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Q.
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?
NEETNEET 2019System of Particles and Rotational Motion
Solution:
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 $=m v^{2}=100 \times$ $0.2^{2} J =4 J$