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Chapter 8: Problem 32

In 9.5 s a fisherman winds \(2.6 \mathrm{~m}\) of fishing line onto a reel whose radius is \(3.0 \mathrm{~cm}\) (assumed to be constant as an approximation). The line is being reeled in at a constant speed. Determine the angular speed of the reel.

Short Answer

Expert verified
The angular speed of the reel is approximately 9.13 rad/s.

Step by step solution

01

Convert Distance to Radians

To find the angular speed, we first need to determine how many radians the reel turns in 9.5 s. The formula that relates arc length \(s\), radius \(r\), and angle \(\theta\) in radians is \(s = r \cdot \theta\). Here, \(s = 2.6\) m and \(r = 0.03\) m. Substitute these into the equation: \(2.6 = 0.03 \cdot \theta\).
02

Solve for Theta

Solve the equation \(2.6 = 0.03 \cdot \theta\) for \(\theta\). Divide both sides by 0.03: \(\theta = \frac{2.6}{0.03}\). Calculate \(\theta = 86.67\ ext{radians}\).
03

Calculate Angular Speed

Angular speed \(\omega\) is the angle turned per time. Using \(\omega = \frac{\theta}{t}\), where \(\theta = 86.67\ ext{radians}\) and \(t = 9.5\ ext{s}\), substitute and solve: \(\omega = \frac{86.67}{9.5}\). Calculate \(\omega = 9.13\ ext{rad/s}\).

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Key Concepts

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

Radians Conversion
Understanding how to convert various measurements into radians is crucial when dealing with angular movements. In our exercise, we found the angular distance in radians to determine angular speed.
To do this, we use the formula that relates arc length (denoted by \(s\)), radius (\(r\)), and angle \(\theta\) in radians:
  • \( s = r \cdot \theta \)
In this scenario, \(s\) represents the length of the fishing line wound onto the reel, and \(r\) is the radius of the reel.
By rearranging the formula, you solve for the angle \(\theta\) by dividing the arc length by the radius:
  • \( \theta = \frac{s}{r} \)
Thus, by plugging in the given values of \(s = 2.6\) m and \(r = 0.03\) m, we calculate the number of radians to be \(86.67\) radians. This conversion is the first step toward finding out how fast the reel spins.
Arc Length
The concept of arc length is a critical component of circular motion. It represents the distance a point on the edge of a circle travels as the circle rotates. In our exercise, the arc length corresponds to the amount of fishing line wound onto the reel.
Arc length in circles is proportional to the radius and the central angle in radians. This relationship is expressed mathematically as:
  • \( s = r \cdot \theta \)
Here, \(s\) is the arc length or distance traveled around the circle; \(r\) is the radius, and \(\theta\) is the central angle in radians.
In practice, if you know any two of these quantities, you can solve for the third. In our problem, we knew \(s\) and \(r\), allowing us to solve for \(\theta\). This illustrates how knowing the arc length can help determine other essential features of circular motion.
Angular Velocity Calculation
Angular velocity measures how fast an object rotates or revolves around a point, often expressed in radians per second. In this exercise, we determined the angular velocity of the fishing reel as it gathered the line.
Angular velocity, represented by \(\omega\), can be calculated using the formula:
  • \( \omega = \frac{\theta}{t} \)
Where \( \theta \) is the angle in radians, and \( t \) is the time over which the rotation occurs. Utilizing the angle \(86.67\) radians we found earlier and noting that the winding happens over \(9.5\) seconds, we calculate:
  • \( \omega = \frac{86.67}{9.5} \)
Solving this gives us an angular velocity of approximately \(9.13\) radians per second.
This calculation is vital for understanding how quickly the reel is working, translating linear movement into rotational motion.

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Most popular questions from this chapter

The front and rear sprockets on a bicycle have radii of 9.00 and \(5.10 \mathrm{~cm}\), respectively The angular speed of the front sprocket is \(9.40 \mathrm{rad} / \mathrm{s}\). Determine (a) the linear speed (in \(\mathrm{cm} / \mathrm{s}\) ) of the chain as it moves between the sprockets and \((\mathrm{b})\) the centripetal acceleration (in \(\mathrm{cm} / \mathrm{s}^{2}\) ) of the chain as it passes around the rear sprocket.

A propeller is rotating about an axis perpendicular to its center, as the drawing shows. The axis is parallel to the ground. An arrow is fired at the propeller, travels parallel to the axis, and passes through one of the open spaces between the propeller blades. The vertical drop of the arrow may be ignored. There is a maximum value for the angular speed \(\omega\) of the propeller beyond which the arrow cannot pass through an open space without being struck by one of the blades. (a) If the arrow is to pass through an open space, does it matter if the arrow is aimed closer to or farther away from the axis (see points \(\mathrm{A}\) and \(\mathrm{B}\) in the drawing, for example)? Explain. (b) Does the maximum value of \(\omega\) increase, decrease, or remain the same with increasing arrow speed \(v ?\) Why? (c) Does the maximum value of \(\omega\) increase, decrease, or remain the same with increasing arrow length \(L\) ? Justify your answer.

Interactive Solution \(\underline{8.61}\) at offers one approach to solving this problem. The drawing shows the blade of a chain saw. The rotating sprocket tip at the end of the guide bar has a radius of \(4.0 \times 10^{-2} \mathrm{~m}\). The linear speed of a chain link at point \(\mathrm{A}\) is \(5.6 \mathrm{~m} / \mathrm{s}\). Find the angular speed of the sprocket tip in rev/s.

The earth orbits the sun once a year \(\left(3.16 \times 10^{7} \mathrm{~s}\right)\) in a nearly circular orbit of radius \(1.50 \times 10^{11} \mathrm{~m} .\) With respect to the sun, determine (a) the angular speed of the earth, (b) the tangential speed of the earth, and (c) the magnitude and direction of the earth's centripetal acceleration.

Two identical dragsters, starting from rest, accelerate side-by- side along a straight track. The wheels on one of the cars roll without slipping, while the wheels on the other slip during part of the time. (a) For which car, he winner or the loser, do the wheels roll without slipping? Why? For the dragster whose wheels roll without slipping, is there (b) a relationship between its linear speed and the angular speed of its wheels, and (c) a relationship between the magnitude of its linear acceleration and the magnitude of the angular acceleration of its wheels? If a relationship exists in either case, what is it? Problem A dragster starts from rest and accelerates down the track. Each tire has a radius of \(0.320 \mathrm{~m}\) and rolls without slipping. At a distance of \(384 \mathrm{~m},\) the angular speed of the wheels is \(288 \mathrm{rad} / \mathrm{s}\). Determine, (a) the linear speed of the dragster and (b) the magnitude of the angular acceleration of its wheels.
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