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A circular ring of mass $10 \mathrm{~kg}$ rolls along a horizontal floor. The center of mass of the ring has a speed $1.5 \mathrm{~m} / \mathrm{s}$. The work required to stop the ring is
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Verified Answer
The correct answer is:
$-22.5 \mathrm{~J}$
Given,
mass of a circular ring, $m=10 \mathrm{~kg}$
speed of centre of mass of the ring
i.e, linear speed of ring, $u=1.5 \mathrm{~m} / \mathrm{s}$
Total initial kinetic energy of rolling ring,
$$
K_i=K_{\text {rotational }}+K_{\text {linear }}=\frac{1}{2} I \omega^2+\frac{1}{2} m v^2
$$
Here, $I=m R^2$ and $\omega=\frac{v}{R}$
$$
\begin{aligned}
& =\frac{1}{2} m R^2\left(\frac{v}{R}\right)^2+\frac{1}{2} m v^2=\frac{1}{2} m v^2+\frac{1}{2} m v^2=m v^2 \\
& =10 \times(1.5)^2=22.5 \mathrm{~J}
\end{aligned}
$$
According to work-energy theorem, work required to stop the ring.
$\omega=$ change in kinetic energy
$$
=K_f-K_i=0-22.5=-22.5 \mathrm{~J}
$$
mass of a circular ring, $m=10 \mathrm{~kg}$
speed of centre of mass of the ring
i.e, linear speed of ring, $u=1.5 \mathrm{~m} / \mathrm{s}$
Total initial kinetic energy of rolling ring,
$$
K_i=K_{\text {rotational }}+K_{\text {linear }}=\frac{1}{2} I \omega^2+\frac{1}{2} m v^2
$$
Here, $I=m R^2$ and $\omega=\frac{v}{R}$
$$
\begin{aligned}
& =\frac{1}{2} m R^2\left(\frac{v}{R}\right)^2+\frac{1}{2} m v^2=\frac{1}{2} m v^2+\frac{1}{2} m v^2=m v^2 \\
& =10 \times(1.5)^2=22.5 \mathrm{~J}
\end{aligned}
$$
According to work-energy theorem, work required to stop the ring.
$\omega=$ change in kinetic energy
$$
=K_f-K_i=0-22.5=-22.5 \mathrm{~J}
$$
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