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:

  1. (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.

$=\frac{1}{2} mv^{2}+\frac{1}{2} 1 \omega^{2}$

$=\frac{1}{2} mv^{2}+\frac{1}{2} \times \frac{mR^{2}}{2} \times(\frac{v}{R})^{2}$

$=\frac{1}{2} mv^{2}+\frac{1}{4} mv^{2}=\frac{3}{4} mv^{2}$

$=\frac{3}{4} \times 100 \times(20 \times 10^{-2})^{2}=3 J$

$|\Delta KE|=3 J$