# Onset of bouncing for rolling hoop with off-centre mass (Irodov ex 1.265)

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A small body $A$ is fixed to the inside of a thin rigid hoop of radius $R$ and the mass equal to that of the body $A$. The hoop rolls without slipping over a horizontal plane, at the moments when the body $A$ gets into lower position, the center of the hoop moves with velocity $v_0$ At what values of $v_0$ will the hoop move without bouncing.

It's very intuitive that bouncing can happen only at the top point.

Let there be normal $N$ at the top between $A$ and hoop and hoop rotating with $\omega$, for centripetal force and condition that $N$ lifts up $mg$ of hoop$$N+mg=m\omega^2R\\ N=mg\\ \implies 2mg=m \omega^2R\implies \omega^2=\dfrac{2g}{R}$$

Now I need to conserve energy, the best option for most cases is from CM but conserving it here from CM makes it tedious, so from where and how should I conserve it to get it easily.

edited Oct 20, 2018
I have altered your title to make it more descriptive of the problem.

There were some errors in my previous answer.

It is not quite true that bouncing can happen only when particle A reaches the top of the hoop. I think what you mean is that bouncing occurs for the lowest value of $v_0$ when particle A is at the top of the hoop. At faster speeds it can occur before A reaches the top.

Your calculation so far is correct. Bouncing starts when the upward centrifugal force caused by the rotation of particle A around its average position exceeds the combined weight of the hoop and particle A. When the velocity of the centre C of the hoop is $v$ the velocity of A relative to C is also $v$. Therefore bouncing starts when $$m\frac{v^2}{R} \gt 2mg$$ $$v^2 \gt 2gR$$ Bouncing can occur before A reaches its highest position if the vertical upward component of the centrifugal force exceeds $2mg$.

As you realise, we have to use conservation of energy to find $v$, the seed of the hoop when A reaches its highest position.

When A is at its lowest point it is stationary and has no KE. The COM of the hoop is moving with velocity $v_0$ so the hoop has translational KE $\frac12 mv_0^2$. The hoop also has rotational KE. In the COM frame every point on the hoop is moving at speed $v_0$ so the hoop has rotational KE of $\frac12 mv_0^2$. The total energy when A is at its lowest point is therefore $mv_0^2$.

When A reaches its highest point the centre of the hoop has some unknown speed $v$ and particle A has speed $2v$. The translational KE of the hoop and particle A are $\frac12 mv^2$ and $2mv^2$ respectively. The hoop also has rotational KE of $\frac12 mv^2$ as calculated above. The total KE in this position is $3mv^2$. The gravitational PE of A has increased by $2mgR$.

Total mechanical energy is the same in both positions. Applying the inequality above we get $$mv_0^2=3mv^2+2mgR$$ $$v_0^2-2gR = 3v^2 \gt 3(2gR) = 6gR$$ $$v_0^2 \gt 8gR$$