Question

A skier starts at the top of 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 (alpha) does a radial line from the center of the snowball to the skier make with the vertical?

Answer 1

Answer:

The answer to the question is

At the instant she loses contact with the snowball, the angle (alpha) a radial line from the center of the snowball to the skier make with the vertical is 48.2 °

Explanation:

At the point where the skier loses contact with the snpwball we have the centripetal force given by

m·g·cos θ - N = $m\frac{v^2}{R}$

Where  N = 0 at the point the skier leaves the snowball

That is

 m·g·cos θ  = $m\frac{v^2}{R}$

The height from which the skier drops from the snowball is given by

h = r - r·cosθ

Therefore the potential energy of the skier just before leaveing the snowball is

m·g·h = m·g·r·(1-cosθ)

From conservation of energy, the total energy of the skier is constant which means that is the potential energy is transformed to kinetic energy of the form

PE = KE That is  

$\frac{1}{2} mv^2$ = m·g·r·(1-cosθ) or

$m\frac{v^2}{R}$ = 2·m·g·(1-cosθ) Howerver since   m·g·cos θ  = $m\frac{v^2}{R}$ then we have

m·g·cos θ  = 2·m·g·(1-cosθ)  which gives

cosθ = 2·(1-cosθ) or 3·cosθ = 2

or cosθ =  $\frac{2}{3}$ and θ = $cos^{-1}\frac{2}{3}$ = 48.1896851 °

≈48.2 °

Answer 2

The angle that a radial line from the centre of the snowball to the skier makes with the vertical is $48.18^\circ$.

Newton's Laws of Motion

This problem can be analysed by drawing a free body diagram.

From Newton's second law of motion, we have;

$\sum F = ma$

i.e.; $mg\,cos \theta - N = \frac{mv^2}{R}$

The normal force is a contact force. So as the skier loses contact, $N=0$.

So, the equation becomes;

$mg\,cos \theta = \frac{mv^2}{R}$

$\implies v^2 = gR \,cos\,\theta$

Applying the law of conservation of energy, we get;

$(KE)_i + (PE)_i=(KE)_f + (PE)_f$

But initially, $\theta = 0^\circ$

$\implies v_i = 0$

$\therefore (KE)_i = 0$

So, the equation becomes;

$0 + mgR = \frac{1}{2}mv^2 + mg\,cos\,\theta \,R$

But $v^2 = gR \,cos\,\theta$

$\therefore gR = \frac{1}{2}(gR\,cos\, \theta) + g\,cos\,\theta \,R$

$\implies cos\, \theta=\frac{2}{3}$

$\therefore \theta = 48.18^ \circ$

Learn more about Newton's laws of motion and freebody diagram here:

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