Question

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? Express your answer to two significant figures and include the appropriate units. * Incorrect; Try Again; One attempt remaining Part B Find an algebraic expression for the ice cube's speed when it is at angle θ, where the angle is measured counterclockwise from the bottom of the circle. Your expression should give 3.0 m/s for θ=0ᵒ and your answer to part a for θ=180ᵒ. Express your answer in terms of θ.

Answer 1

The ice cube's speed at the top of the circle is 1.4 m/s. The speed when it is at angle θ is given by v(θ) = v_b * cos(θ).

Step-by-step explanation:

To find the ice cube's speed at the top, we can use the principle of conservation of mechanical energy. At the bottom, the ice cube has kinetic energy equal to its potential energy at the top. So we can set up the equation:

0.5 * m * vb2 = m * g * h

where vb is the speed at the bottom, m is the mass of the ice cube, g is the acceleration due to gravity, and h is the height. Solving for the speed at the top, we have:

vt = sqrt(2 * g * h)

Plugging in the values, we get:

vt = sqrt(2 * 9.8 * 0.1) = 1.4 m/s

For part b, the ice cube's speed when it is at angle θ can be expressed as:

v(θ) = vb * cos(θ)

where θ is the angle measured counterclockwise from the bottom of the circle.