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Chapter 4: Problem 59

A woman rides a carnival Ferris wheel at radius \(15 \mathrm{~m}\), completing five turns about its horizontal axis every minute. What are (a) the period of the motion, the (b) magnitude and (c) direction of her centripetal acceleration at the highest point, and the (d) magnitude and (e) direction of her centripetal acceleration at the lowest point?

Short Answer

Expert verified
(a) 12 seconds (b) 4.11 m/s虏 (c) downwards (d) 4.11 m/s虏 (e) upwards.

Step by step solution

01

Determine the Period of Motion

Given that the Ferris wheel completes five turns every minute, we first find the number of seconds per minute, which is 60 seconds. We then calculate the period \( T \), which is the time taken for one complete revolution. Since there are 5 revolutions in 60 seconds, the period \( T\) is given by: \[ T = \frac{60 \text{ seconds}}{5} = 12 \text{ seconds} \] Thus, the period (a) of the motion is 12 seconds.
02

Calculate the Centripetal Acceleration

The formula for centripetal acceleration \( a_c \) is \( a_c = \frac{v^2}{r} \), where \( v \) is the linear velocity, and \( r \) is the radius. First, we need the linear velocity. The circumference of the Ferris wheel is \( 2\pi r = 2\pi \times 15 = 30\pi \ \text{meters} \). With one revolution taking 12 seconds, the linear velocity \( v \) is given by: \[ v = \frac{30\pi}{12} = \frac{5\pi}{2} \ \text{m/s} \] Now we calculate the centripetal acceleration \( a_c \) at a point on the wheel: \[ a_c = \frac{(\frac{5\pi}{2})^2}{15} = \frac{(25\pi^2 / 4)}{15} = \frac{25\pi^2}{60} = \frac{5\pi^2}{12} \approx 4.11 \ \text{m/s}^2 \] So, the magnitude of centripetal acceleration (b) is approximately \(4.11 \ \text{m/s}^2\).
03

Direction of Centripetal Acceleration at the Highest Point

At the highest point, centripetal acceleration points towards the center of the circle. Since the woman is at the top, the direction of the centripetal acceleration is downwards (c).
04

Direction of Centripetal Acceleration at the Lowest Point

At the lowest point, centripetal acceleration again points towards the center of the circle. However, since the woman is at the bottom, the direction of the centripetal acceleration is upwards (e).

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Key Concepts

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

Period of Motion
The period of motion refers to the time it takes for one complete cycle of movement. In this exercise, it's about how long it takes for the Ferris wheel to make one full revolution. Since the Ferris wheel completes five turns every minute, we start calculating the period by recognizing there are 60 seconds in a minute. This means for every five revolutions, it takes 60 seconds.

To find the period, we need to determine how many seconds one full revolution takes. We divide the total seconds by the number of revolutions:
  • Period, \( T = \frac{60 \text{ seconds}}{5} \)
  • Therefore, \( T = 12 \text{ seconds} \)
Hence, the period of motion in this scenario is 12 seconds. This means every 12 seconds the woman completes one full turn around the Ferris wheel.
Linear Velocity
Linear velocity is the speed at which a point moves along its path. For circular motion like a Ferris wheel, calculating linear velocity involves understanding how fast a point on the wheel's edge is moving in a straight line.

Using the radius of the wheel, 15 meters, we calculate the distance traveled in one revolution鈥攖he circumference:
  • Circumference, \( = 2\pi r = 2\pi \times 15 = 30\pi \text{ meters} \)
Given one revolution takes 12 seconds (the period of motion), the linear velocity \( v \) is:
  • \( v = \frac{30\pi}{12} = \frac{5\pi}{2} \ \text{m/s} \)
This tells us that any point on the edge of the Ferris wheel moves at approximately \( \frac{5\pi}{2} \ \text{m/s} \) along its circular path.
Centripetal Force
Centripetal force is the force that keeps an object moving in a circular path, always directed towards the center of the circle. For the Ferris wheel, this force is responsible for continuously changing the direction of the woman's motion, keeping her in her circular path.

