Mechanics - mass on a circular ramp

1. Sep 28, 2005

dusty8683

the problem we got in class was that an object of mass m was sliding down a quarter-circle ramp (or "quarter pipe) that had a radius a, and a coefficient of kinetic friction "mu". if the mass began down the ramp from rest and continued to the bottom, what would it's final velocity be?

i had no idea how to solve it by the time the teacher worked it out on the board. she simplified it converting it to a flat ramp at an angle of 45 deg above the horizontal. i understand more than well how to work problems of that sort... but how would you go about it the "correct" way??

i've tried integrating the normal force of m*g*cos("theta") WRT theta from "pi" to 3*"pi"/2... which obviously didn't work. i've also tried calculating the unit normal vector of the line r("theta")=a*cos("theta")i+a*sin("theta")j and ("pi" <= "theta" <= 3*"pi"/2) and using that with the normal force which also didn't work.

i don't need this answer for class... i just don't like only knowing how to figure this out the "simplified" way. if you could explain how to do figure this out, i'd appreciate it. thanks.

2. Sep 29, 2005

mukundpa

Consider the poaition where line joining m to the center of the circular path makes an angle $$\theta$$ with the horizontal. Tengential acceleration of m at this position will be g$$(cos \theta - \mu sin \theta ).$$
Now if the small tengential displacement is dl = R $$d \theta$$ then we can write

$$a = v \frac {dv}{dl} = g(cos \theta - \mu sin \theta ).$$
gives

$$v dv = g(cos \theta - \mu sin \theta ).R d \theta$$

Integrate this for 0 to pi/2

3. Sep 29, 2005

mukundpa

If we write
as

$$v dv = g(cos \theta - \mu sin \theta ). dl$$

then

$$cos \theta dl = dy$$ and $$sin \theta dl = dx$$

Integrate this for any arbitrary curved path, x = 0 to d and y = 0 to h, I think, you will get a wonderful result.

Just try !!!