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I'll do a followup question as an example to explain why I'm confused here.

|dw:1443076205726:dw|

Wow, sorry about the formatting in that post. I think that 1 is wrong and 2 is right.

Work done "by a force" = change in kinetic energy "due to that force"

|dw:1443076331186:dw|

Is that thing spinning and falling freely under gravitation ?

Yes.

|dw:1443076456909:dw|

This is like pendulum/SHO right ?

SHO = simple harmonic oscillation

Yes, I drew that arrow in the diagram too...

Yup.

Kind of like how a gransfather clockworks.

do u want to apply this kinda concept -
\[\omega _{f}^2 =\omega_{i}^2 +2\alpha \theta\]?

|dw:1443076723125:dw|

helps if you needed to understand the relationships I guess? Haha.

OK, so (2) is right, right?

but with pendulums, we assume the balls are not spinning

depends, if you assume the ball is spinning, add \(\frac{1}{2}I\omega^2\)
otherwise forget it

that is accounted as translational kinetic energy along circumferencce of circle, right ?

That's the point - it's not...

Or is it?

If the force is perpendicular to the motion, it does no work. so you can forget about it

But there has to be a theta, right? According to this, there is none.

|dw:1443077457196:dw| lmao... it's an easy one I guess.

Forgot something in it

|dw:1443077907086:dw|

are you taking 0 for potential energy at L/2 ?

Yeah, and I'm concentrating the entire system at its center of mass which is legal in physics.

|dw:1443078285888:dw|

It's been so many years, let me try and catch up on these..

potential energy = \(mg\left(\dfrac{L}{2} - \dfrac{L}{2}\cos\phi\right) \)
|dw:1443078517457:dw|

Fixed:\[mgL/2 (1 + \cos \theta) = 1/2 m(3/2)^2 + 1/2 \cdot mL^2/3 \cdot 3^2/L^2 \]

Still dunno.

Something doesn't feel right about this. Ugh.

That gives \(\sin \theta = 0.65\)

Ok so we're good with potential energy
let me try and understand whats going with rotational ke

are you sure the moment of inertia of that rod is \(\dfrac{mL^2}{3}\) ?

Yes.

also the speed of center of mass will not be 3 m/s
it should be way less than that

Yeah, I wrote that correction above. It's 3/2 and not 3.

And so \(\sin \theta = -1/10\)
Which is invalid.

So we definitely have to add rotational kinetic energy to that!

yes replace \(m\) by \(I\) and \(v\) by \(\omega\)

so is \(mgL = mv^2\) correct in my first question?

Total mechanical energy, \(mgL/2(1+\sin\theta) + \frac{1}{2} I\omega^2\), is conserved

\[mgL/2(1 + \sin \theta) = \frac{1}{2}I \omega^2 + \frac{1}{2}mv^2\]

|dw:1443079851015:dw|

see, im rusty on these, im just going by intuition, i can be wrong...

Yup. And \(I = mL^2/3\).

so it is the property of that rotating rod about that point A
it should not change

Yes, indeed.

wait, what if this problem is wrong?

let's reverse the thing and find the maximum velocity at the bottom

LMAO!! We just ignored @Jhannybean

\[2mgL = 4mv_{CM}^2/3\]\[3gL/2 = v_{CM}^{2}\]

\[v_{bottom}^2 = 4 v^2_{CM} = 6gL = 6\cdot 10 \cdot 0.25 = \cdots\]So I guess we're fine.

\(3g(1+\sin\theta) =L\omega^2\)

What's that? Just the rotational kinetic energy?

So \(v_{bottom} = \sqrt{15} > 3\)

Yeah, that's a good idea... what if we just include the rotational kinetic energy?

So absolutely nothing works at this point. =_=_=_=_=_=

Could you solve the rest? I just am not able to.

plugin \(\omega = \dfrac{v}{L} =\dfrac{3}{25} \)

I'm getting
\[1+\sin\theta = \dfrac{9}{125g}\]

so that's wrong, isn't it?

not so sure
http://www.wolframalpha.com/input/?i=solve+1%2Bsinx+%3D+9%2F%28125*10%29%2C0%3Cx%3C2pi

But 0

sorry missed the 1/2 in the middle for the E, it should be \(E= \frac 1 2 I \omega^2\) :(

Amazing. I wonder why I forgot that.

Thank you so very much!

It a pleasure ;)

I mean you didn't really need to type all of that, but thanks!

But we do need to add the linear component in plane motion.

umm what do you mean by the plane motion ?

Nothing, just rotational+translational motion.

Ok. Btw to which group do you take this original question?

Oh, it's pure rotation. I got it. :)

Basically in plane motion, the axis of rotation is translating.

you mean the third catagory where you should sum the linear motion energy and the ration energy?

Exactly. But here, the axis is stationary and the original question is a kind of pure rotation.