Falling at an angle

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adjacent

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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##
 
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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

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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
 
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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

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Oj.Then it's ##mg0.6=mg\sin\gamma x+\frac{1}{2}mv^2+\frac{Mv^2}{4}##?
 
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Like I said, ##m = M##. Just choose one of them and use it everywhere! Then simplify.
 

adjacent

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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.
 
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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

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

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