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Chapter 9: Problem 20

It is desired that the outer edge of a grinding wheel \(9.0 \mathrm{~cm}\) in radius move at a constant rate of \(6.0 \mathrm{~m} / \mathrm{s}\). (a) Determine the angular speed of the wheel. (b) What length of thin thread could be wound on the rim of the wheel in \(3.0 \mathrm{~s}\) when it is turning at this rate?

Short Answer

Expert verified
(a) Angular speed is approximately \(66.67\,\text{rad/s}\). (b) Thread length is \(18.0\,\text{m}\).

Step by step solution

01

Introduction to Linear and Angular Speeds

The linear speed \(v\) is related to the angular speed \(\omega\) by the relationship \(v = r \cdot \omega\), where \(r\) is the radius of the wheel. To find the angular speed, we can rearrange this formula to solve for \(\omega\): \(\omega = \frac{v}{r}\).
02

Convert Units Consistently

The radius is given in centimeters (\(9.0\ \mathrm{cm}\)), while the linear speed is in meters per second (\(6.0\ \mathrm{m/s}\)). To ensure consistency, convert the radius to meters: \(9.0 \mathrm{~cm} = 0.09 \mathrm{~m}\).
03

Calculate Angular Speed

Substitute the values into the formula: \(\omega = \frac{6.0 \mathrm{~m/s}}{0.09 \mathrm{~m}}\). Calculating gives: \[\omega \approx 66.67 \mathrm{~rad/s}\]
04

Calculate Length of Thread Wound

The length of thread wound around the wheel is equal to its linear speed times the time: \(\text{Length} = v \cdot t\). Given \(v = 6.0 \mathrm{~m/s}\) and \(t = 3.0 \mathrm{~s}\), substituting these values gives: \[\text{Length} = 6.0 \mathrm{~m/s} \times 3.0 \mathrm{~s} = 18.0 \mathrm{~m}\]

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Key Concepts

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

Understanding Linear Speed
Linear speed is the rate at which an object travels over a distance in a given period of time. It's a measure of how fast something is moving in a straight path. In physics, we often represent linear speed with the symbol \( v \). For a wheel or any rotating object, linear speed can be seen at its outer edge or circumference. In our example, the grinding wheel's linear speed is given as \( 6.0 \ \text{m/s} \).

This means every second, the point on the edge of the wheel moves 6 meters. This measurement helps us understand how quickly something is moving linearly across a specific distance.
Defining Angular Velocity
Angular velocity refers to how fast an object turns or rotates around a center point or axis. It's the rate of change of the angle through which an object rotates, usually measured in radians per second \( \text{rad/s} \). While linear speed describes straight-line motion, angular velocity describes circular motion.

To calculate angular velocity \( \omega \), we use the formula \( \omega = \frac{v}{r} \), where \( v \) is the linear speed and \( r \) is the radius of the object. This formula helps you understand how quickly the whole object spins based on how fast a point on it is moving linearly. In our problem, substituting the values into the equation gives an angular velocity of approximately \( 66.67 \ \text{rad/s} \).
Importance of Unit Conversion
Unit conversion is crucial in physics because maintaining consistent units makes calculations accurate and understandable. In our problem, the linear speed is given in meters per second, while the radius is initially in centimeters. To find the angular speed accurately, we first convert the radius from \( 9.0 \ \text{cm} \) to meters, which is \( 0.09 \ \text{m} \).

Converting units ensures that when we use the formula \( \omega = \frac{v}{r} \), both the linear speed \( v \) and radius \( r \) share the same unit system, allowing our calculation of \( \omega \), or angular velocity, to be correct and consistent.
Understanding the Radius of the Wheel
The radius of a wheel is the distance from its center point to the edge or rim. This measurement is crucial because it directly impacts the wheel's angular speed and linear speed. In our scenario, the grinding wheel has a radius of \( 9.0 \ \text{cm} \), which we convert to \( 0.09 \ \text{m} \) for ease in calculations.

