A frictionless uniform ring of mass m has 2 beads A and B of mass m ea...
Problem
A frictionless uniform ring of mass m has 2 beads A and B of mass m each; both being able to slide on the ring. Initially, the ring lies on a frictionless horizontal table, and both beads are given velocities v towards the x-axis. The relative velocity of the beads wrt to each other just before they collide is? Explain in detail.
Solution
Let's assume that the ring has a radius R.
Step 1: Find the velocity of the center of mass of the system
The velocity of the center of mass of the system can be found by applying the conservation of momentum.
Initial momentum = Final momentum
2mv = (m + 2m)VCM
VCM = v/3
Step 2: Find the velocity of the beads wrt to the center of mass
Let the velocities of bead A and B wrt to the center of mass be vA and vB respectively.
Using the relative velocity formula, we get:
vrel = vB - vA
Step 3: Find the velocities of the beads
As the ring is a uniform circular motion, the velocity of the beads can be found by using the formula:
v = ωR
where ω is the angular velocity of the ring.
Initial velocity of bead A = v
Velocity of bead A after time t = vA = v - ωt
Initial velocity of bead B = v
Velocity of bead B after time t = vB = v + ωt
Step 4: Find the time taken for the beads to collide
The beads will collide when they meet each other on the ring. Let the time taken for the beads to collide be tc.
Distance traveled by bead A = Distance traveled by bead B
∴ (v - ωtc)tc = (v + ωtc)tc
∴ tc = R/ω
Step 5: Find the relative velocity of the beads just before they collide
Substituting tc in the expression for vrel, we get:
vrel = vB - vA
= (v + ωtc) - (v - ωtc)
= 2ωtc