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Chapter 8: Problem 37

A string trimmer is a tool for cutting grass and weeds; it utilizes a length of nylon "string" that rotates about an axis perpendicular to one end of the string. The string rotates at an angular speed of 47 rev/s, and its tip has a tangential speed of \(54 \mathrm{m} / \mathrm{s}\). What is the length of the rotating string?

Short Answer

Expert verified
The length of the string is approximately 0.183 meters.

Step by step solution

01

Understand the Relationship Between Angular and Tangential Speed

The problem involves angular speed \((\omega)\) and tangential speed \((v_t)\) of the rotating string. The relationship between these speeds can be expressed as: \[ v_t = \omega \cdot r \] where \( r \) is the radius of the rotation, in this case, the length of the string. Here, \( \omega \) is given in revolutions per second, and \(v_t\) is the tangential speed.
02

Convert Angular Speed to Radians per Second

Angular speed \(\omega\) is given in revolutions per second \((47 \text{ rev/s})\). To use the formula, we need angular speed in radians per second, since \(1 \text{ revolution} = 2\pi \text{ radians}\). \[ \omega = 47 \text{ rev/s} \times 2\pi \text{ rad/rev} = 94\pi \text{ rad/s} \]
03

Rearrange the Formula and Solve for the String Length

We know \( v_t = 54 \text{ m/s} \) and \( \omega = 94\pi \text{ rad/s}\). Substitute these values into the formula to solve for \( r \), the length of the string:\[ r = \frac{v_t}{\omega} = \frac{54 \text{ m/s}}{94\pi \text{ rad/s}} \]
04

Calculate the Length of the String

Perform the division to find the length of the string:\[ r = \frac{54}{94\pi} \approx 0.183 \text{ meters} \] So, the length of the string is approximately 0.183 meters.

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Key Concepts

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

Angular Speed
Angular speed, often represented by \( \omega \), describes how fast an object is rotating. Think of it as the rate of rotation, similar to how linear speed is the rate of movement in a straight path. Angular speed is usually measured in revolutions per second (rev/s) or radians per second (rad/s).
  • In the original exercise, the angular speed given is 47 rev/s.
  • To work in a standard unit, it is often necessary to convert angular speed from rev/s to rad/s.
  • This conversion is important because many formulas in physics use radians as the angle unit.
Each revolution is equivalent to \(2\pi\) radians. Thus, to convert revolutions to radians, multiply by \(2\pi\). In this scenario, we have \(47 \text{ rev/s}\times 2\pi \text{ rad/rev} = 94\pi \text{ rad/s}\).
Tangential Speed
Tangential speed refers to the speed at which a point on the edge of a rotating object is moving along its path. This type of speed is related to rotational motion but describes the linear motion of any point located on the radius of rotation.
  • In the problem, the tip of the string was moving with a tangential speed of 54 m/s.
  • This speed essentially tells us how fast the tip of the string is cutting through the grass and weeds.
  • Tangential speed is dependent on both the angular speed and the radius of the rotation.
Tangential speed \((v_t)\) is calculated using the formula \(v_t = \omega \cdot r\), where \(r\) is the radius or length of the string. By knowing \(v_t\) and \(\omega\), we can solve for \(r\).
Unit Conversion
In physics, converting units is a critical step to ensure consistency and accuracy in calculations. For this specific problem, we dealt with converting angular speed from revolutions per second to radians per second.
  • Conversion is needed because most rotational calculations use radians.
  • By using the conversion factor, \(1 \text{ revolution} = 2\pi \text{ radians}\), you can seamlessly switch units.
Remember, having consistent units allows the formulas to work correctly and provides meaningful results. Converting angular speed was necessary for applying the formula \(v_t = \omega \cdot r\) to find the length of the string.
Rotational Motion
Rotational motion is the motion of an object around a central point or axis. It is a fundamental concept that helps us analyze objects that spin, orbit, or rotate in some manner.
  • Every point in a rotating body moves in a circular path around a central axis.
  • Key parameters in rotational motion include angular speed (how fast something rotates) and tangential speed (how fast a point on the edge moves through space).
In the context of this problem, the rotating string of the trimmer embodies these principles. The concept of rotational motion allows us to describe how the string rotates (angular speed) and the actual path movement of the string (tangential speed). Understanding this concept is essential for solving various physics problems involving rotations and translating between linear and rotational speeds.

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Most popular questions from this chapter

A motorcyclist is traveling along a road and accelerates for \(4.50 \mathrm{s}\) to pass another cyclist. The angular acceleration of each wheel is \(+6.70 \mathrm{rad} / \mathrm{s}^{2},\) and, just after passing, the angular velocity of each wheel is \(+74.5 \mathrm{rad} / \mathrm{s},\) where the plus signs indicate counterclockwise directions. What is the angular displacement of each wheel during this time?

The earth orbits the sun once a year \(\left(3.16 \times 10^{7}\right.\) s) 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.

An automobile, starting from rest, has a linear acceleration to the right whose magnitude is 0.800 m/s2 (see the figure). During the next 20.0 s, the tires roll and do not slip. The radius of each wheel is 0.330 m. At the end of this time, what is the angle through which each wheel has rotated?

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 angular open spaces between the three propeller blades are each \(\pi / 3\) rad \(\left(60.0^{\circ}\right) .\) The vertical drop of the arrow may be ignored. There is a maximum value \(\omega\) for the angular speed of the propeller, beyond which the arrow cannot pass through an open space without being struck by one of the blades. Find this maximum value when the arrow has the lengths \(L\) and speeds \(v\) shown in the following table. $$ \begin{array}{lll} & L & v \\ (\mathrm{a}) & 0.71 \mathrm{m} & 75.0 \mathrm{m} / \mathrm{s} \\ \hline(\mathbf{b}) & 0.71 \mathrm{m} & 91.0 \mathrm{m} / \mathrm{s} \\ \hline(\mathbf{c}) & 0.81 \mathrm{m} & 91.0 \mathrm{m} / \mathrm{s} \\ \hline \end{array} $$

A thin rod (length \(=1.50 \mathrm{m}\) ) is oriented vertically, with its bottom end attached to the floor by means of a frictionless hinge. The mass of the rod may be ignored, compared to the mass of an object fixed to the top of the rod. The rod, starting from rest, tips over and rotates downward. (a) What is the angular speed of the rod just before it strikes the floor? (Hint: Consider using the principle of conservation of mechanical energy.\()\) (b) What is the magnitude of the angular acceleration of the rod just before it strikes the floor?
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