How Fast Does the Center of Mass of a Falling Stick Move?

[ehild]

So, $$ ω^2 = \frac{12g(1-cosθ)}{L(3sin^2θ+1)} $$ looks alright to you ?

I like it

And then

$$ v_{cm} = -\frac{L}{2}(\dotθsinθ) $$

gives $$ ω = -[\frac{3gL(1-cosθ)(sinθ)}{(3sin^2θ+1)}]^\frac{1}{2} $$

Is this fine ?

why minus? And you miss a factor of 2. $$ ω =2\sqrt{ \frac{3g(1-cosθ)}{L(3sin^2θ+1)} }$$

and Vcm=-ωsinθ L/2.

 

[Tanya Sharma]

Oh ! that was a typo..I meant

$$ v_{cm} = -[\frac{3gL(1-cosθ)(sinθ)}{(3sin^2θ+1)}]^\frac{1}{2} $$

Minus makes perfect sense...

 

[ehild]

Oh ! that was a typo..I meant

$$ v_{cm} = -[\frac{3gL(1-cosθ)(sinθ)}{(3sin^2θ+1)}]^\frac{1}{2} $$

Minus makes perfect sense...

$$ v_{cm} = -[\frac{3gL(1-cosθ)(sin^2θ)}{(3sin^2θ+1)}]^\frac{1}{2} $$
sinθ has to be squared if you pull it under the square root. But otherwise it is perfect:thumbs: Have a good rest, you deserve it.

ehild

 

[Tanya Sharma]

ehild...I am falling short of words to express my gratitude...All I will say is "THANK YOU ehild" .

You rock!

 

[ehild]

ehild...I am falling short of words to express my gratitude...All I will say is "THANK YOU ehild" .

You rock!

You are welcome. That was a challenging problem. I am also exhausted

ehild