Conceptual problem in rotation mechanics [closed]

Question

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 = $\frac{3g}{8R}$

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.