Substituting differentials in physics integrals.

[subsonicman]

Today I tried to show that rotational kinetic energy was equivalent to standard translational kinetic energy.

So I started with kinetic energy, T = ∫dT. Then, because T=1/2mv^2, I substituted dT=1/2v^2dm and then because m=ρV, I substituted dm=ρdV. Then, after substituting v=ωr, I got the equation for rotational kinetic energy, 1/2Iω^2.

The problem I have is with the substituting differentials. Shouldn't dT=1/2v^2dm+vdvdm because both v and m are varying? Also, shouldn't dm=ρdV+Vdρ? I remember seeing this substitution made when calculating the mass of some shape from its density but I can't seem to justify it from the knowledge I have.

Any help would be appreciated.

 

[Simon Bridge]

Today I tried to show that rotational kinetic energy was equivalent to standard translational kinetic energy.

So I started with kinetic energy, T = ∫dT. Then, because T=1/2mv^2, I substituted dT=1/2v^2dm and then because m=ρV, I substituted dm=ρdV. Then, after substituting v=ωr, I got the equation for rotational kinetic energy, 1/2Iω^2.

The problem I have is with the substituting differentials. Shouldn't dT=1/2v^2dm+vdvdm because both v and m are varying?

1. dT = (1/2)v^2.dm + mv.dv
2. what is dv/dm ?

 

[dauto]

You're confused. You're not doing an integration by parts. You're just doing a change of variable of integration. The product rule makes no sense here.

 

[subsonicman]

Yeah, I was being stupid. Thanks for the help!

 

[PJC01]

I would first commend you on your attempt to show the equivalence between rotational and translational kinetic energy. This is an important concept in physics and it is great that you are exploring it further.

In terms of your concerns about the substitutions in the integral, I can understand your confusion. It is important to note that the differential notation in physics can often be misleading and requires careful interpretation.

In this case, when you substitute dT=1/2v^2dm, you are essentially saying that a small change in kinetic energy (dT) is equal to 1/2 times the square of the velocity (v^2) times a small change in mass (dm). This is a valid substitution because we are assuming that the velocity is constant and only the mass is changing.

Similarly, when you substitute dm=ρdV, you are saying that a small change in mass (dm) is equal to the density (ρ) times a small change in volume (dV). Again, this is a valid substitution because we are assuming that the density is constant and only the volume is changing.

To address your concern about the varying variables, it is important to remember that when we are dealing with differentials, we are taking infinitesimally small changes. So even though both v and m may be varying, in the infinitesimal limit, we can assume them to be constant.

In the case of the mass substitution, it is important to note that the density (ρ) is also a function of volume (V). So when we take the derivative of this function, we get dρ/dV, which is why we only use dV and not Vdρ.

I hope this helps clarify your doubts. Remember, in physics, it is important to carefully interpret the notation and assumptions made in order to make valid substitutions. Keep up the good work in exploring and understanding these concepts!