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Question: A small beam of mass m moving with velocity v gets threaded on a stationary semicircular ring of mas...

A small beam 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 rotates freely about its center o. The bead comes to rest relative to the ring. Then, the final angular velocity of the system,
A. v/Rv/R
B. 2v/R2v/R
C. v/2Rv/2R
D. 3v/R3v/R

Explanation

Solution

First, we should use the equation of linear angular momentum of the system, Li=mvR{L_i} = mvRto calculate the initial angular velocity of the system and I=mR2I = m{R^2} to calculate the inertia of the system. Then calculate the final angular momentum as Lf=Iω{L_f} = I\omega and put it in I=mR2I = m{R^2}. Then, apply the law of conservation of angular momentum to calculate final angular momentum

Complete step by step solution:


Given,
The velocity of the small beam is vv{\rm{ }}
The mass of the small beam is mm
The radius of the ring is RR
The initial angular velocity is ωi=0{\omega _i} = 0\,.
The final angular velocity is ωf{\omega _f}.
The equation of initial angular momentum of the beam of massmm and velocity vv{\rm{ }}moving in a circular ring of radius RRcan be written as,
Li=mvR{L_i} = mvR …… (1)
The net inertia of the system can be written as,
Isystem=Ibead+Iring{I_{system}} = {I_{bead}} + {I_{ring}} …… (2)
The inertia of the bead and ring can be written as,
Ibead=Iring=mR2{I_{bead}} = {I_{ring}} = m{R^2} …… (3)
Substituting equation (3) in equation (2) we get,
Isystem=mR2+mR2=2mR2{I_{system}} = m{R^2} + m{R^2} = 2m{R^2} …… (4)
Again, we have,
Lf=Isystemωf{L_f} = {I_{system}}{\omega _f} …… (5)
Applying law of conservation of angular momentum, we have,
Li=Lf{L_i} = {L_f} …… (6)
Substituting equation (1) in equation (6) and equation (5) in equation (6) we have,
mvR=IsystemωfmvR = {I_{system}}{\omega _f} …… (7)
Substituting value of Isystem{I_{system}}from equation (4) in equation (7) we get,
mvR=(2mR2)ωf ωf=v2R\begin{array}{l}mvR = (2m{R^2}){\omega _f}\\\ \Rightarrow {\omega _f} = \dfrac{v}{{2R}}\end{array}

Hence, the correct answer is (C).

Note: The beam moves with a linear velocity vv{\rm{ }} and its angular velocity is zero. The ring rotates with angular velocity ωf{\omega _f}. In the solution, the students can use the law of conservation of angular momentum where the initial momentum is equal to the final momentum.