RoyalCat
Aug7-09, 06:03 AM
A mass m is traveling inside a horizontal circular tube of radius r, starting with an initial velocity v_0
There is friction between the outer wall of the tube and the mass (Friction between the floor and the mass is negligible)
EDIT:
What is the mass' velocity as a function of the distance it's traveled v(s)
(Sorry for the confusion if anyone happened to read what was written here earlier)
This is a problem that is listed as a solved example in my book, so I only really needs help with one little calculus step.
The first solution is an energy-based one.
The initial energy of the mass is \tfrac{1}{2}mv_0^2
The friction force acting on the mass : f_{friction}=\mu N= \mu \frac{mv^2}{r}
At the limit of a very small fraction of the distance along the curve, S, the force is constant, and so the work of friction along that distance is:
dW_f=\mu\frac{mv^2}{r}\,dS
This quantity is also equal to the kinetic energy lost by the mass due to friction.
\mu\frac{mv^2}{r}\,dS=-dE_k=-d(\tfrac{1}{2}mv^2})
Now according to the book: -d(\tfrac{1}{2}mv^2})=-mv\,dv
Now I can see why that's true, an integration of that expression over dv would provide us with the infinitesimal difference in kinetic energy, but I'm really have trouble understanding what the author did here to reach his conclusion, and why it is valid.
If anyone's got a webpage explaining switching between integrands (At least that's what I think it's called) I'll gladly read up on the link.
With thanks in advance,
Anatoli
There is friction between the outer wall of the tube and the mass (Friction between the floor and the mass is negligible)
EDIT:
What is the mass' velocity as a function of the distance it's traveled v(s)
(Sorry for the confusion if anyone happened to read what was written here earlier)
This is a problem that is listed as a solved example in my book, so I only really needs help with one little calculus step.
The first solution is an energy-based one.
The initial energy of the mass is \tfrac{1}{2}mv_0^2
The friction force acting on the mass : f_{friction}=\mu N= \mu \frac{mv^2}{r}
At the limit of a very small fraction of the distance along the curve, S, the force is constant, and so the work of friction along that distance is:
dW_f=\mu\frac{mv^2}{r}\,dS
This quantity is also equal to the kinetic energy lost by the mass due to friction.
\mu\frac{mv^2}{r}\,dS=-dE_k=-d(\tfrac{1}{2}mv^2})
Now according to the book: -d(\tfrac{1}{2}mv^2})=-mv\,dv
Now I can see why that's true, an integration of that expression over dv would provide us with the infinitesimal difference in kinetic energy, but I'm really have trouble understanding what the author did here to reach his conclusion, and why it is valid.
If anyone's got a webpage explaining switching between integrands (At least that's what I think it's called) I'll gladly read up on the link.
With thanks in advance,
Anatoli