Question

An object of radius ${}^{\prime }{R}^{\prime }$ and mass ${}^{\prime }{M}^{\prime }$ is rolling horizontally without slipping with speed ${}^{\prime }{V}^{\prime }$. It then rolls up the hill to a maximum height h$=3v^2/4g$. The moment of inertia of the object is ($g=$ acceleration due to gravity)

• A. $\frac{2}{5}M{R}^2$
• B. $\frac{M{R}^2}{2}$
• C. $M{R}^2$
• D. $\frac{3}{2}M{R}^2$

Answer 1

The correct option is B $\frac{M{R}^2}{2}$

From conservation of total energy, we get that initial total kinetic energy of the object is equal to the final potential energy of the object at maximum height.

Initial kinetic energy of the object $K.E_i=\frac{1}{2}M{v}^2+\frac{1}{2}I{w}^2$

As the object is rolling on the surface, thus we get $v=Rw$

$⟹$ $K.E_i=\frac{1}{2}M{v}^2+\frac{1}{2}I\frac{v^2}{R^2}$

Final potential energy of the object $P.E_f=Mgh=\frac{3}{4}M{v}^2$

$\therefore$ $\frac{1}{2}M{v}^2+\frac{1}{2}I\frac{v^2}{R^2}=\frac{3}{4}M{v}^2$

Or $\frac{1}{2}I\frac{v^2}{R^2}=\frac{1}{4}M{v}^2$

$⟹$ $I=\frac{1}{2}M{R}^2$