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Chapter 9: Problem 1

(a) What angle in radians is subtended by an arc \(1.50 \mathrm{~m}\) long on the circumference of a circle of radius \(2.50 \mathrm{~m} ?\) What is this angle in degrees? (b) An arc \(14.0 \mathrm{~cm}\) long on the circumference of a circle subtends an angle of \(128^{\circ} .\) What is the radius of the circle? (c) The angle between two radii of a circle with radius \(1.50 \mathrm{~m}\) is 0.700 rad. What length of arc is intercepted on the circumference of the circle by the two radii?

Short Answer

Expert verified
The angle of the given arc in the first circle is 0.60 rad or 34.38掳. The radius of the second circle is 6.25 cm. The arc length intercepted by two radii on the third circle is 1.05 m.

Step by step solution

01

Calculating Angle in Radians for the First Circle

Given the arc length \(1.50 \mathrm{~m}\) and radius \(2.50 \mathrm{~m}\), using the formula Arc Length = Radius * Angle, we get Angle = Arc Length / Radius = \(1.50 \mathrm{~m} / 2.50 \mathrm{~m} = 0.60\) rad.
02

Converting the Angle from Radians to Degrees

The relationship between degrees and radians is \(180^\circ = \pi\) radians. So, the angle in degrees is \(0.60 * 180/ \pi = 34.38^\circ\).
03

Finding the Radius of the Second Circle

Given the arc length \(14.0 \mathrm{~cm}\) and angle \(128^\circ\), first convert the angle to radians using the formula: Radians = Degrees * \(\pi /180\). So we get \(128^\circ = 2.24\) rad. Using the formula from step 1, we find the radius = Arc Length / Angle = \(14.0 \mathrm{~cm} / 2.24\) rad \(= 6.25 \mathrm{~cm}\).
04

Calculating Arc Length Intercepted by Two Radii on Third Circle

Given the angle between two radii is 0.700 rad and the radius is \(1.50 \mathrm{~m}\), now use the formula Arc Length = Radius * Angle to find the intercepted arc length. So, the arc length = \(1.50 \mathrm{~m} * 0.700\) rad \(= 1.05 \mathrm{~m}\).

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Key Concepts

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

Radians to Degrees Conversion
Understanding the relationship between radians and degrees is crucial when working with angles in circle geometry. While degrees are commonly used in everyday applications, radians are a fundamental unit in mathematics and physics.

One complete revolution around a circle is equal to 360 degrees or \(2\pi\) radians. Thus, to convert an angle from radians to degrees, you would use the proportion: \[ \text{Degrees} = \text{Radians} \times \frac{180}{\pi} \].

This formula comes in handy when you have an angle in radians and need to express it in degrees鈥攁 common requirement in various math and engineering tasks. For example, if you have an angle of 0.60 radians and want to convert it to degrees, you would calculate \(0.60 \times \frac{180}{\pi}\), giving you approximately 34.38 degrees.

Remember, since \(\pi\) is an irrational number, the decimal value in degrees will often be an approximation.
Calculating Arc Length
The arc length of a circle is the distance along the curved line that makes up the arc. It can be found if you know the radius of the circle and the angle the arc subtends at the center, measured in radians.

The formula to calculate the arc length is: \[ \text{Arc Length} = \text{Radius} \times \text{Angle (in radians)} \].

This formula is derived from the circumference of a circle, which is \(2\pi r\), where \(r\) is the radius. Since the circumference represents the arc length for an angle of \(2\pi\) radians (a full revolution), for a smaller angle, you simply multiply the radius by the angle in radians to get the proportionate arc length.

For instance, to calculate the arc length intercepted by an angle of 0.700 radians on a circle with a radius of 1.50 meters, you would do \(1.50 \times 0.700\) to find an arc length of 1.05 meters.
Circle Radius Calculation
In some problems, you might need to solve for the radius of a circle when given the arc length and the angle it subtends. To isolate the radius in the arc length formula, you rearrange it to \[ \text{Radius} = \frac{\text{Arc Length}}{\text{Angle (in radians)}} \].

