Find r/R to perform SHM

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A small object is mounted to the perimeter of a hoop of radius r . The mass of the object and the hoop is same . The hoop is placed into a fixed semi-cylinder shape rough through of radius R , such that the small is at the top . Find the least r/R ratio such that the object performs SHM . My try : I am trying to use energy method .
But in this how to find the value of h .

asked May 5, 2018
edited May 6, 2018
You have got the right idea of trying to find $h(\theta)$.

It is not obvious what solution the question is looking for. The easiest solution is to note that for SHM the PE increases with displacement from the equilibrium position. An increase in PE means that there will be a restoring force $F=-dV/d\theta$ where $V=mgh$. You could also verify that the restoring force is proportional to displacement.

If $R\gg r$ you can see that PE will decrease for displacements either side of equilibrium. You have to choose $r/R$ such that PE will increase  for small displacements and be proportional to displacement. You can decide this from the geometry of the situation rather than the dynamics. So you do not need to consider KE. You only need to consider the potential function $V(\theta)$.

Another possible condition is that the ring must not slip as it rolls. But this depends on coefficient of friction which you have not been given. Does the source of the question give you an answer?

For SHM we require that the restoring force (or torque) should be directed toward the equilibrium position and be proportional to the displacement $\theta$.

The height of the centre O of the ring, taking its lowest point at $\theta=0$ as origin, is
$h(\theta)=(R-r)(1-\cos\theta)$.

When the centre O of the ring moves anticlockwise through angle $\theta$ from the vertical, the point of contact P between the ring and cylinder moves a distance $s=R\theta$ to the right. Relative to the line COP joining the centre of the cylinder C to point of contact P, the ring turns clockwise through an angle $\phi=s/r=(R/r)\theta$. Relative to the vertical the ring turns clockwise through angle
$\phi-\theta=(R/r-1)\theta=k\theta$
So the height of the bead above the origin is
$H(\theta)=h+r\cos(k\theta)$

The ring and bead each have mass $m$, so the potential energy of the system (ring plus bead) is
$V(\theta)= mgh+mgH=mg[2(R-r)(1-\cos\theta)+r\cos(k\theta)]$.

Without loss of generality we can take $mg=1$. Then the restoring torque is
$\tau=-\frac{dV}{d\theta}= -[2(R-r)\sin\theta-kr\sin(k\theta)] \approx -[2(R-r)-rk^2]\theta$
for small angles $\theta$.

This shows that, for small angles, the restoring force $\tau$ is approximately proportional to displacement $\theta$, as required. We also require that $\tau$ should be -ve when $\theta$ is +ve, ie that potential energy should increase away from the equilibrium position : $\frac{dV}{d\theta}\gt 0$.

Using $R/r=\rho$ we require :
$2(\rho - 1) - k^2 = 2\rho -2 - (\rho^2-2\rho+1) > 0$
$0 > 3 - 4\rho + \rho^2 =(1-\rho)(3-\rho)$
ie
$1 < \rho=R/r < 3$

The least ratio for SHM is $r/R=1/3$. (Note that physically we cannot have $r/R>1$ anyway.)

Note : It might also be possible for the ring to perform SHM if it starts in other positions, ie not at the lowest point of the cylinder where $\theta=0$. See Equilibrium and movement of a cylinder with asymmetric mass centre on an inclined plane for an example in which an object oscillates on an inclined plane. (However, I did not prove that those oscillations are SHM.)

answered May 6, 2018 by (28,806 points)
edited May 6, 2018