. Here, since the center of rotation is at A then the moment of inertia is $I_A=1/2MR^2 + MR^2 $.

Finally, this is the equation used by the author where a is the angular acceleration : $$ΣΤ=Ι'a=>\\T2R - k(x/2)R=3/2MR^2 \frac{\ddot x}{R} $$

Since we are taking the torques with respect to A, shouldn't we also have a torque coming from the mass of the pulley? Why doesn't it show up in the equation?

Update: Moving on, for the object we have: $$ΣF=m\ddot x\\T=-m\ddot x$$ Plugging this in torque equation we can reach the final equation : $$-m\ddot x (2R) - k(x/2)R=3/2MR \ddot x \\=>(3M/2 +2m)\ddot x + (k/2)x=0$$

Which is the standard form of an SHM. Now back to my concern. If we had included the weight's torque in the torque equation the new term wouldn't lead to the standard form of an SHM. Adding the weight torque, the torque equation with respect to A would be: $$ΣΤ=Ι'a=>\\T2R - k(x/2)R + Mgr=3/2MR^2 \frac{\ddot x}{R} $$

Should I or should I not include the weight torque? Why can't I reach a standard form if I include the weight?

  • 1
    $\begingroup$ It seems like you are right about everything. The moment of inertia is not $\frac 12 MR^2$ about point A, and the pulley needs to have mass (so that $I$ is not $0$), so its weight does have a non-zero torque about point $A$. $\endgroup$ Aug 23, 2018 at 11:21
  • $\begingroup$ @AaronStevens Is it possible to take the sum of torques with respect to another point? Will that affect the moment of inertia in the new equation? $\endgroup$ Aug 23, 2018 at 11:25
  • 2
    $\begingroup$ At equilibrium, the all downward forces are cancelled out by the spring force. Note that the spring force and the weight of the pulley act at same point along opposite direction. So, the force and torque due to gravity is always cancelled by a part of the force and torque due to sprint. Generally, in these type of spin block systems, gravity only changes the equilibrium position, it does not affect the frequency $\endgroup$ Aug 23, 2018 at 11:32
  • 1
    $\begingroup$ Ok so you are looking at the pulley oscillating. If you are interested in the angular acceleration about point A, then you have to use point A as your axis. Therefore your moment of inertia and torques must be about point A. If you were just wondering about equilibrium, then you could choose a different point however. $\endgroup$ Aug 23, 2018 at 11:37
  • 1
    $\begingroup$ @ArchismanPanigrahi did provide you with the solution. Stated in other words, you need to examine your definition of $x$. Most likely, this is deviation from equilibrium point, not from relaxation point of the spring. You can add $mgr$ as a torque if you like, but than you must define $x$ from relaxation point. $\endgroup$
    – npojo
    Aug 23, 2018 at 16:06

1 Answer 1


To answer your question in short, yes, you have to consider the torque of gravity.

Suppose, the acceleration of the block (mass $m$) is $a$ (upward) and that of the pulley(mass $M$) is $a/2$. Suppose, you displace the block by $x$, so the elongation of spring is $x/2$. Let the tension in left string be $T$ and in the right string be $S$.

Thus, $T = mg+ ma$. (Newton's 2nd law on the block).

$kx/2 - T -S - Mg = Ma/2$ (2nd law on pulley)

And, the torque equation is,

$kxR/2 - MgR - 2TR = \frac{3}{2} MR^2 \frac{a}{2R}$ ,

where the angular acceleration of pulley about point A is $\frac{a}{2R}$ (why?). (Note: it would be much easier to take torques about centre of mass of pulley, but since you used point A, I too did it)

Eliminating $T,S$ we get,

$a (8m + 3M) = 2kx - 2Mg - 4mg$

where $ a = -\frac{d^2 x}{dt^2}$.

The time period of the oscillation is $2 \pi \sqrt {\frac{8m+3M}{2k}}$, which is independent of $g$. Note: If you had calculated displacements from equilibrium, then the terms $2Mg - 4mg$ would have been absorbed in it. We can write $2Mg - 4mg =2kx_0$, where $x_0$ is the equilibrium elongation.

  • $\begingroup$ The textbook I'm studying completely ignored gravity forces without mentioning why. Everything is clear with your answer Archisman, thank you. $\endgroup$ Aug 23, 2018 at 18:10
  • $\begingroup$ @JohnKatsantas Which textbook? $\endgroup$ Aug 23, 2018 at 18:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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