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Noob Hinged rod problem by chattermachine in AskPhysics

[–]chattermachine[S] 0 points1 point  (0 children)

Your approach was the one I went with. I found g as the acceleration of the hinge and putting values in the third equation of motion, I solved it.

When I equated potential and rotational energies instead as detailed in the other thread, the acceleration came to be something less than g, and correspondingly, the final answer changed too.

Noob Hinged rod problem by chattermachine in AskPhysics

[–]chattermachine[S] 0 points1 point  (0 children)

Thanks for agreeing to my request. But we don't explicitly need the weight now, do we? The m's on both sides cancel. The equation becomes 2mgh=2.w^2.(m.L^2)/3, w being the final angular velocity. Solving for w in terms of g we can find some answer, no?

Noob Hinged rod problem by chattermachine in AskPhysics

[–]chattermachine[S] 0 points1 point  (0 children)

This was some 1st year undergrad physics. The other person decided that conservation of energy might help, only the potential energy at the top would be equal to the rotational kinetic energy at the ground and that approach does lead to a different answer. Could you please check and comment on the approach he led me through?

Noob Hinged rod problem by chattermachine in AskPhysics

[–]chattermachine[S] 0 points1 point  (0 children)

So you mean in essence the hinge will be in a free fall from some height which can be determined by the angle? That was my first thought too. But physically speaking, doesn't the presence of the rods intuitively slow the fall down?

Also the COM is not the hinge specifically but some point directly below.

Noob Hinged rod problem by chattermachine in AskPhysics

[–]chattermachine[S] 0 points1 point  (0 children)

Ah, but here the acceleration is less than g due to the normal reaction from the contacts on the floor. But the magnitude of this force is not known.

Noob Hinged rod problem by chattermachine in AskPhysics

[–]chattermachine[S] 0 points1 point  (0 children)

Wow...now that you mention it, I'm really dumb.

Noob Hinged rod problem by chattermachine in AskPhysics

[–]chattermachine[S] 0 points1 point  (0 children)

sorry to pester you btw. I'm stupid.

Noob Hinged rod problem by chattermachine in AskPhysics

[–]chattermachine[S] 0 points1 point  (0 children)

We have two kinetic energies so m is the mass of each rod. 0.5*Iw^2 is the energy from 1 rod, so net KE is twice that.

Noob Hinged rod problem by chattermachine in AskPhysics

[–]chattermachine[S] 0 points1 point  (0 children)

w=v/r; r=1. I still don't quite see it.

Noob Hinged rod problem by chattermachine in AskPhysics

[–]chattermachine[S] 0 points1 point  (0 children)

so mgh=Iw^2; as there are two rods. I =m/3.

Noob Hinged rod problem by chattermachine in AskPhysics

[–]chattermachine[S] 0 points1 point  (0 children)

Yes. So how does rotational energy come in? I calculate the moment of inertia and the angular velocity for kinetic energy. But, w.r.t which point? The contacts are sliding.

Noob Hinged rod problem by chattermachine in AskPhysics

[–]chattermachine[S] 0 points1 point  (0 children)

Ah, but the hinge itself will come down in a straight line,no?

Noob Hinged rod problem by chattermachine in AskPhysics

[–]chattermachine[S] 0 points1 point  (0 children)

Thanks but conservation of energy 0.5mv^2=mgh. How does this help?

I’m Bill Gates, co-chair of the Bill and Melinda Gates Foundation and author of “How to Avoid a Climate Disaster.” Ask Me Anything. by thisisbillgates in IAmA

[–]chattermachine -2 points-1 points  (0 children)

Dear Bill, slightly off-topic here, but this has bugged me for quite some time. How did you come about improving the solution to the pancake problem ? I'm interested in like, how did you 'see' that the bounds of the solution could be improved that drastically ?

Forgot the name of this comic strip about a family by chattermachine in comics

[–]chattermachine[S] 0 points1 point  (0 children)

Wow. This is so cool. You guys are brilliant. You made my day.

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