System of Particles and Rotational Motion - Result Question 71
78. A disc of radius $2 m$ and mass $100 kg$ rolls on a horizontal floor. Its centre of mass has speed of $20 cm / s$. How much work is needed to stop it ?
[2019]
(a) $3 J$
(b) $30 kJ$
(c) $2 J$
(d) $1 J$
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Answer:
Correct Answer: 78. (a)
Solution:
- (a) Work done to stop the disc $=$ change in total kinetic energy of disc
Final $KE=0$
Initial $KE=$ Translational K.E. + Rotational K.E.
$=\frac{1}{2} mv^{2}+\frac{1}{2} 1 \omega^{2}$
$=\frac{1}{2} mv^{2}+\frac{1}{2} \times \frac{mR^{2}}{2} \times(\frac{v}{R})^{2}$
$=\frac{1}{2} mv^{2}+\frac{1}{4} mv^{2}=\frac{3}{4} mv^{2}$
$=\frac{3}{4} \times 100 \times(20 \times 10^{-2})^{2}=3 J$
$|\Delta KE|=3 J$
