Mechanics - mass on a circular ramp

[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.

 

[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

 

[mukundpa]

If we write

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

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 !