an irodov (simple)

see in hte question itself we are given that 2 horz forces act..one at the top od OO' and othere at botton of OO'

assume them to be F1 and F2....

now can u finish it ???

but wont a force act at the COM of the axis (though it is not given but doesnt the force on a body act at its centre of mass?)

mW^2L0/2 = F1 + F2

(F1-F2)(55/2) = mgL0/2 //

thats wat im saying.. the force acting on a body (here axis) acts at the COM. (the midpt.) So thats y i gave that F

can u tell me the ques no??

ashish post 8 is partly correct

ur concept is right

now u need to add verticle forces at O and O' to balance mg

Celestine, at end A there will be vertical forces also, in general. You don't know how it has been connected to the axle; it might have been welded as well.

The problem can be easily tackled if we take our system as the rod + the axis. Then the reaction at A becomes an internal force. Take the reference frame that rotates at the same rate as the rod. In this reference frame our system is in rotational equilibrium. So the net torque should be zero about any point.
Taking the moments about the upper point (point O in #8), we see that the torque due to gravity must be balanced by the torque of the (pseudo) centrifugal one (since there is no horizontal reaction at the bottom). This gives
mg\dfrac{\ell_0}{2}=m\omega^2\dfrac{\ell_0}{2}\cdot \dfrac{\ell}{2}
from where we get the required angular velocity as
\omega =\sqrt{\dfrac{2g}{\ell}}
In this case (coming back to the inertial frame) the centripetal force required by the rod to rotate in supplied by the horizontal reaction at the top, and hence
F_{h,\,\mathrm{top}}=m\omega^2\dfrac{\ell_0}{2}=mg\dfrac{\ell_0}{\ell}