Fall Speed Based on Angle: 0°, 45°, 90°

[voko]

Yes.

 

[adjacent]

So the total kinetic energy = potential energy at the top?
But that does not give any increase in speed at any given point on the path AD

 

[voko]

No. Total mechanical energy at any point = total mechanical energy at the top.

 

[adjacent]

No. Total mechanical energy at any point = total mechanical energy at the top.

Really?
Shouldn't it be total kinetic+ total potential energy (At that position)= Total potential energy at the top.

 

[voko]

What is total mechanical energy? And what is it at the top?

 

[adjacent]

Mechanical energy is sum of Potential and kinetic energy
(I will be learning those names in Gr.11)

Now back to the question
$P_E=mg\sin\gamma x$
$E_K=\frac{1}{2}mv^2+\frac{Mv^2}{4}$
so,
$mg0.6=mg\sin\gamma x+\frac{1}{2}mv^2+\frac{Mv^2}{4}*2$

 

[voko]

If I say something you do not know, say that right away :)

I think you are using two variables for the same thing, mass of the system: $m = M$. Use just one for simplicity. Then, why do you have *2 in the final equation, which you did not have in the equation for $E_K$? $E_K$ already includes the energy of both wheels.

 

[adjacent]

If I say something you do not know, say that right away :)

I think you are using two variables for the same thing, mass of the system: $m = M$. Use just one for simplicity. Then, why do you have *2 in the final equation, which you did not have in the equation for $E_K$? $E_K$ already includes the energy of both wheels.

m=M/2 So linear kinetic energy + rotational kinetic energy
Rotational kinetic energy is $\frac{Mv^2}{2}$
And you said m =M/2
So mass of two wheels should be 2M

 

[voko]

m=M/2 So linear kinetic energy + rotational kinetic energy
Rotational kinetic energy is $\frac{Mv^2}{2}$
And you said m =M/2

I could have been sloppy in my notation. I used $m$ as a dummy variable when talking about the moment of inertia of an abstract cylinder. So it is not the $m$ that we have in the linear kinetic energy. I used $M$ to denote the total mass of the system, which is the $m$ in the linear kinetic energy. It is the mass of the entire system, i.e., the two wheels and the infinitely light rod. The total rotational kinetic energy is $mv^2 \over 4$ or $Mv^2 \over 4$ depending on whether you prefer $m$ or $M$ for the total mass of the system; whichever your preference is, use it in the linear kinetic energy and the potential energy.

 

[adjacent]

Oj.Then it's $mg0.6=mg\sin\gamma x+\frac{1}{2}mv^2+\frac{Mv^2}{4}$?

 

[voko]

Like I said, $m = M$. Just choose one of them and use it everywhere! Then simplify.

 

[adjacent]

Oh.My problem is solved now.Did you have all this in mind from the beginning? or Did you solve the problem at first?
-Just curious-
Time to thank you.

 

[voko]

Oh.My problem is solved now.Did you have all this in mind from the beginning? or Did you solve the problem at first?

Let's put it this way: I knew the general character of motion (uniformly accelerated motion in a straight line - did you get that?) when I initially responded. Then as you were getting stuck at various points I made sure that I had a clear picture of the relevant details.

I suggest that you pay attention to the following: how the 3D problem was reduced to the 2D problem. That is very important in physics! Then how it was simplified further by the choice of the coordinate $x$. And finally, how conservation of energy was used instead of the tedious force/torque analysis. Reduction of dimensionality, convenient coordinates and conservation laws are the most powerful tools in physics, master them.

 

[adjacent]

Ok.Thank you.Since you are so intelligent, What's your education level?

 

[voko]

What's your education level?

Masters in maths.