Question

The velocity of centre of mass of a disc rolling on an inclined plane changes from $V$ to $2 V$ at any instant of time. If $M$ is the mass of disc then increases in its kinetic energy will be

• A. $(9 / 4) MV ^{2}$
• B. $M V^{2} / 2$
• C. MV $^{2}$
• D. $3 M V^{2}$

Answer 1

Answer: (A)

Total $KE =\frac{1}{2} MV ^{2}+\frac{1}{2} I \omega^{2}=\frac{3}{4} MV ^{2}$
$\Delta K E=\frac{3}{4} M\left(4 V^{2}-V^{2}\right)=\frac{9}{4} M V^{2}$