/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 94 A rubber ball with a radius of \... [FREE SOLUTION] | 91影视

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Chapter 10: Problem 94

A rubber ball with a radius of \(3.2 \mathrm{cm}\) rolls along the horizontal surface of a table with a constant linear speed \(v\). When the ball rolls off the edge of the table, it falls \(0.66 \mathrm{m}\) to the floor below. If the ball completes 0.37 revolution during its fall, what was its linear speed, \(v ?\)

Short Answer

Expert verified
The linear speed of the ball was approximately 0.203 m/s.

Step by step solution

01

Calculate Time of Fall

We use the equation for free fall: \[ s = \frac{1}{2}gt^2 \]where \( s = 0.66 \) m is the distance the ball falls and \( g = 9.8 \) m/s\(^2\) is the acceleration due to gravity. Rearranging the equation to solve for time \( t \):\[ t = \sqrt{\frac{2s}{g}} = \sqrt{\frac{2 \times 0.66}{9.8}} \approx 0.366 \text{ seconds}\]
02

Calculate Distance Rolled

A complete revolution of a ball covers a distance equal to its circumference. Given the radius \( r = 3.2 \) cm, we calculate the circumference \( C \):\[ C = 2\pi r = 2\pi \times 3.2 \times 10^{-2} \text{ m} = 0.2011 \text{ m} \]For 0.37 revolutions, the distance rolled \( d \) is:\[ d = 0.37 \times 0.2011 \approx 0.0744 \text{ m} \]
03

Calculate Linear Speed

Linear speed \( v \) is the distance rolled divided by the time of fall:\[ v = \frac{d}{t} = \frac{0.0744}{0.366} \approx 0.2033 \text{ m/s} \]

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Key Concepts

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

Linear Speed
Linear speed measures how fast an object travels along a path. It differs from angular speed, which describes how fast something rotates. Here, we're focusing on a ball rolling off the table. As it rolls and falls, it travels a certain distance in a given time.
During this process, linear speed is important. It links the distance a ball moves to time taken in a direct line. A simple formula for linear speed is:
  • \( v = \frac{d}{t} \)
In this equation, \( v \) is linear speed, \( d \) is the distance traveled, and \( t \) is the time taken. Understanding linear speed helps solve problems involving motion along a straight path.
Free Fall
Free fall refers to when an object moves only under the influence of gravity. It鈥檚 like when a ball rolls off a table's edge and plummets to the ground. No other forces (like air resistance) are acting significantly on it.
The time taken for an object to hit the ground can be calculated using the formula for free fall:
  • \( s = \frac{1}{2}gt^2 \)
Here, \( s \) is the distance fallen and \( g = 9.8 \: \text{m/s}^2 \) is the acceleration due to gravity. By rearranging this formula, you can solve for the time, \( t \), it takes to fall. Free fall is a fundamental concept for understanding motion in physics without initial vertical speed.
Circumference Calculation
Circumference is the distance around a circle. Calculating it is crucial when dealing with rolling objects, like our rubber ball example.
The formula for finding a circle's circumference is:
  • \( C = 2\pi r \)
where \( r \) is the radius of the circle and \( \pi \) is approximately 3.14159. In our example, the ball rotates, and each revolution equates to its circumference being the traveled distance. When part of a revolution is completed鈥攍ike 0.37 of a full turn鈥攖he traveled distance is that fraction of the circumference. Understanding circumference helps in linking rotational movements to linear distances in real-world examples.

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Most popular questions from this chapter

A person rides on a 12-m-diameter Ferris wheel that rotates at the constant rate of 8.1 rpm. Calculate the magnitude and direction of the force that the seat exerts on a \(65-\mathrm{kg}\) person when he is (a) at the top of the wheel, (b) at the bottom of the wheel, and (c) halfway up the wheel.

A yo-yo moves downward until it reaches the end of its string, where it "sleeps." As it sleeps-that is, spins in place-its angular speed decreases from \(35 \mathrm{rad} / \mathrm{s}\) to \(25 \mathrm{rad} / \mathrm{s}\) During this time it completes 120 revolutions. (a) How long did it take for the yo-yo to slow from \(35 \mathrm{rad} / \mathrm{s}\) to \(25 \mathrm{rad} / \mathrm{s} ?\) (b) How long does it take for the yo-yo to slow from \(25 \mathrm{rad} / \mathrm{s}\) to 15 rad / s? Assume a constant angular acceleration as the yoyo sleeps.

The rotor in a centrifuge has an initial angular speed of \(430 \mathrm{rad} / \mathrm{s} .\) After \(8.2 \mathrm{s}\) of constant angular acceleration, its angular speed has increased to 550 rad/s. During this time, what were (a) the angular acceleration of the rotor and (b) the angle through which it turned?

One of the most studied objects in the night sky is the Crab nebula, the remains of a supernova explosion observed by the Chinese in \(1054 . \ln 1968\) it was discovered that a pulsar-a rapidly rotating neutron star that emits a pulse of radio waves with each revolution-lies near the center of the Crab nebula. The period of this pulsar is 33 ms. What is the angular speed (in rad/s) of the Crab nebula pulsar?

A carousel at the local carnival rotates once every 45 seconds. (a) What is the linear speed of an outer horse on the carousel, which is \(2.75 \mathrm{m}\) from the axis of rotation? (b) What is the linear speed of an inner horse that is \(1.75 \mathrm{m}\) from the axis of rotation?
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