# Falling at an angle

Gold Member
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.

Gold Member
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.

Gold Member
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.

Gold Member
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.

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

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