The radius plays a vital role in determining the angular velocity using the formula \( \omega = \frac{v}{r} \). A larger radius would mean slower rotation for the same linear speed, whereas a smaller radius would lead to faster rotation, illustrating how size affects motion dynamics.

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Most popular questions from this chapter

A fan tums at a rate of 900 rpm (i.e., rev/min). ( \(a\) ) Find the angular speed of any point on one of the fan blades. (b) Find the tangential speed of the tip of a blade if the distance from the center to the tip is \(20.0 \mathrm{~cm}\). (a) \(f=900 \frac{\mathrm{rev}}{\min }=15.0 \frac{\mathrm{rev}}{\mathrm{s}}=15.0 \mathrm{~Hz}\) Since \(\omega=2 \pi f\) $$ \omega=2 \pi(15.0 \mathrm{~Hz}) $$ and \(s 0\) $$ \omega=94.2 \frac{\mathrm{rad}}{\mathrm{s}} $$ for all points on the fan blade. (b) The tangential speed is \(r \omega\), where \(\omega\) must be in \(\mathrm{rad} / \mathrm{s}\). Therefore, $$ v=r \omega=(0.200 \mathrm{~m})(94.2 \mathrm{rad} / \mathrm{s})=18.8 \mathrm{~m} / \mathrm{s} $$ Notice that the rad does not carry through the equations properly -we insert it or delete it as needed.

The human body can safely tolerate a vertical acceleration \(9.00\) times that due to gravity. With what minimum radius of curvature may a pilot safely turn the plane upward at the end of a dive if the plane's speed is \(770 \mathrm{~km} / \mathrm{h}\) ?

What is the maximum speed at which a car can round a curve of \(25-\mathrm{m}\) radius on a level road if the coefficient of static friction between the tires and road is \(0.80\) ? The radial force required to keep the car in the curved path (the centripetal force) is supplied by friction between the tires and the road. If the mass of the car is \(m\), the maximum friction force (which is the centripetal force) equals \(\mu, F_{N}\) or \(0.80 \mathrm{mg}\), this arises when the car is on the verge of skidding sideways. Therefore, the maximum speed is given by $$ \frac{m v^{2}}{r}=0.80 \mathrm{mg} \quad \text { or } \quad v=\sqrt{0.80 \mathrm{gr}}=\sqrt{(0.80)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(25 \mathrm{~m})}=14 \mathrm{~m} / \mathrm{s} $$

A roller coaster is just barely moving as it goes over the top of the hill. It rolls nearly without friction down the hill and then up over a lower hill that has a radius of curvature of \(15 \mathrm{~m}\). How much higher must the first hill be than the second if the passengers are to exert no forces on their seats as they pass over the top of the lower hill?

A \(200-\mathrm{g}\) object is tied to the end of a cord and whirled in a horizontal circle of radius \(1.20 \mathrm{~m}\) at a constant \(3.0 \mathrm{rev} / \mathrm{s}\). Assume that the cord is horizontal - that is, that gravity can be neglected. Determine ( \(a\) ) the centripetal acceleration of the object and ( \(b\) ) the tension in the cord. (a) The object is not accelerating tangentially to the circle but is undergoing a radial, or centripetal, acceleration given by $$ a_{c}=\frac{v^{2}}{r}=r \omega^{2} $$ where \(\omega\) must be in \(\mathrm{rad} / \mathrm{s}\). Since \(3.0 \mathrm{rev} / \mathrm{s}=6.0 \pi \mathrm{rad} / \mathrm{s}\), $$ a_{C}=(1.20 \mathrm{~m})(6.0 \pi \mathrm{rad} / \mathrm{s})^{2}=426 \mathrm{~m} / \mathrm{s}^{2}=0.43 \mathrm{~km} / \mathrm{s}^{2} $$ (b) To cause the acceleration found in \((a)\), the cord must pull on the \(0.200-\mathrm{kg}\) mass with a centripetal force given by $$ F_{C}=m a_{C}=(0.200 \mathrm{~kg})\left(426 \mathrm{~m} / \mathrm{s}^{2}\right)=85 \mathrm{~N} $$ This is the tension in the cord
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