Log out Go to App
Go to App

Learning Materials

Features

Discover

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 \(h = \frac{R}{3}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Josie and Charlotte push a 12 -kg bag of playground sand for a sandbox on a frictionless, horizontal, wet polyvinyl surface with a constant, horizontal force for a distance of \(8.0 \mathrm{m},\) starting from rest. If the final speed of the sand bag is \(0.40 \mathrm{m} / \mathrm{s},\) what is the magnitude of the force with which they pushed?
(a) How much work does a major-league pitcher do on the baseball when he throws a \(90.0 \mathrm{mi} / \mathrm{h}(40.2 \mathrm{m} / \mathrm{s})\) fastball? The mass of a baseball is 153 g. (b) How many fastballs would a pitcher have to throw to "burn off' a 1520 -Calorie meal? (1 Calorie \(=1000\) cal \(=1\) kcal.)Assume that \(80.0 \%\) of the chemical energy in the food is converted to thermal energy and only \(20.0 \%\) becomes the kinetic energy of the fastballs.
A block of wood is compressed 2.0 nm when inward forces of magnitude $120 \mathrm{N}$ are applied to it on two opposite sides. (a) Assuming Hooke's law holds, what is the effective spring constant of the block? (b) Assuming Hooke's law still holds, how much is the same block compressed by inward forces of magnitude \(480 \mathrm{N} ?\) (c) How much work is done by the applied forces during the compression of part (b)?
Show that 1 kilowatt-hour (kW-h) is equal to 3.6 MJ.
An airline executive decides to economize by reducing the amount of fuel required for long-distance flights. He orders the ground crew to remove the paint from the outer surface of each plane. The paint removed from a single plane has a mass of approximately \(100 \mathrm{kg} .\) (a) If the airplane cruises at an altitude of \(12000 \mathrm{m},\) how much energy is saved in not having to lift the paint to that altitude? (b) How much energy is saved by not having to move that amount of paint from rest to a cruising speed of $250 \mathrm{m} / \mathrm{s} ?$
See all solutions

Recommended explanations on Physics Textbooks

Engineering Physics

Read Explanation

Modern Physics

Read Explanation

Oscillations

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Scientific Method Physics

Read Explanation

Famous Physicists

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free