Rotational Motion Question 30
Question: A hoop of radius r and mass m rotating with an angular velocity ω0 is placed on a rough horizontal surface. The initial velocity of the centre of the hoop is zero. What will be the velocity of the centre of the hoop when it ceases to slip?
[JEE MAIN 2013]
Options:
A) $ \frac{r{\omega_0}}{4} $
B) $ \frac{r{\omega_0}}{3} $
C) $ \frac{r{\omega_0}}{2} $
D) $ r{\omega_0} $
Show Answer
Answer:
Correct Answer: C
Solution:
- [c]
$\frac{ω}{2}$
Radius of hoop = r, Initial angular velocity =ω0,
Initial velocity (u) = 0,
velocity of centre of hoop = ?
By momentum conservation, initial momentum of hoop will be same as final momentum
i.e., $L_i=L_f$
$mr^2ω0 = mvr + \frac{mr^2v}{r}$ rω0 = v+v v=$\frac{rω0}{2}$