The magnitude of centripetal force can be understood by calculating centripetal acceleration \( a_c \), which is given by the formula:
  • \( a_c = \frac{v^2}{r} \)
Here, we've already calculated the linear velocity as \( \frac{5\pi}{2} \ \text{m/s} \) and the radius \( r \) as 15 meters. Plug these into the equation:
  • \( a_c = \frac{(\frac{5\pi}{2})^2}{15} = \frac{5\pi^2}{12} \ \approx 4.11 \ \text{m/s}^2 \)
The direction varies, depending on the point of the ferris wheel.
  • At the highest point, the direction is downwards towards the center.
  • At the lowest point, centripetal acceleration points upwards, still towards the center.
This force is vital for maintaining circular motion, ensuring the woman stays safely on the Ferris wheel's path.

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Most popular questions from this chapter

An electron having an initial horizontal velocity of magnitude \(1.00 \times 10^{9} \mathrm{~cm} / \mathrm{s}\) travels into the region between two horizontal metal plates that are electrically charged. In that region, the electron travels a horizontal distance of \(2.00 \mathrm{~cm}\) and has a constant downward acceleration of magnitude \(1.00 \times 10^{17} \mathrm{~cm} / \mathrm{s}^{2}\) due to the charged plates. Find (a) the time the electron takes to travel the \(2.00 \mathrm{~cm}\), (b) the vertical distance it travels during that time, and the magnitudes of its (c) horizontal and (d) vertical velocity components as it emerges from the region.

A cat rides a merry-go-round turning with uniform circular motion. At time \(t_{1}=2.00 \mathrm{~s},\) the cat's velocity is \(\vec{v}_{1}=\) \((3.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}+(4.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}},\) measured on a horizontal \(x y\) coordinate system. At \(t_{2}=5.00 \mathrm{~s},\) the cat's velocity is \(\vec{v}_{2}=(-3.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}+\) \((-4.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}} .\) What are (a) the magnitude of the cat's centripetal acceleration and (b) the cat's average acceleration during the time interval \(t_{2}-t_{1},\) which is less than one period?

A baseball is hit at Fenway Park in Boston at a point \(0.762 \mathrm{~m}\) above home plate with an initial velocity of \(33.53 \mathrm{~m} / \mathrm{s}\) directed \(55.0^{\circ}\) above the horizontal. The ball is observed to clear the 11.28 -m-high wall in left field (known as the "green monster") \(5.00 \mathrm{~s}\) after it is hit, at a point just inside the left-field foulline pole. Find (a) the horizontal distance down the left-field foul line from home plate to the wall; (b) the vertical distance by which the ball clears the wall; (c) the horizontal and vertical displacements of the ball with respect to home plate \(0.500 \mathrm{~s}\) before it clears the wall.

The fast French train known as the TGV (Train 脿 Grande Vitesse) has a scheduled average speed of \(216 \mathrm{~km} / \mathrm{h}\). (a) If the train goes around a curve at that speed and the magnitude of the acceleration experienced by the passengers is to be limited to \(0.050 \mathrm{~g}\), what is the smallest radius of curvature for the track that can be tolerated? (b) At what speed must the train go around a curve with a \(1.00 \mathrm{~km}\) radius to be at the acceleration limit?

A carnival merry-go-round rotates about a vertical axis at a constant rate. A man standing on the edge has a constant speed of \(3.66 \mathrm{~m} / \mathrm{s}\) and a centripetal acceleration \(\vec{a}\) of magnitude \(1.83 \mathrm{~m} / \mathrm{s}^{2}\) Position vector \(\vec{r}\) locates him relative to the rotation axis. (a) What is the magnitude of \(\vec{r} ?\) What is the direction of \(\vec{r}\) when \(\vec{a}\) is directed (b) due east and (c) due south?
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