Maximum Angular Velocity of a quarter circle

[Yasin]

Homework Statement

Finding the general formula for max angular velocity ( answers say 0.839*(g/b)) but I do not understand how

Homework Equations

0.839*(g/b)

The Attempt at a Solution

 

[kuruman]

You have to show some effort before you ask for help. The answer itself is not a relevant equation.

 

[Yasin]

Sorry, I am new and my attempt I tried converting mgh(gravitational potential energy) into (mv^2)/2 (Kinetic energy)
b being the height i get mgb=mv^2/2 so i get 2gb=v^2
v^2/r^2 is w^2 so i get 2gb/b^2=w^2 which is sqr(2g/b) as the answer
but somehow the answer in the book i use to study is different so i must be wrong
Now i see this is wrong because this is valid for a string and the question has a segment quarter circle.
Same method but using point mass at line where angle is 45 degrees i now get
w=sqr(2g(1-sin45)/b)

 

[CWatters]

Regarding the potential energy... Does the whole semicircle fall a distance b?

 

[Yasin]

Regarding the potential energy... Does the whole semicircle fall a distance b?

The fall is any distance but here I assume when the part where 'm' is located is nearest to bottom that will be the max speed thus max angular velocity

 

[CWatters]

What I was hinting at is the available PE cannot be mgb because the centre of mass cannot fall a distance "b".

 

[Yasin]

What I was hinting at is the available PE cannot be mgb because the centre of mass cannot fall a distance "b".

Yes you are right and i did change the last one by editing the first post
and i think i have solved it thanks for all support the solution is as follows in picture for any interested
Some reason the textbook solution is wrong

 

[kuruman]

Mechanical energy conservation is the way to go, but your solution is incorrect. Here you have an extended object that is acted upon by the external force of gravity. If you were to pretend that the entire mass of this contraption were concentrated at one special point, where would that special point be? Once you find that, then you can say $\frac{1}{2}mv^2=\frac 1 2 m (\omega r)^2$ where $r$ is not $b$, but the distance from the pivot to the aforementioned special point. What you have here is a physical pendulum.