Question

A hoop of mass $m$ is projected on a floor with linear velocity $v_0$ and reverse spin ${\omega }_0$. The coefficient of friction between the hoop and the ground is $\mu$.
a. Under what condition will the hoop return back?
b. How far will it go?
c. How long will it continue to slip when its centre of mass becomes stationary?
d. What is the velocity of return?

• A. a. ${\omega }_0>\frac{v_0}{R}$.
  b. $\frac{v_0^2}{R}$,
  c. $\frac{R}{3{\mu }_g}({\omega }_0-\frac{v_0}{R})$,
  d. $\frac{{\omega }_{0}R}{2}-\frac{v_0}{2}$.
• B. a. ${\omega }_0>\frac{v_0}{R}$.
  b. $\frac{v_0^3}{R}$,
  c. $\frac{R}{2{\mu }_g}({\omega }_0-\frac{v_0}{R})$,
  d. $\frac{{\omega }_{0}R}{4}-\frac{v_0}{2}$.
• C. a. ${\omega }_0>\frac{v_0}{R}$.
  b. $\frac{v_0^2}{R}$,
  c. $\frac{R}{2{\mu }_g}({\omega }_0-\frac{v_0}{R})$,
  d. $\frac{{\omega }_{0}R}{3}-\frac{v_0}{2}$.
• D. a. ${\omega }_0>\frac{v_0}{R}$.
  b. $\frac{v_0^2}{R}$,
  c. $\frac{R}{2{\mu }_g}({\omega }_0-\frac{v_0}{R})$,
  d. $\frac{{\omega }_{0}R}{2}-\frac{v_0}{2}$.

Answer 1

The correct option is D a. ${\omega }_0>\frac{v_0}{R}$.

b. $\frac{v_0^2}{R}$,
c. $\frac{R}{2{\mu }_g}({\omega }_0-\frac{v_0}{R})$,
d. $\frac{{\omega }_{0}R}{2}-\frac{v_0}{2}$.