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Chapter 6: Problem 99

A skier starts from rest at the top of a frictionless slope of ice in the shape of a hemispherical dome with radius \(R\) and slides down the slope. At a certain height \(h,\) the normal force becomes zero and the skier leaves the surface of the ice. What is \(h\) in terms of \(R ?\)

Short Answer

Expert verified
Answer: The skier leaves the surface of the ice at a height of \(h = \frac{R}{3}\), where R is the radius of the hemisphere.

Step by step solution

01

Analyze the forces on the skier

When the skier is at height h, there are two forces acting on the skier: the gravitational force (mg) and the normal force (N). At the point where the skier leaves the surface of the ice, the normal force becomes zero (N = 0). At this point, the gravitational force provides the required centripetal force for the circular motion, so we can write: \(mv^2 / r = mg\).
02

Conservation of energy

As the skier slides down the slope, the gravitational potential energy is converted into kinetic energy. We can use the conservation of energy principle to write the equation: \(mgh = \frac{1}{2}mv^2\), where h is the height at which the skier leaves the surface, and v is the velocity of the skier at height h.
03

Relate height and radius

Since the skier is moving along the hemispherical dome, we can relate the height and the radius of the curvature using the equation: \(r = R - h\).
04

Solve for velocity

From the energy conservation equation, we can solve for the velocity of the skier when the normal force becomes zero: \(v^2 = 2gh\).
05

Substitute velocity and radius

Now, we substitute the expressions for the velocity and radius in the first equation (centripetal force): \(m(2gh) / (R - h) = mg\).
06

Solve for height

Cancelling out the mass (m) and gravitational acceleration (g) from the equation, we get: \(2h = R - h\). Solving for h, we find: \(h = \frac{R}{3}\). So, the skier leaves the surface of the ice at a height of \(h = \frac{R}{3}\).

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