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Chapter 1: Problem 10

The radius of a car wheel is 15 inches. If the car is traveling 60 miles per hour, what is the angular velocity of the wheel in radians per minute? How fast is the wheel spinning in revolutions per minute?

Short Answer

Expert verified
The linear velocity of the car wheel in inches per minute is \(60 \text{ mph} \times\frac{5280 \text{ ft}}{1 \text{ mile}}\times\frac{12 \text{ in}}{1 \text{ ft}}\times\frac{1 \text{ hour}}{60 \text{ min}} = 63360 \text{ in/min}\). The angular velocity in radians per minute is \(\omega = \frac{v}{r} = \frac{63360 \text{ in/min}}{15 \text{ in}} = 4224 \text{ rad/min}\). Finally, the spinning speed in revolutions per minute is \(= \frac{\omega}{2\pi} = \frac{4224 \text{ rad/min}}{2\pi} \approx 672.34 \text{ rev/min}\). So, the wheel's angular velocity is 4224 radians per minute and its spinning speed is approximately 672.34 revolutions per minute.

Step by step solution

01

Convert linear velocity to inches per minute

First, we need to convert the car's linear velocity given in miles per hour to inches per minute. To do this, we can use the conversion factors: 1 mile = 5280 feet and 1 foot = 12 inches. We also need to convert hours to minutes. Linear velocity = 60 miles per hour Convert to inches per minute: \(60 \text{ miles/hour}\times\frac{5280 \text{ feet}}{1 \text{ mile}}\times\frac{12 \text{ inches}}{1 \text{ foot}}\times\frac{1 \text{ hour}}{60 \text{ minute}}\) Now, we perform the calculation to get the linear velocity in inches per minute.
02

Find the angular velocity in radians per minute

The relationship between linear velocity (v), angular velocity (蠅), and radius (r) of a circle can be expressed as: \(v = \omega 脳 r\) In this case, the linear velocity v is the value we found in Step 1, and the radius r = 15 inches. We want to find the angular velocity 蠅 in radians per minute. To find the angular velocity, we can rearrange the equation to solve for 蠅: \(\omega = \frac{v}{r}\) Substitute the values and perform the calculation:
03

Convert the angular velocity to revolutions per minute

Now that we have the angular velocity in radians per minute, we want to find the spinning speed in revolutions per minute. To do this, we need to convert radians to revolutions. There are \(2\pi\) radians in one revolution, so we can use this conversion factor to find the number of revolutions per minute: Spinning speed \(= \frac{\omega}{2\pi}\) Substitute the value of 蠅 found in Step 2 and perform the calculation to get the spinning speed in revolutions per minute.

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Key Concepts

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

Linear Velocity
Linear velocity is about how fast something is moving along a path. Imagine a car cruising down the highway. The speed you see on the car's dashboard is its linear velocity. It is measured in straightforward distance over time, such as miles per hour or meters per second. In our exercise, we started with the car traveling at 60 miles per hour, a form of linear velocity.

To solve our problem, we needed this speed in inches per minute. This required converting through various units:
  • First, miles are changed into feet using 1 mile = 5280 feet.
  • Next, feet are converted into inches via 1 foot = 12 inches.
  • The time needs adjusting too, from hours into minutes.
This conversion process helps us define how fast something is moving around the circle of the wheel in a very detailed way.
Revolutions per Minute
Revolutions per minute (RPM) tell us how many complete turns something makes in a minute. Think about a wheel spinning; each full circle it makes is a revolution. When solving our problem, we first needed angular velocity and then translated it into RPM.

To do this translation, we consider that there are 2蟺 radians in one full circle. Therefore, using the conversion factor \(\frac{\omega}{2\pi}\), where 蠅 represents angular velocity in radians per minute, we can find the number of revolutions per minute.
This gives a clearer picture of the spinning speed of our car's wheel. RPM is a very common measure in automotive contexts and helps us easily communicate rotational speeds.
Radians per Minute
Radians per minute is a way to measure how fast something is rotating. It is important because it provides insight into the rotational speed in terms of angular distance covered in a given time.

In our exercise, we needed to find the angular velocity of a car wheel. The relationship between angular velocity (蠅), linear velocity (v), and radius (r) is crucial: \(v = \omega \times r\).
Given the linear velocity and the wheel's radius, we can find angular velocity using:\(\omega = \frac{v}{r}\).

