# Homework Help: Skiing Problems

1. Dec 19, 2005

A skier starts at the top of a a very large frictionless snowball, with a very small initial speed, and skis straight down the side. At what point does she lose contact with the snowball and fly off at a tangent? That is, at the instant she loses contact with the snowball, what angle $$\theta$$ does a radial line from the center of the snowball to the skier make with the vertical?

So at the top, the velocity is 0. We are interested in the point where the skier falls off. The force acting on the skier is his weight. Its components are $$mg\cos\theta$$ and $$mg\sin\theta$$. So $$\Sigma F = m\frac{v^{2}}{R}$$ or $$v = \sqrt{gR\cos\theta}$$. Set the potential energy at A to the kinetic energy at our point of interest which we will call B.

$$U_{A} = K_{B}$$
$$mgh = \frac{1}{2}mv^{2}$$

I know I have to express the height as a function of $$r$$ and $$\theta$$. How would I do this?

A 0.5 kg ball is tied to a string 2 meters in length, and the other end is tied to a rigid support. The ball is held straight out horizontally from the point of support, with the string pulled taut, and is then released. (a) What is the speed of the ball at the lowest point of its motion? (b) What is the tension in the string at this point?

(a) Would it just be $$\frac{1}{2}mv^{2}$$
(b) $$F = mgh$$

Thanks

Last edited: Dec 19, 2005
2. Dec 19, 2005

### andrewchang

for the first, problem, try to draw a triangle that will allow you to find the change in height (i.e. the change in the radius)

3. Dec 19, 2005

### Astronuc

Staff Emeritus
The first problem, andrewchang addressed.

In the second problem, the gravitational potential energy (mgh) is converted into kinetic energy at the lowest point of motion, which is h (or r) below the starting point.