A small particle of mass 'm' is attached at B to a hoop of mass m and radius r, whole system is placed on the rough horizontal ground. The system is released from rest when B is directly above A and rolls without slipping. Angular acceleration of system when AB becomes horizontal to the ground is? In the solution to this problem, they have equated the potential and kinetic energies, differentiated the equation and have got angular acceleration. Why can't I just write torque on the system and then find angular acceleration from it? (Answer does not come out to be the same) Here's what I did

  • $2mg\frac{R}{2}=(mR^2+mR^2)+[m(√2R)^2]$

  • Angular acceleration=$\frac{g}{4R}$

  • Equating torque on the COM of the system to the moment of inertia at the point of contact of the hoop and the ground (Instantaneous axis of rotation)

  • Here's what the book does (x is the angle between the vertical and line joining center of hoop with the position of B): $$mgR(1-\cos x)=\frac{1}{2}(2mR^2)w^2+\frac{1}{2}[2R\cos(x/2)w]^2$$

  • Differentiating and putting $x=\frac{\pi}{2}$

  • Angular acceleration = $\frac{3g}{8R}$

  • $\begingroup$ Can you show their calculation and yours? $\endgroup$ Oct 31, 2019 at 8:23
  • $\begingroup$ @MarcoOcram sure just a second $\endgroup$
    – John Smith
    Oct 31, 2019 at 8:24
  • $\begingroup$ @MarcoOcram the website isn't letting me submit the pictures $\endgroup$
    – John Smith
    Oct 31, 2019 at 8:35
  • $\begingroup$ Perhaps you could just show you calculation, and say how your answer differed from the other. $\endgroup$ Oct 31, 2019 at 8:37
  • $\begingroup$ @MarcoOcram I have updated the answer with the calculations $\endgroup$
    – John Smith
    Oct 31, 2019 at 8:46

1 Answer 1


The angular momentum (as you calculated) with respect to the instantaneous point of contact $P$ of the ring with the ground, at the instant when the line $AB$ is horizontal, is ${\vec{L}_P}_{\text{|when the line AB is horizontal}}=-4mR^2\omega \hat{k}$, where $\hat{k}$ is the unit vector pointing out of the paper if the ring is rolling to the right without slipping (i.e., rotation is clockwise).

But the rate of change of angular momentum about the instantaneous point of contact $P$ when the line AB is horizontal, is tricky.

$$\frac{d\vec{L}_P}{dt}_{\text{|when the line AB is horizontal}} \neq -4mR^2 \alpha \hat{k}$$

This is because the bead-particle $m$ attached to the ring, has two components of acceleration. (In the calculation below $\hat{r}$ and $\hat{\theta}$ are the polar coordinate unit vectors with the origin located at the center of the ring)

$$\frac{d\vec{v}_m}{dt} = \frac{d(\vec{v}_{\text{of the ring's frame of reference}} + R \omega \hat{\theta})}{dt} = \frac{d(R \omega \hat{i} + R \omega \hat{\theta})}{dt} = R \alpha \hat{i} + R \alpha \hat{\theta} - R \omega^2 \hat{r}$$

Therefore, when you calculate the $\frac{d\vec{L}_P}{dt}_{\text{|when the line AB is horizontal}}$, you get the following,

$$\frac{d\vec{L}_P}{dt}_{\text{|when the line AB is horizontal}}= (4mR^2 \alpha - mR^2 \omega^2)(-\hat{k}) = \vec{\tau}_P = -mgR \hat{k}$$

I hope this clears your confusion. I think you should be able to do the rest of the problem.

  • $\begingroup$ This answer turned out to be useful even two years on ! The power of stackexchange $\endgroup$ Jun 27, 2021 at 10:18
  • $\begingroup$ Few remarks for easier read for future people: Ajay has written torque equation about bottom most point P, in the differentiation for velocity, the whole quantity $\frac{d}{dt}( \omega \hat{\theta})$ was derived (omega changes and also unit vector changes, the part about unit vector changing can be found in kleppner and kolenkow). $\endgroup$ Jun 27, 2021 at 10:32
  • $\begingroup$ @Buraian, a small clarification required, the formula stated in the OP, it considers the rotational kinetic energy of both the hoop and the particle and the next term includes the translational kinetic energy of the particle, so why didn't we add the translational kinetic energy of hoop or am I mistaken somewhere? $\endgroup$
    – UNAN
    Jun 29, 2021 at 15:13
  • $\begingroup$ Hi! Take note that OP has written the energy about the point of contact directly, if you expand it out, you'll find the rotation energy term about com and translation energy about com contained in it @Community $\endgroup$ Jun 29, 2021 at 15:19
  • $\begingroup$ $2MR^2 = (MR^2) + (MR^2)$ the first is inertia of hoop, second the parallel axis theorem. Inertia of hoop is the rotation and parallel axis corresponds to ktranslational energy. For more details , consult Resnick Halliday, they show a proof where the term coming from parallel axis theorem is identified with translation KE @Community $\endgroup$ Jun 29, 2021 at 15:20

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