Two rods, Two hinges.

Ok try doing it by energy method..

Write total enery = cons and differentiate ..

Anyone interested ?

qwerty is solving

o.O

lol

i wil giv it a try though

\tau _{c}=I_{c}\alpha
(F-F_{x})\frac{l}{2}sin\theta +F_{y}lcos\theta =I_{c}\alpha............................(1)

F-F_{x}=2ma_{x}

2mg+F_{y}=2ma_{y}

a_{c}=a_{x}(\hat{i})-a_{y}(j)

now Point P is ICR .

Hence COM performs pure rotation abt pt P.

thus ,

i.e,\vec{a_{c}}=PCsinA\alpha (\hat{i})-PCcosA(\hat{j})

i.e,\vec{a_{c}}=\frac{l}{2}sin\theta \alpha (\hat{i})-lcos\theta \alpha (\hat{j})

hence
F-F_{x}=2ma_{x}=2m\frac{l}{2}sin\theta \alpha =ml\alpha sin\theta

2mg+F_{y}=2ma_{y}=2ml\alpha cos\theta

putting the values of F-F_x and F_y in torque eqn

(ml\alpha sin\theta )\frac{l}{2}sin\theta +(2ml\alpha cos\theta -2mg)lcos\theta =I_{c}\alpha

\frac{ml^{2}sin^{2}\theta \alpha }{2} +2ml^{2}cos^{2}\theta \alpha -I_{c}\alpha =2mglcos\theta

\alpha = \frac{2mglcos\theta }{\frac{ml^{2}sin^{2}\theta }{2} +2ml^{2}cos^{2}\theta -I_{c}}

I_{c}=\frac{2ml^{2}}{12}+\frac{2ml^{2}cos^{2}\theta }{4}

\alpha = \frac{2mglcos\theta }{\frac{ml^{2}sin^{2}\theta }{2} +2ml^{2}cos^{2}\theta -(\frac{2ml^{2}}{12}+\frac{2ml^{2}cos^{2}\theta }{4}) }

\alpha = \frac{6gcos\theta }{l(3cos^{2}\theta +1) }

it can be done even widout ICR

system rotates abt com with alpha

so a_p = a_c/p+a_p=CPsinA(i) - CPcosA(j) + a_x(i) -a_y(j) = 0

hence a_x=CPsinA , a_y=CPcosA

correction in last post ,
ap = ac/p+ap=CPsinA(-i) + CPcosA(j) + ax(i) -ay(j) = 0

no one?

philip plz post the solution

Nice for reviving the topic!

subho has done it correct

energy method se hi karna chahiye like subho

btw subho nice solution yaar

never thought abt doing it the energy way

btw subho pls point out da mistake in my solution

i got my mistake ,the 2 rod system isnt rigid , so some eqns go wrng

and also you did not notice that |α| should be same for both :( even though i mentioned it in post #13 itself

the hinge you see in the middle is not fixed

i sense that this sum will involve the basic definition of angular velocity .... dθ /dt !
....or is it ....?

First resolve the force F one perpendicular to the rod and another along the rod.

Therefore Fp = Fsinθ
Fa=Fcosθ
Now Fa will produce x acceleration to the rod in its direction. Therefore
Fa = 2mx
x=Fa/2m
Now this mx will be the force that will provide the torque to the left hand side rod.

Therefore now angular acceleration
For the left one
mxl = ml²/3*α
α=3x/l=3Fa/2ml=3Fcosθ/2ml

For the right one
FpL = mL²/3*α
α=3Fp/mL
α = 3Fsinθ/ml

@virang1 : did not want to discourage the only person who tried....

but honestly u are not even close.