Rotational MotionHard
Question
A uniform bar of length 6a and mass 8m lies on a smooth horizontal table. Two point masses m and 2m moving in the same horizontal plane with speed 2v and v respectively, strike the bar (as shown in the fig.) and stick to the bar after collision. Denoting angular velocity (about the centre of mass), total energy and centre of mass velocity by ω, E and vc respectively, we have after collision :
Options
A.vc = 0
B.
C.
D.E = 3
Solution
Two point masses and bar taken together as a system the angular momentum about centre of bar is (J = Iw)
2mva + 2mv × 2a =
= 30 aω = 6v ⇒
& applied force on bar
2mv - m × 2v = 0 ⇒ vc = 0
E =
Iω2 = 
× 30ma2 × 
⇒ E = 
2mva + 2mv × 2a =
= 30 aω = 6v ⇒
& applied force on bar2mv - m × 2v = 0 ⇒ vc = 0
E =
Iω2 = × 30ma2 × ⇒ E = Create a free account to view solution
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