A hoop of radius r and mass m rotating with an angular velocity ω_0 is placed on a rough horizontal surface. The initial velocity of the centre of the - Physics MCQ JEE Main 2026 | ExamTest.live

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A hoop of radius r and mass m rotating with an angular velocity ω_{0} 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?

Solution & Step-by-step Explanation

Use conservation of angular momentum about a point on the ground.Initially: L_{i} = I_{c m} ω_{0} = (m r^{2}) ω_{0} .Finally (rolling without slipping): v = ω r . L_{f} = m v r + I_{c m} ω = m (r ω) r + m r^{2} ω = 2 m r^{2} ω . L_{i} = L_{f} \Rightarrow m r^{2} ω_{0} = 2 m r^{2} ω \Rightarrow ω = \frac{ω _{0}}{2} .Velocity of center v = ω r = \frac{r ω _{0}}{2} .

Try it yourself before checking the explanation above.

A hoop of radius r and mass m rotating with an angular velocity ω_{0} 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?

\frac{r\omega_0}{3}

\frac{r\omega_0}{2}

r\omega_0

\frac{r\omega_0}{4}

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