Rotational Motion Question 2
Question 2 - 2024 (01 Feb Shift 2)
A disc of radius $R$ and mass $M$ is rolling horizontally without slipping with speed $v$. It then moves up an inclined smooth surface as shown in figure. The maximum height that the disc can go up the incline is :
(1) $\frac{v^{2}}{g}$
(2) $\frac{3}{4} \frac{v^{2}}{g}$
(3) $\frac{1}{2} \frac{v^{2}}{g}$
(4) $\frac{2}{3} \frac{v^{2}}{g}$
Show Answer
Answer: (3)
Solution:
Only the translational kinetic energy of disc changes into gravitational potential energy. And rotational KE remains unchanged as there is no friction.
$\frac{1}{2} m v^{2}=m g h$
$h=\frac{v^{2}}{2 g}$
