Two rods, Two hinges.

qwerty : justify P being thre ICR.

Ok try doing it by energy method..

Write total enery = cons and differentiate ..

Philip plz post ur soluton. And do it by force method.

Anyone interested ?

qwerty is solving

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

it can be done widout ICR .........

the hinge at P is fixed , so its accn shud be zero ,see #29,30

no one?

philip plz post the solution

Nice for reviving the topic!

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

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