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Chapter 3: Problem 32

A roller coaster car moves in a vertical circle of radius \(R\). At the top of the circle the car has speed \(v_{1}\), and at the bottom of the circle it has speed \(v_{2},\) where \(v_{2}>v_{1} .\) (a) When the car is at the top of its circular path, what is the direction of its radial acceleration, \(a_{\mathrm{rad}, \text { top }} ?\) (b) When the car is at the bottom of its circular path, what is the direction of its radial acceleration, \(a_{\mathrm{rad}, \text { bottom }} ?\) (c) In terms of \(v_{1}\) and \(v_{2}\), what is the ratio \(a_{\text {rad, bottom }} / a_{\text {rad, top }} ?\)

Short Answer

Expert verified
The direction of radial acceleration at the top is downward and at the bottom is upward. The ratio of the radial accelerations at the bottom and the top is \(\frac{v_{2}^{2}}{v_{1}^{2}}\).

Step by step solution

01

Understand radial acceleration

By definition, radial acceleration is the component of acceleration of a body moving in a circular path that points towards the center of the circle on which the body is moving. The magnitude of radial acceleration, also called centripetal acceleration, can be calculated using formula \(a_{\text{rad}} = \frac{v^2}{r}\), where \(v\) is the speed of the body and \(r\) is the radius of the circle.
02

Determine the direction of radial acceleration at the top and bottom points

(a) At the top of the vertical circular path, the direction of the radial acceleration (\(a_{\text{rad, top}}\)) is downward, towards the center of the path. (b) At the bottom of the circular path, the direction of the radial acceleration (\(a_{\text{rad, bottom}}\)) is upward, towards the center of the path.
03

Find the ratio of radial accelerations

(c) The ratio of the radial accelerations at the bottom and the top can be found by applying the formula for radial acceleration. So, \(a_{\text{rad, bottom}}/a_{\text{rad, top}} = \frac{v_{2}^{2}/R}{v_{1}^{2}/R} = \frac{v_{2}^{2}}{v_{1}^{2}}\). This means the ratio of the radial accelerations is equal to the ratio of the squares of the speeds.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Centripetal Acceleration
is the radius of the circular path. Constant speed does not mean the object is not accelerating; the constant change in direction means there鈥檚 a continuous change in velocity, thus an acceleration. For every point along the circular path, the acceleration vector points directly towards the center, illustrating the radial nature of this acceleration.
Circular Motion
Circular motion refers to the movement of an object along a circular path. This motion is typically uniform if the speed of the object remains constant and it follows the circumference of the circle smoothly. During circular motion, two forces are acting upon the object: a centripetal force that pulls it towards the center, and a centrifugal force, which is the object's inertia, that acts in the opposite direction.

Despite often feeling like we're being pushed outwards when going around a curve, what we're really experiencing is our body's resistance to the change in direction, as the centripetal force acts towards the center. Understanding the balance of these forces is key to analyzing problems involving circular motion, such as turning cars, orbiting satellites, and spinning rides at amusement parks.
Motion in a Vertical Circle
The motion in a vertical circle, like a roller coaster loop, adds complexity due to the influence of gravity. At different points in the loop, the object experiences changes in speed due to gravitational potential energy converting into kinetic energy and vice versa.

At the top of the circle, the object has less kinetic energy and moves slower, meaning the centripetal acceleration is smaller compared to the bottom of the circle where the object moves faster due to increased kinetic energy. Thus, gravity affects the speed (and therefore the radial acceleration) of the object differently at various points in the loop.
Acceleration Ratio in Circular Motion
. This clearly illustrates that the acceleration at the bottom of the loop will be greater due to the higher speed, fundamentally affecting how riders feel at different points of a roller coaster.

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Most popular questions from this chapter

Canada geese migrate essentially along a north-south direction for well over a thousand kilometers in some cases, traveling at speeds up to about \(100 \mathrm{~km} / \mathrm{h}\). If one goose is flying at \(100 \mathrm{~km} / \mathrm{h}\) relative to the air but a \(40 \mathrm{~km} / \mathrm{h}\) wind is blowing from west to east, (a) at what angle relative to the north-south direction should this bird head to travel directly southward relative to the ground? (b) How long will it take the goose to cover a ground distance of \(500 \mathrm{~km}\) from north to south? (Note: Even on cloudy nights, many birds can navigate by using the earth's magnetic field to fix the northsouth direction.)

Henrietta is jogging on the sidewalk at \(3.05 \mathrm{~m} / \mathrm{s}\) on the way to her physics class. Bruce realizes that she forgot her bag of bagels, so he runs to the window, which is \(38.0 \mathrm{~m}\) above the street level and directly above the sidewalk, to throw the bag to her. He throws it horizontally \(9.00 \mathrm{~s}\) after she has passed below the window, and she catches it on the run. Ignore air resistance. (a) With what initial speed must Bruce throw the bagels so that Henrietta can catch the bag just before it hits the ground? (b) Where is Henrietta when she catches the bagels?

A physics professor did daredevil stunts in his spare time. His last stunt was an attempt to jump across a river on a motorcycle (Fig. \(\mathbf{P 3 . 6 3 )}\). The takeoff ramp was inclined at \(53.0^{\circ},\) the river was \(40.0 \mathrm{~m}\) wide, and the far bank was \(15.0 \mathrm{~m}\) lower than the top of the ramp. The river itself was \(100 \mathrm{~m}\) below the ramp. Ignore air resistance. (a) What should his speed have been at the top of the ramp to have just made it to the edge of the far bank? (b) If his speed was only half the value found in part (a), where did he land?

A Ferris wheel with radius \(14.0 \mathrm{~m}\) is turning about a horizontal axis through its center (Fig. E3.31). The linear speed of a passenger on the rim is constant and equal to \(6.00 \mathrm{~m} / \mathrm{s}\) What are the magnitude and direction of the passenger's acceleration as she passes through (a) the lowest point in her circular motion and (b) the highest point in her circular motion? (c) How much time does it take the Ferris wheel to make one revolution?

At \(t=0\) a rock is projected from ground level with a speed of \(15.0 \mathrm{~m} / \mathrm{s}\) and at an angle of \(53.0^{\circ}\) above the horizontal. Neglect air resistance. At what two times \(t\) is the rock \(5.00 \mathrm{~m}\) above the ground? At each of these two times, what are the horizontal and vertical components of the velocity of the rock? Let \(v_{0 x}\) and \(v_{0 y}\) be in the positive \(x\) - and \(y\) -directions, respectively.
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