Question

A small body $A$ is fixed to the inside of a thin rigid hoop of radius $R$ and 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 the lower position, the centre of the hoop moves with velocity $v_0$ (figure shown above). If the hoop moves without bouncing for $v_0=\sqrt{xgR}$, then find the value of $x$.

Answer 1

Clearly the tendency of bouncing of the hoop will be maximum when the small body $A$, will be at the highest point of the hoop during its rolling motion. Let the velocity of C.M. of the hoop equal $v$ at thisposition. The static friction does no work on the hoop, so from conservation of mechanical energy; $E_1=E_2$
or, $0+\frac{1}{2}m{v}_0^2+\frac{1}{2}m{R}^2{\left(\frac{v_0}{R}\right)}^2-mgR=\frac{1}{2}m{\left(2v\right)}^2+\frac{1}{2}m{v}^2+\frac{1}{2}m{R}^2{\left(\frac{v}{R}\right)}^2+mgR$
or, $3v^2=v_0^2-2gR$ (1)
From the equation $F_n=m{w}_n$ for body $A$ at final position 2,
$mg+N^{\prime }=m{\omega }^{2}R=m{\left(\frac{v}{R}\right)}^{2}R$ (2)
As the hoop has no acceleration in vertical direction, so for the hoop,
$N+N^{\prime }=mg$ (3)
From equations (2) and (3),
$N=2mg-\frac{m{v}^2}{R}$ (4)
As the hoop does not bounce, $N\ge 0$ (5)
So from equations (1), (4) and (5),
$\frac{8gR-v_0^2}{3R}\ge 0$ or $8gR\ge v_0^2$
Hence $v_0\le \sqrt{8gR}$