A hoop with radius r and mass m is rotating with an initial angular velocity omega_0 and is placed on a rough horizontal floor. If the initial linear velocity of its center is zero, what will be the final linear velocity of the center when the hoop stops slipping and begins pure rolling?

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EEJ MAIN Physics QUESTION #7016

Question 1

A hoop with radius $r$ and mass $m$ is rotating with an initial angular velocity $\omega_0$ and is placed on a rough horizontal floor. If the initial linear velocity of its center is zero, what will be the final linear velocity of the center when the hoop stops slipping and begins pure rolling?

$\frac{r\omega_0}{4}$
$\frac{r\omega_0}{3}$
$\frac{r\omega_0}{2}$✔️
$r\omega_0$

Correct Answer Explanation

Using the principle of conservation of angular momentum about the point of contact: $L_i = L_f$. Initial angular momentum $L_i = I \omega_0 = (mr^2) \omega_0$. Final angular momentum $L_f = I \omega + mrv = (mr^2)(\frac{v}{r}) + mrv = 2mrv$. Equating $mr^2\omega_0 = 2mrv$ gives $v = \frac{r\omega_0}{2}$.

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