Question

A small hoop of mass $m$ is given an initial velocity of magnitude $v_0$ on the horizontal circular ring of radius `$r$'. If the coefficient of kinetic friction is ${\mu }_k$ the tangential acceleration of the hoop immediately after its release is (assume that horizontal ring to be fixed and not in contact with any supporting surface)

• A. ${\mu }_{k}g$
• B. ${\mu }_k\frac{v_0^2}{r}$
• C. ${\mu }_k\sqrt{g^2+\frac{v_0^2}{r}}$
• D. ${\mu }_k\sqrt{g^2+\frac{v_0^4}{r^2}}$

Answer 1

The correct option is D ${\mu }_k\sqrt{g^2+\frac{v_0^4}{r^2}}$

Force acting at the center of hoop are :

• Normal force $N$ upwards
• $mg$ downwards.

Centripetal acceleration, $a_c=\frac{v_0^2}{r}$

Frictional force, $F={\mu }_{k}N$

$N=\sqrt{m^2{g}^2+\frac{m^2{v}_0^4}{r^2}}$

Tangential acceleration, $a_t=\frac{F}{m}$

$⟹a_t={\mu }_k\sqrt{g^2+\frac{v_0^4}{r^2}}$