Arun
Last Activity: 6 Years ago
The forces acting on the rod are the weight mg, normal force FN and the frictional force Ffr.From the conservation of energy we have
Ei = Ef;
Ei = mgl .
Ef = mglcosq + 1/2Iw2 = mglcosq + 1/2[1/3m(2l)2]w2.
mgl = mglcosq + 1/2Iw2, which gives
w2 = 3g/2l(1 - cosq ), and w = [3g/2l(1 - cosq )]1/2.
We find the angular acceleration from differentiating w2 = 3g/2l(1 - cosq )
2wdw/dt = 3g/2lsinq dq/dt, where w/dt = a, and dq/dt = w.
then we have a = (3g/4l)sinq .
We write SF = ma for the rod:
y-comment: mg - FN = may;
but we have ay = asinq = lasinq = (3g/4)sin2q .
then we have FN = mg - m(3g/4)sin2q = (mg/4)(4 - 3sin2q).
and
x-component: Ffrmax = msFN = max = mla cosq = (3/4)mgsinq cosq .
