Question

A small mass m is attached to the inside of a rigid ring of the same mass m and radius R. The ring rolls without slipping over a horizontal plane. At the moment when the mass m gets into the lowest position, the centre of the ring moves with velocity $v_0$. lf $(v_0{)}_{max}=\sqrt{ngR}$ , when the ring moves without bouncing, find n.

Answer 1

According to the concept of conservation of energy and vertical circular motion, we will consider that when the ring rolls without slipping then, conservation of energy at the highest and the lowest point (total energy = translation energy + rotational energy).
Now at the lowest point, total energy = $\frac{1}{2}\left(m{v}_0^2\right)+\frac{1}{2}\left(2m{R}^2{\omega }_0^2\right)$ (Total $I=2m{R}^2$)
At the highest point, total energy = $\frac{1}{2}\left(m{v}^{\prime 2}\right)+\frac{1}{2}\left(2m{R}^2{\omega }^{\prime 2}\right)+\frac{1}{2}m(v^{\prime }+R{\omega }^{\prime }{)}^2+2mgR$
Now as $v=R\omega ,v^{\prime }=R{\omega }^{\prime }$
$\omega =\frac{v}{R},{\omega }^{{}^{\prime }}=\frac{v^{\prime }}{R}$.
After substituting these values in the energy equation,
we will get,
$\frac{1}{2}m{v}_0^2+m{v}_0^2=\frac{1}{2}\left(m{v}^{\prime 2}\right)+\frac{1}{2}\left(2m{v}^{\prime 2}\right)+\frac{1}{2}\left(m4v^{\prime 2}\right)+2mgR$
$\frac{3}{2}\left(m{v}_0^2\right)=\frac{1}{2}\left(m{v}^{\prime 2}\right)+3m{v}^{\prime 2}+2mgR$
$\frac{3}{2}\left(v_0^2\right)=\frac{7}{2}\left(v^{\prime 2}\right)+2gR$
For substituiting the value of v', we will solve the equation
$2mg-N=m\frac{v^{\prime 2}}{R}$
N = 0,
$2mg=m\frac{v^{\prime 2}}{R}$
$2gR=v^{\prime 2}$
$v^{\prime }=\sqrt{2gR}$
Now we will substitute this value of "v'" in the main equation, therefore
$\frac{3}{2}\left(v_0^2\right)=\frac{7}{2}\left(2gR\right)+2gR$
$\frac{3}{2}\left(v_0^2\right)=9gR$
$v_0=\sqrt{6gR}$
Hence, $n=6$