When the angle is given in degrees, remember to first convert it to radians before performing the calculation. Returning to the example from the textbook solutions, if an arc of 14.0 cm long subtends an angle of 128 degrees, convert the angle to radians by multiplying 128 degrees by \(\frac{\pi}{180}\), which gives approximately 2.24 radians. Then, the radius is computed as \(\frac{14.0 \text{ cm}}{2.24}\) resulting in a radius of about 6.25 cm.

Knowing how to calculate the radius from arc length and the subtended angle is especially useful in fields like architecture, engineering, and even in crafting items like jewelry or constructing segments of race tracks where precise curvature is required.

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Most popular questions from this chapter

\(9.88^{\circ}\) DATA You are analyzing the motion of a large flywheel that has radius \(0.800 \mathrm{~m}\). In one test run, the wheel starts from rest and turns in a horizontal plane with constant angular acceleration. An accelerometer on the rim of the flywheel measures the magnitude of the resultant acceleration \(a\) of a point on the rim of the flywheel as a function of the angle \(\theta-\theta_{0}\) through which the wheel has turned. You collect these results: $$ \begin{array}{l|cccccccc} \boldsymbol{\theta}-\boldsymbol{\theta}_{\mathbf{0}}(\mathbf{r a d}) & 0.50 & 1.00 & 1.50 & 2.00 & 2.50 & 3.00 & 3.50 & 4.00 \\ \hline \boldsymbol{a}\left(\mathbf{m} / \mathbf{s}^{\mathbf{2}}\right) & 0.678 & 1.07 & 1.52 & 1.98 & 2.45 & 2.92 & 3.39 & 3.87 \end{array} $$ Construct a graph of \(a^{2}\left(\right.\) in \(\left.\mathrm{m}^{2} / \mathrm{s}^{4}\right)\) versus \(\left(\theta-\theta_{0}\right)^{2}\) (in rad \(^{2}\) ). (a) What are the slope and \(y\) -intercept of the straight line that gives the best fit to the data? (b) Use the slope from part (a) to find the angular acceleration of the flywheel. (c) What is the linear speed of a point on the rim of the flywheel when the wheel has turned through an angle of \(135^{\circ} ?\) (d) When the flywheel has turned through an angle of \(90.0^{\circ},\) what is the angle between the linear velocity of a point on its rim and the resultant acceleration of that point?

A circular saw blade with radius \(0.120 \mathrm{~m}\) starts from rest and turns in a vertical plane with a constant angular acceleration of \(2.00 \mathrm{rev} / \mathrm{s}^{2} .\) After the blade has turned through \(155 \mathrm{rev},\) a small piece of the blade breaks loose from the top of the blade. After the piece breaks loose, it travels with a velocity that is initially horizontal and equal to the tangential velocity of the rim of the blade. The piece travels a vertical distance of \(0.820 \mathrm{~m}\) to the floor. How far does the piece travel horizontally, from where it broke off the blade until it strikes the floor?

Two metal disks, one with radius \(R_{1}=2.50 \mathrm{~cm}\) and mass \(M_{1}=0.80 \mathrm{~kg}\) and the other with radius \(R_{2}=5.00 \mathrm{~cm}\) and mass \(M_{2}=1.60 \mathrm{~kg},\) are welded together and mounted on a frictionless axis through their common center (Fig. \(\mathbf{P 9 . 7 9}\) ). (a) What is the total moment of inertia of the two disks? (b) A light string is wrapped around the edge of the smaller disk, and a \(1.50 \mathrm{~kg}\) block is suspended from the free end of the string. If the block is released from rest at a distance of \(2.00 \mathrm{~m}\) above the floor, what is its speed just before it strikes the floor? (c) Repeat part (b), this time with the string wrapped around the edge of the larger disk. In which case is the final speed of the block greater? Explain.

Three small blocks, each with mass \(m\), are clamped at the ends and at the center of a rod of length \(L\) and negligible mass. Compute the moment of inertia of the system about an axis perpendicular to the rod and passing through (a) the center of the rod and (b) a point onefourth of the length from one end.

A uniform disk has radius \(R_{0}\) and mass \(M_{0}\). Its moment of inertia for an axis perpendicular to the plane of the disk at the disk's center is \(\frac{1}{2} M_{0} R_{0}^{2}\). You have been asked to halve the disk's moment of inertia by cutting out a circular piece at the center of the disk. In terms of \(R_{0}\), what should be the radius of the circular piece that you remove?
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