Question

A hoop of radius $r$ weights $m\, kg$. It rolls without sliding along a horizontal floor so that its centre of mass has a speed $v \,m / s$. How much work has to be done to stop it?

• A. $\frac{1}{2} m v^{2}$
• B. $m v^{2}$
• C. $2 m v^{2}$
• D. $\frac{1}{3} m v^{2}$

Answer 1

Answer: (B)

Total kinetic energy of hoop is
$K=\frac{1}{2} I \omega^{2}+\frac{1}{2} m v^{2}$
$\Rightarrow K=\frac{1}{2} m r^{2} \cdot \frac{v^{2}}{r^{2}}+\frac{1}{2} m v^{2}$
$\Rightarrow K=\frac{1}{2} m v^{2}+\frac{1}{2} m v^{2}$
$\Rightarrow K=m v^{2}$
Thus, a work of $m v^{2}$ has to be done to stop it.