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 3.0 m/s. What is the ice cube's speed at the top?

Given that,

Diameter of pipe

d = 20cm

Then, radius = d/2 = 20/2

r = 10cm = 0.1m

The speed at the bottom is

Vi = 3m/s

Speed at the top Vf?

At the bottom the cube is at a height of 0m

Then, y1 = 0m

At the top the cube is at a height which is the same as the diameter of the pipe

y2 = 0.2m

Now, let us consider, the energy conservation equation, which is the sum of kinetic energy and gravitational potential energy, given by,

K2 + U2 = K1 + U1

½m•Vf² + m•g•y2 = ½m•Vi² + m•g•y1

Divide all through by m

½•Vf² + g•y2 = ½•Vi² + g•y1

Since y1 = 0

So we have,

½•Vf² + g•y2 = ½•Vi²

½•Vf² = ½•Vi² - g•y2

Multiply through by 2

Vf² = Vi² - 2g•y2

Vf = √ (Vi²-2g•y2)

g is a constant = 9.81m/s2

Vf = √ (3²-2*9.81*0.2)

Vf = √ (9-0.981)

Vf = √8.019

Vf = 2.83m/s

The speed of the ice cubes at the top of the pipe is 2.83m/s