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 whill be the velocity of the centre of the hoop when it ceases to slip ? [JEE main 2013]​

Answer 1

Given:

Radius of hoop = r

Mass of hoop = m

Initial angular velocity = ω₀

Initial linear velocity of center of hoop = 0

To Find:

Final linear velocity of center of the hoop when it creases to slip

Answer:

Let final angular velocity be 'ω' and final linear velocity of center of the hoop be 'v'.

By conservation of angular momentum about point of contact with the surface we have;

$\rm I_{cm} \omega_0 = mvr + I_{cm} \omega$

As slipping has stopped. So, v = rω

For hoop (ring) $\sf I_{cm}$ = mr²

So,

$\rm \implies m {r}^{2} \omega_0 = mvr + m {r}^{2} \omega \\ \\ \rm \implies m {r}^{2} \omega_0 = mvr + m (r \omega)r \\ \\ \rm \implies m {r}^{2} \omega_0 = mvr + m vr \\ \\ \rm \implies \cancel{m }{r}^{ \cancel{2}} \omega_0 = 2 \cancel{m}v \cancel{r} \\ \\ \rm \implies 2v = r \omega_0 \\ \\ \rm \implies v = \dfrac{r\omega_0}{2}$

$\therefore$ Final linear velocity of center of the hoop when it creases to slip (v) = $\rm \dfrac{r \omega_0}{2}$

Answer 2

• Explanation :- From the principe of conservation of angular momentum , se have

⟹ ΣIW = constant

Where , I is moment of inertia of hoop = (I = mr² )

• By conservation of law of angular of mementum (⃗L)

⟹ ⃗L = ⃗Lcm + Mr₀⃗× ⃗v₀

• ∴ mr²w₀ = mvr + mr²× v/r
• ⟹mr²w₀= mr( v+v)
• ⟹ v = w₀r/2 Answer

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