Question

A hoop of radius $2m$ weighs $100kg$. It rolls along a horizontal floor so that its centre of mass has a speed of $20cm{s}^{-1}$ How much work has to be done to stop it?

• A. $2J$
• B. $4J$
• C. $6J$
• D. $8J$

Answer 1

Answer: (B)

Total energy of the hoop=Translational Kinetic energy+ Rotational Kinetic energy $=\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2$
For hoop, I $=m{r}^2$
Also $v=r\omega$
Thus total energy $=\frac{1}{2}m{v}^2+\frac{1}{2}m{r}^2{\left(\frac{v}{r}\right)}^2=m{v}^2$
Hence the work required to stop the hoop $=$ Its total energy $=m{v}^2=100\times$ $0.2^{2}J=4J$