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

[adjacent]

:zzz: just got free

Let $x$ be the distance from point D (in the direction of A) where the system is located. Can you find the potential energy of the system at $x$?

yes.
$P_E=mg\sin\gamma x$

That is correct for a point mass. But we have a rod and wheels, which rotate. Their rotation also has kinetic energy. Are you familiar with that?

Rotational kinetic energy?It's an A level subject.

 

[voko]

Have you studied moments of force (torques) and moments of inertia?

 

[adjacent]

I have studied moment of force but not inertia(That's also an A-level topic)

You just tell me how moment of inertia is used(I am a fast learner)

 

[voko]

For a point mass at distance $d$ away from the axis of rotation, the moment of inertia is $I = md^2$. When it is rotating with angular speed $\omega$, its linear speed is $\omega d$, so its kinetic energy is $\frac {mv^2} 2 = \frac {m (\omega d)^2 } 2 = \frac {I \omega^2} 2$. The same is true for a system of point masses $m_i$ and distances $d_i$, so the kinetic energy is given by the same formula, where $I = \sum m_i d_i^2$. For solid bodies, we replace the sum with an integral. See http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html for more details.

This particular system has two wheels and one rod. I suggest that we treat the rod as very thin and very light, in which case its moment of inertia is negligible, so the total moment of inertia is twice the moment of inertia of a wheel. The wheel is a cylinder, whose moment of inertia about its axis of symmetry is given by $\frac {mR^2} 2$, where $m$ is the mass and $R$ is the radius. In this case $m = \frac M 2$, so the total moment of inertia is $MR^2 \over 2$, and the rotational kinetic energy is then $\frac {MR^2 \omega^2} 4 = \frac {MR^2} 4 \left( \frac v R \right)^2 = \frac {Mv^2} 4$.

 

[adjacent]

Ok.Now the wheel system has kinetic energy of linear kinetic energy + rotational kinetic energy?

 

[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.