JEE MAIN - Physics (2013 (Offline) - No. 25)
A hoop of radius $$r$$ and mass $$m$$ rotating with an angular velocity $${\omega _0}$$ is placed on a rough horizontal surface. The initial velocity of the center of the hoop is zero. What will be the velocity of the center of the hoop when it cases to slip?
$${{r{\omega _0}} \over 4}$$
$${{r{\omega _0}} \over 3}$$
$${{r{\omega _0}} \over 2}$$
$${r{\omega _0}}$$
Explanation
_en_25_1.png)
From conservation of angular momentum at point of contact,
$$m{r^2}{\omega _0} = mvr + m{r^2}\omega $$
$$m{r^2}{\omega _0} = mvr + m{r^2}\left( {{v \over r}} \right)$$ [ as $$v = r\omega $$ ]
$$m{r^2}{\omega _0} = mvr + mvr$$
$$m{r^2}{\omega _0} = 2mvr$$
$$ v = {{{\omega _0}r} \over 2}$$
Comments (0)
