a uniform rod AB of mass m n length l is at rest on a smooth horizontal surface. an impulse J is applied to the end B perpendicular to the rod in horizontal direction . speed of partical at a dist. l/6 frm the centre towards A of the rod after time t=Ï€ml/12J is.
1.2J/m
2.j/√2m
3.J/m
4.√2J/m
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2 Answers
106J = mVcm ... (i)
Jl/2 = ml2/12*ω ... (ii)
==> ω = 6J/ml and Vcm =J/m
at t=Ï€ml/12J,
θ = ωt = π/2
So the rod is perpendicular at that moment wrt initial condition.
So,
v = √(J/m)2 + (J/m)2
= √2*J/m
1velocity of particle = √Vcm2 + (wr)2 + 2Vcm*wr*cosθ
where θ is angle between Vcm and wr....
Here finally, wr and Vcm are perpendicular
Hence speed = √Vcm2 + (wr)2
Vcm = J/m and wr = 6J/ml*l/6 = J/m
put the values and get the answer
