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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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Most popular questions from this chapter

Human feet and legs store elastic energy when walking or running. They are not nearly as efficient at doing so as kangaroo legs, but the effect is significant nonetheless. If not for the storage of elastic energy, a \(70-\mathrm{kg}\) man running at \(4 \mathrm{m} / \mathrm{s}\) would lose about \(100 \mathrm{J}\) of mechanical energy each time he sets down a foot. Some of this energy is stored as elastic energy in the Achilles tendon and in the arch of the foot; the elastic energy is then converted back into the kinetic and gravitational potential energy of the leg, reducing the expenditure of metabolic energy. If the maximum tension in the Achilles tendon when the foot is set down is \(4.7 \mathrm{kN}\) and the tendon's spring constant is $350 \mathrm{kN} / \mathrm{m},$ calculate how far the tendon stretches and how much elastic energy is stored in it.
A \(1500-\mathrm{kg}\) car coasts in neutral down a \(2.0^{\circ}\) hill. The car attains a terminal speed of \(20.0 \mathrm{m} / \mathrm{s} .\) (a) How much power must the engine deliver to drive the car on a level road at $20.0 \mathrm{m} / \mathrm{s} ?$ (b) If the maximum useful power that can be delivered by the engine is \(40.0 \mathrm{kW},\) what is the steepest hill the car can climb at \(20.0 \mathrm{m} / \mathrm{s} ?\)
A 75.0 -kg skier starts from rest and slides down a 32.0 -m frictionless slope that is inclined at an angle of \(15.0^{\circ}\) with the horizontal. Ignore air resistance. (a) Calculate the work done by gravity on the skier and the work done by the normal force on the skier. (b) If the slope is not frictionless so that the skier has a final velocity of \(10.0 \mathrm{m} / \mathrm{s},\) calculate the work done by gravity, the work done by the normal force, the work done by friction, the force of friction (assuming it is constant), and the coefficient of kinetic friction.
The tension in the horizontal towrope pulling a waterskier is \(240 \mathrm{N}\) while the skier moves due west a distance of \(54 \mathrm{m}\). How much work does the towrope do on the water-skier?

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