Question

(6) A body is rolling horizontally without slipping with speed v. It then rolls up a hill to
a maximum height h. If h = 3v2/4g, what might the body be?
Applicati925)​

Answer 1

The body can be a uniform disc or a solid cylinder

Explanation:

If the angular velocity of the body is $\omega$, its moment of inertia $I$ and mass $M$

Then

The initial Kinetic Energy of the body

$K=\frac{1}{2}Mv^2+\frac{1}{2}I\omega^2$

But

$v=R\omega$      

$\implies \omega=\frac{v}{R}$

Therefore,

$K=\frac{1}{2}Mv^2+\frac{1}{2}I(\frac{v}{R})^2$

$\implies K=\frac{1}{2}Mv^2+\frac{1}{2}I\frac{v^2}{R^2}$

With this kinetic energy, the body goes to height $=\frac{3v^2}{4g}$

Therefore,

$K=Mgh$

$\frac{1}{2}Mv^2+\frac{1}{2}I\frac{v^2}{R^2}=Mg\frac{3v^2}{4g}$

$\implies M+\frac{I}{R^2}=\frac{3}{2}M$

$\implies\frac{I}{R^2}=\frac{3}{2}M-M$

$\implies \frac{I}{R^2}=\frac{1}{2}M$

$\implies I=\frac{1}{2}MR^2$

The body can be a disc or a solid cylinder whose mass is M and radius is R

Hope this answer is helpful.

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