a uniform rod of mass m and lenght l is hinged at its end, is released from rest when it is in horizontal position.The normal reaction at the hinge when the rod becomes vertcal is...

A) mg/2   B) 3mg/2  C)5mg/2   D) 2mg

    pzl solve urgent.....i am gettin the answer as 5mg/4.....pzl solve with procedure.....thanx....

take torque due to mg at any angle theta use

Ia=mgcos(theta)

integrate it to find omega at vertical position.

normal rxn due to hing will be in

+y direction.since rod is in circular motion.

N-mg=m(omega square)l*0.5

I * alpha = torque = mg *(L/2) *sin

therefore (ML^2)/3 *dw/dt = Mg*(L/2)*sin 

but dw/dt = w *dw/d               (as dw/dt = d/dt * dw/d = wdw/d)

 

i'm posting my solution, plz tell if its correct or not...

if its not correct then plz tell me wats wrong..

solution::

since, there is just rotatory motion, therefore, the loss in potential energy goes to the rotational kinetic energy

=> loss in P.E. of center of mass = gain in rotational K.E. of rod

=> mg L/2 = (1/2) ((m L^2) /3) w^2

=> w^2 = 3g/L

where w is angular velocity

=>reaction at hinge (N) = mg + mw^2 L/2

=> N = mg + 3mg/2

=> N = 5mg/2

 

 

if you find it convincing & correct then plz do rate....