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Chapter 4: Problem 29

A tire \(0.500 \mathrm{m}\) in radius rotates at a constant rate of 200 rev/min. Find the speed and acceleration of a small stone lodged in the tread of the tire (on its outer edge).

Short Answer

Expert verified
The speed of the stone is approximately \(10.47 \: m/s\) and the acceleration is approximately \(209.44 \: m/s^2\).

Step by step solution

01

Convert Rotational Speed to Radians Per Second

The rotational speed is given in revolutions per minute. To use this speed in calculations involving the circumference of a circle or the radius (since the stone is on the circumference of the tire), it needs to be converted into radians per second. Use the formula: \[\omega=\frac{(200 \: rev/min) * 2\pi}{60 s/min}= \frac{20\pi}{3} \: rad/s\]
02

Calculate the Speed of the Stone

The speed of the stone (assuming it is stuck to the tread) will be the same as the speed of the point on the outer edge of the tire. The linear speed (v) can be calculated using the formula: \(v=r*\omega\). Substituting the given values, \[v=(0.500 \: m)*\frac{20\pi}{3} \: rad/s = \frac{10\pi}{3} \: m/s\]
03

Calculate the Acceleration of the Stone

The linear acceleration (a) of the stone can be calculated using the formula: \(a=r*\omega^{2}\) . Substituting the given values, \[a=(0.500 \: m)*(\frac{20\pi}{3} \: rad/s)^{2}=\frac{200\pi^{2}}{3} \: m/s^2\]

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Key Concepts

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

Understanding Angular Velocity
Angular velocity is a term used in rotational motion to describe how fast an object spins around a central point. In the context of the exercise, the tire is rotating, and we need to know how quickly it turns. Angular velocity is usually expressed in radians per second. This unit might seem odd at first, but it鈥檚 essential for expressing how an object rotates. Radians are a type of angle measurement that relate to the radius of a circle. To convert the rotational speed from revolutions per minute to radians per second, you use the formula:
  • Convert revolutions to radians by multiplying with \(2\pi\), since a complete revolution is \(2\pi\) radians.
  • Convert minutes to seconds, knowing there are 60 seconds in a minute.
So 200 revolutions per minute becomes \(\frac{20\pi}{3}\) radians per second.
Exploring Linear Speed
Linear speed, often just called speed, tells us how fast an object is moving through space. It's slightly different from angular velocity because it focuses on straight-line movement, rather than rotation. For the stone in the tire tread, its linear speed is simply how fast it would be moving if you were watching it fly past in its circular track.To determine linear speed from angular velocity, you multiply the angular velocity by the radius of the path. The formula is \(v = r\omega\), where:
  • \(v\) is the linear speed
  • \(r\) is the radius of the circle
  • \(\omega\) is the angular velocity
From our example, a radius of 0.5 meters and an angular velocity of \(\frac{20\pi}{3}\) rad/s gives a linear speed of \(\frac{10\pi}{3}\) m/s. This means the stone is zipping around the tire track at this impressive speed!
Demystifying Tangential Acceleration
Tangential acceleration is all about how quickly something鈥檚 speed is changing as it spins. Even if the overall rotation speed doesn't change, a point on the tire is constantly changing its direction, making this concept crucial.Tangential acceleration is calculated using the formula \(a = r\omega^2\), where:
  • \(a\) is the tangential acceleration
  • \(r\) is the radius of the circle
  • \(\omega\) is the angular velocity
In our problem, with a radius of 0.5 meters and an angular velocity of \(\frac{20\pi}{3}\) radians per second, the acceleration becomes \(\frac{200\pi^2}{3}\) m/s虏. This value indicates the rate of change of speed for the stone, attached to the tire's edge, as it follows the circular path.

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Most popular questions from this chapter

An automobile whose speed is increasing at a rate of \(0.600 \mathrm{m} / \mathrm{s}^{2}\) travels along a circular road of radius \(20.0 \mathrm{m} .\) When the instantaneous speed of the automobile is \(4.00 \mathrm{m} / \mathrm{s},\) find \((\mathrm{a})\) the tangential acceleration component, (b) the centripetal acceleration component, and (c) the magnitude and direction of the total acceleration.

Heather in her Corvette accelerates at the rate of \((3.00 \hat{\mathbf{i}}-2.00 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s}^{2},\) while Jill in her Jaguar accelerates at \((1.00 \hat{\mathbf{i}}+3.00 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s}^{2} .\) They both start from rest at the origin of an \(x y\) coordinate system. After \(5.00 \mathrm{s},\) (a) what is Heather's speed with respect to Jill, (b) how far apart are they, and (c) what is Heather's acceleration relative to Jill?

One strategy in a snowball fight is to throw a snowball at a high angle over level ground. While your opponent is watching the first one, a second snowball is thrown at a low angle timed to arrive before or at the same time as the first one. Assume both snowballs are thrown with a speed of \(25.0 \mathrm{m} / \mathrm{s} .\) The first one is thrown at an angle of \(70.0^{\circ}\) with respect to the horizontal. (a) At what angle should the second snowball be thrown to arrive at the same point as the first? (b) How many seconds later should the second snowball be thrown after the first to arrive at the same time?

A fisherman sets out upstream from Metaline Falls on the Pend Oreille River in northwestern Washington State. His small boat, powered by an outboard motor, travels at a constant speed \(v\) in still water. The water flows at a lower constant speed \(v_{w}\) He has traveled upstream for \(2.00 \mathrm{km}\) when his ice chest falls out of the boat. He notices that the chest is missing only after he has gone upstream for another 15.0 minutes. At that point he turns around and heads back downstream, all the time traveling at the same speed relative to the water. He catches up with the floating ice chest just as it is about to go over the falls at his starting point. How fast is the river flowing? Solve this problem in two ways. (a) First, use the Earth as a reference frame. With respect to the Earth, the boat travels upstream at speed \(v-v_{w}\) and downstream at \(v+v_{w \cdot}\) (b) \(\mathrm{A}\) second much simpler and more elegant solution is obtained by using the water as the reference frame. This approach has important applications in many more complicated problems; examples are calculating the motion of rockets and satellites and analyzing the scattering of subatomic particles from massive targets. PICTURE CANT COPY

A train slows down as it rounds a sharp horizontal turn, slowing from \(90.0 \mathrm{km} / \mathrm{h}\) to \(50.0 \mathrm{km} / \mathrm{h}\) in the \(15.0 \mathrm{s}\) that it takes to round the bend. The radius of the curve is \(150 \mathrm{m} .\) Compute the acceleration at the moment the train speed reaches \(50.0 \mathrm{km} / \mathrm{h} .\) Assume it continues to slow down at this time at the same rate.
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