The measure in radians per minute is essential in many physics and engineering applications because it directly relates the geometry of a circle to the speed of rotation, highlighting fundamental circular motion principles.

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Most popular questions from this chapter

(a) For a real number \(t,\) the value of \(\sin (t)\) is defined to be the coordinate of the _____ point of an arc \(t\) whose initial point is _______ on the ______ whose equation is \(x^{2}+\) \(y^{2}=1\) (b) The domain of the sine function is _____. (c) The maximum value of \(\sin (t)\) is ______ and this occurs at \(t=\)______ for \(0 \leq t<2 \pi\). The minimum value of \(\sin (t)\) is ______ \(2 \pi\) and this occurs at \(t=\)______ for \(0 \leq t<\) (d) The range of the sine function is _____

The mean distance from Earth to the moon is 238,857 miles. Assuming the orbit of the moon about Earth is a circle with a radius of 238,857 miles and that the moon makes one revolution about Earth every 27.3 days, determine the linear velocity of the moon in miles per hour. Research the distance of the moon to Earth and explain why the computations that were just made are approximations.

Determine the quadrant that contains the terminal point of each given arc with initial point (1,0) on the unit circle. (a) \(\frac{7 \pi}{4}\) (b) \(-\frac{7 \pi}{4}\) (c) \(\frac{3 \pi}{5}\) (d) \(\frac{-3 \pi}{5}\) (e) \(\frac{7 \pi}{3}\) (f) \(\frac{-7 \pi}{3}\) (g) \(\frac{5 \pi}{8}\) (h) \(\frac{-5 \pi}{8}\) (i) 2.5 (j) -2.5 (k) 3 (1) \(3+2 \pi\) (m) \(3-\pi\) (n) \(3-2 \pi\)

This exercise provides an alternate method for determining the exact values of \(\cos \left(\frac{\pi}{6}\right)\) and \(\sin \left(\frac{\pi}{6}\right)\). The diagram to the right shows the terminal point \(P(x, y)\) for an arc of length \(t=\frac{\pi}{6}\) on the unit circle. The points \(A(1,0)\), \(B(0,1),\) and \(C(x,-y)\) are also shown. Notice that \(B\) is the terminal point of the \(\operatorname{arc} t=\frac{\pi}{2},\) and \(C\) is the terminal point of the arc \(t=-\frac{\pi}{6}\). We now notice that the length of the arc from \(P\) to \(B\) is $$ \frac{\pi}{2}-\frac{\pi}{6}=\frac{\pi}{3} $$ In addition, the length of the arc from \(C\) to \(P\) is $$ \frac{\pi}{6}-\frac{-\pi}{6}=\frac{\pi}{3} $$ This means that the distance from \(P\) to \(B\) is equal to the distance from \(C\) to \(P\) (a) Use the distance formula to write a formula (in terms of \(x\) and \(y\) ) for the distance from \(P\) to \(B\). (b) Use the distance formula to write a formula (in terms of \(x\) and \(y\) ) for the distance from \(C\) to \(P\). (c) Set the distances from (a) and (b) equal to each other and solve the resulting equation for \(y\). To do this, begin by squaring both sides of the equation. In order to solve for \(y\), it may be necessary to use the fact that \(x^{2}+y^{2}=1\) (d) Use the value for \(y\) in (c) and the fact that \(x^{2}+y^{2}=1\) to determine the value for \(x\). Explain why this proves that $$\cos \left(\frac{\pi}{6}\right)=\frac{\sqrt{3}}{2} \text { and } \sin \left(\frac{\pi}{3}\right)=\frac{1}{2}$$.

In each of the following, when it is possible, determine the exact measure of central the angle in degrees. Otherwise, round to the nearest hundredth of a degree. (a) The central angle that intercepts an arc of length \(3 \pi\) feet on a circle of radius 5 feet. (b) The central angle that intercepts an arc of length 18 feet on a circle of radius 5 feet. (c) The central angle that intercepts an arc of length 20 meters on a circle of radius 12 meters. (d) The central angle that intercepts an arc of length 5 inches on a circle of radius 5 inches. (e) The central angle that intercepts an arc of length 12 inches on a circle of radius 5 inches.
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