Suppose you are riding on a roller coaster, which moves through a vertical circular loop. Show that your apparent weight at the bottom of the loop is six times your weight when you experience weightlessness at the top, independent of the size of the loop. Assume that friction is negligible.

Short Answer

Expert verified
Question: Show that the apparent weight of a rider at the bottom of a circular vertical loop on a roller coaster is six times their weight when they experience weightlessness at the top, regardless of the loop's size. Answer: The apparent weight at the bottom of the loop is twice the weight when the rider experiences weightlessness at the top, not six times. This ratio is independent of the loop's size and is equal to 2, not 6.

Step by step solution

01

Circular motion and centripetal force

Since the roller coaster is moving in a circular loop, its motion is a combination of a vertical and a horizontal movement. It experiences a centripetal force towards the center of the loop, which must balance the gravitational force when an object is in weightlessness at the top of the loop. The centripetal force F_c acting on a body of mass m and moving with a speed v in a circle of radius r can be expressed as: \[F_c = mr\omega^2\] where \(\omega\) is the angular velocity of the body.
02

Considering the forces at the bottom of the loop

At the bottom of the loop, two forces act on the rider: the gravitational force acting downward and the upward centripetal force provided by the rails, which together make up the apparent weight. Summing the forces acting on the rider, we get: \[F_{app,bottom} = F_c + mg\] where \(F_{app,bottom}\) is the apparent weight at the bottom of the loop and mg is the gravitational force.
03

Considering the forces at the top of the loop

At the top of the loop, the rider experiences weightlessness, which means the downward gravitational force must be equal to the upward centripetal force: \[F_c = mg\]
04

Substituting centripetal force in step 2

From step 3, we get the value of centripetal force \(F_c = mg\), substituting it in the expression obtained in step 2, we get: \[F_{app,bottom} = mg + mg\] \[F_{app,bottom} = 2mg\]
05

Calculating the ratio of apparent weight at the bottom and top

We found that the apparent weight at the bottom is \(2mg\), and at the top, the rider experiences weightlessness (\(mg\)). Now we can calculate the ratio of the apparent weight at the bottom(\(F_{app,bottom}\)) to the apparent weight at the top(\(F_{app,top}\)): \[\frac{F_{app,bottom}}{F_{app,top}} = \frac{2mg}{mg}\]
06

Final result

Simplifying the expression from Step 5, we get: \[\frac{F_{app,bottom}}{F_{app,top}} = 2\] The apparent weight at the bottom of the loop is twice the weight when the rider experiences weightlessness at the top. This contradicts the claim that it is six times, showing that the original statement in the exercise is incorrect. The ratio is indeed independent of the loop's size and is equal to 2, not 6.

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.

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

A ball having a mass of \(1.00 \mathrm{~kg}\) is attached to a string \(1.00 \mathrm{~m}\) long and is whirled in a vertical circle at a constant speed of \(10.0 \mathrm{~m} / \mathrm{s}\) a) Determine the tension in the string when the ball is at the top of the circle. b) Determine the tension in the string when the ball is at the bottom of the circle. c) Consider the ball at some point other than the top or bottom. What can you say about the tension in the string at this point?

Two skaters, \(A\) and \(B,\) of equal mass are moving in clockwise uniform circular motion on the ice. Their motions have equal periods, but the radius of skater A's circle is half that of skater B's circle a) What is the ratio of the speeds of the skaters? b) What is the ratio of the magnitudes of the forces acting on each skater?

A race car is making a U-turn at constant speed. The coefficient of friction between the tires and the track is \(\mu_{\mathrm{s}}=1.20 .\) If the radius of the curve is \(10.0 \mathrm{~m},\) what is the maximum speed at which the car can turn without sliding? Assume that the car is performing uniform circular motion.

In a department store toy display, a small disk (disk 1) of radius \(0.100 \mathrm{~m}\) is driven by a motor and turns a larger disk (disk 2) of radius \(0.500 \mathrm{~m}\). Disk 2 , in turn, drives disk 3 , whose radius is \(1.00 \mathrm{~m}\). The three disks are in contact and there is no slipping. Disk 3 is observed to sweep through one complete revolution every \(30.0 \mathrm{~s}\) a) What is the angular speed of disk \(3 ?\) b) What is the ratio of the tangential velocities of the rims of the three disks? c) What is the angular speed of disks 1 and \(2 ?\) d) If the motor malfunctions, resulting in an angular acceleration of \(0.100 \mathrm{rad} / \mathrm{s}^{2}\) for disk 1 , what are disks 2 and 3's angular accelerations?

Mars orbits the Sun at a mean distance of 228 million \(\mathrm{km},\) in a period of 687 days. The Earth orbits at a mean distance of 149.6 million \(\mathrm{km},\) in a period of 365.26 days. a) Suppose Earth and Mars are positioned such that Earth lies on a straight line between Mars and the Sun. Exactly 365.26 days later, when the Earth has completed one orbit, what is the angle between the Earth-Sun line and the Mars-Sun line? b) The initial situation in part (a) is a closest approach of Mars to Earth. What is the time, in days, between two closest approaches? Assume constant speed and circular orbits for both Mars and Earth. c) Another way of expressing the answer to part (b) is in terms of the angle between the lines drawn through the Sun, Earth, and Mars in the two closest approach situations. What is that angle?

See all solutions

Recommended explanations on Physics Textbooks

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