# Conservation of Energy Ice Cube Problem

1. Homework Statement
A very slippery ice cube slides in a vertical plane around the inside of a smooth, 20 cm diameter horizontal pipe. The ice cube's speed at the bottom of the circle is 3 m/s.

a) What is the ice cube's speed at the top?

b) Find an algebraic expression for the ice cube's speed when it is at angle theta where the angle is measured counterclockwise from the bottom of the circle. Your expression should give 3 m/s for 0 degrees and your answer to part a for 180 degrees.

2. Homework Equations
KE= 1/2mv^2
PE=mgy

3. The Attempt at a Solution

Ok, I got part a

(.5)*m*(3m/s)^2=m(9.80)(.2m)+(.5)*m*(vf)^2
vf= 2.25 m/s

But I am really not sure of how to approach part b. I thought about the equations for rotational kinematics, but the acceleration is not constant. Any hints?

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You are applying the same principle as before-- conservation of mechanical energy. In part (a) you specifically set the height of the top to be the diameter of the pipe. But now, you need to generalize what you did before to express the height of the pipe as a function of the angle.

To solve this problem you should sketch a picture of the cross-section of the pipe (it should be a circle), label the bottom and top, as well as a random point and clearly label the angle as measured from the bottom.

Once you have constructed that picture you can then use simple trig to find the height in terms of the radius of the pipe and the angle.

I could show you more, but instead of showing you the solution, I would like to see if you can solve it from that advise, I bet you can, good luck.

Ok, I got it. The height = r(1-cos theta)

So

r(1-cos theta)mg +.5*m*(v1)^2 = .5*m*(vo)^2

Solve for v1

v1= sqrt[(vo)^2+2rg(costheta-1)]

where v1 is the velocity at any position, vo is the velocity at the bottom, and r is the radius of the pipe

Thank you very much!

Awesome! I'm glad you solved it!

sorry this maybe a dumb question but how did you get the height
as r(1-cos theta) ?