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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 centre of the hoop is zero. What will be the velociy of the centre of the hoop when it ceases to slip ?

Option 1)
$r\omega _{0}\;$

Option 2)
$\frac{r\omega _{0}}{4}$

Option 3)
$\frac{r\omega _{0}}{3}$

Option 4)
$\frac{r\omega _{0}}{2}$

Answers (1)

best_answer

As we have learned

Law of conservation of angular moment -

$\vec{\tau }= \frac{\vec{dL}}$

- wherein

If net torque is zero

i.e. $\frac{\vec{dL}}= 0$

$\vec{L}= constant$

angular momentum is conserved only when external torque is zero .

When loop ceases to slip $V_{em}= rw ......(1)$

angular momentum is conserved about point of contact P

$L_i = L_f$

$L_i =$ L of c.m + L about C.M

= 0 + Iw = $Mr ^ 2 w_0$

$L_f =$ L of cm + L about C.M

= Mvr + $(Mr^2)w = 2 Mvr (\because V_{cm}=rw )$

$\Rightarrow 2 M V_{cm} r = M r^2 w_0$

$\Rightarrow v_{cm }= \frac{w_0 r }{2}$

Option 1)

$r\omega _{0}\;$

Option 2)

$\frac{r\omega _{0}}{4}$

Option 3)

$\frac{r\omega _{0}}{3}$

Option 4)

$\frac{r\omega _{0}}{2}$

Posted by

SudhirSol

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