Question
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/R
B. 2v/R
C. v/2R
D. 3v/R
Solution
First, we should use the equation of linear angular momentum of the system, Li=mvRto calculate the initial angular velocity of the system and I=mR2 to calculate the inertia of the system. Then calculate the final angular momentum as Lf=Iω and put it in I=mR2. 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 v
The mass of the small beam is m
The radius of the ring is R
The initial angular velocity is ωi=0.
The final angular velocity is ωf.
The equation of initial angular momentum of the beam of massm and velocity vmoving in a circular ring of radius Rcan be written as,
Li=mvR …… (1)
The net inertia of the system can be written as,
Isystem=Ibead+Iring …… (2)
The inertia of the bead and ring can be written as,
Ibead=Iring=mR2 …… (3)
Substituting equation (3) in equation (2) we get,
Isystem=mR2+mR2=2mR2 …… (4)
Again, we have,
Lf=Isystemωf …… (5)
Applying law of conservation of angular momentum, we have,
Li=Lf …… (6)
Substituting equation (1) in equation (6) and equation (5) in equation (6) we have,
mvR=Isystemωf …… (7)
Substituting value of Isystemfrom equation (4) in equation (7) we get,
mvR=(2mR2)ωf ⇒ωf=2Rv
Hence, the correct answer is (C).
Note: The beam moves with a linear velocity v and its angular velocity is zero. The ring rotates with angular velocity ωf. In the solution, the students can use the law of conservation of angular momentum where the initial momentum is equal to the final momentum.