Question

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

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

Answer 1

Answer: (B)

$4\,J$

Answer 2

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