1lets resolve the impusle delivered into 2 directions...horizontax (X) and vertical (Y)...we know that Jx=Psin\phi
as for Py we have Pcos\phi - Py = P'.....1)
and also for angular impulse about the hinge, we have
Pcos\phi * 3L/4
=I w.....2) {where I is the moment of inertia about hinge pt}
and finally we have P'/m = wL/4......3) {constraint eqn.}
jus gievn the brief input...
cheers!!!
3gud job gordo....................................[new to the site???????]
106completely solving..
Conserving ang momentum,
P0cosθ*3l/4 = (ml2/12 + ml2/16)ω
==> ω = 3P0/ml ... putting cosθ = 7/12
Now, P0sinθ = Jx = 9P0/12
Vcom = l/4*ω = 3P0/4m
==> P0cosθ + Jy = m*3P0/4m
==> 7P0/12 + Jy = 3P0/4
==> Jy = 2P0/12
So hinge impulse = √Jx2 + Jy2
= P0/12*√85
