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Standard XI

Physics

The Principle

A small bead ...

Question

A small bead of mass $m$ moving with velocity $v$ gets threaded on a stationary semicircular ring of mass $m$ and radius $R$ kept on a horizontal table. The ring can freely rotate about its centre. The bead comes to rest relative to the ring. What will be the final angular velocity of the system?

\frac{v}{R}

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\frac{2 v}{R}

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\frac{v}{2 R}

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\frac{3 v}{R}

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Solution

The correct option is C $\frac{v}{2 R}$
Initial momentum $\to L_{i} = m v R$
Final momentum $\to L_{f} = (I_{r} + I_{m}) ω$
Here, no external torque is applied so,
$L_{i} = L_{f} m v R = (m R^{2} + m R^{2}) ω v = 2 R^{2} ω ω = \frac{v}{2 R}$

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A fixed ring of mass m has two beads A, B of mass m each; both being able to slide on the ring without friction. Initially the ring lies on a frictionless horizontal table and the beads are given velocities $ν_{0}$ , relative to the ring, in the same direction. The initial relative acceleration of one of the beads with respect to the other is

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A small bead of mass m moving with velocity v gets threaded on a stationary semicircular ring of mass m and radius R kept on a horizontal table. The ring can freely rotate about its centre. The bead comes to rest relative to the ring. What will be the final angular velocity of the system?

The correct option is C v2RInitial momentum →Li=mvR Final momentum →Lf=(Ir+Im)ω Here, no external torque is applied so, Li=LfmvR=(mR2+mR2)ωv=2R2ωω=v2R

A small bead of mass $m$ moving with velocity $v$ gets threaded on a stationary semicircular ring of mass $m$ and radius $R$ kept on a horizontal table. The ring can freely rotate about its centre. The bead comes to rest relative to the ring. What will be the final angular velocity of the system?

The correct option is C $\frac{v}{2 R}$
Initial momentum $\to L_{i} = m v R$
Final momentum $\to L_{f} = (I_{r} + I_{m}) ω$
Here, no external torque is applied so,
$L_{i} = L_{f} m v R = (m R^{2} + m R^{2}) ω v = 2 R^{2} ω ω = \frac{v}{2 R}$