Question 350: Centre of Mass and Rotational Motion

Question

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?

Accepted Answer

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 $$v = r\omega $$ ]
$$m{r^2}{\omega _0} = mvr + mvr$$
$$m{r^2}{\omega _0} = 2mvr$$
$$ v = {{{\omega _0}r} \over 2}$$

Answer

$${{r{\omega _0}} \over 4}$$

Answer

$${{r{\omega _0}} \over 3}$$

Answer

$${{r{\omega _0}} \over 2}$$

Answer

$${r{\omega _0}}$$

Centre of Mass and Rotational Motion

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?

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