Mechanics Question

ll

What have you tried?

normally you would equate potential energy to kinetic but it wants the end of the rod and it is a rotation. So it wouldnt be mgh = 0.5mv²

Yep, so you need to consider rotational K.E.

How much GPE does the rod lose in the process of motion?

This must be equal to the rotational KE when the rod is vertical, and this is given by $\dfrac{1}{2}I\omega^2$, where $I$ is the moment of intertia of the rod about the end which is fixed to the ceiling, and $\omega$ is the angular speed.

Rearrange this equality for $\omega$. Then clearly, velocity of the other end of the rod is $v = l\omega$

In E = mgh

would h = L or would it be 0.5L

It's a uniform rod, so the CoM is halfway along the rod.

Ok thanks.

Not rlly sure on this on either:

1/total resistance for the LHS would be (1/R1+R2)+1/R3 and so you equate that to Rtot of the RHS which is (1/Rp+2R3), make Rp the subject

I got the LHS but for RHS did u mean 1/(Rp + 2R3) or (1/Rp) + 2R3