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Chapter 8: Problem 30

An automobile traveling at \(97 \mathrm{~km} / \mathrm{h}\) has wheels of diameter \(76 \mathrm{~cm} .(a)\) Find the angular speed of the wheels about the axle. \((b)\) The car is brought to a stop uniformly in 30 turns of the wheels. Calculate the angular acceleration. \((c)\) How far does the car advance during this braking period?

Short Answer

Expert verified
The angular speed of the wheels is \(709.06 \, \text{rad/s}\). The angular acceleration is \(-3.76 \, \text{rad/s}^2\). The car travels about \(71.12 \, \text{m}\) during the stopping period.

Step by step solution

01

Find the angular speed

The linear speed \(v\) can be converted to angular speed \(\omega\) through the relation: \(\omega = \frac{v}{r}\), where \(r\) is the radius of the wheel. Convert speed to cm/s: \(v = 97 \, \text{km/h} * \frac{1000 \, \text{m}}{1 \, \text{km}} * \frac{100 \, \text{cm}}{1 \, \text{m}} * \frac{1 \, \text{h}}{3600 \, \text{s}} = 26944.44 \, \text{cm/s}\). Afterward, take \(r = \frac{76 \, \text{cm}}{2} = 38 \, \text{cm}\). Substituting the values, we have: \(\omega = \frac{26944.44 \, \text{cm/s}}{38 \, \text{cm}} = 709.06 \, \text{rad/s}\).
02

Calculate the angular acceleration

Given that the car came to a stop uniformly in 30 turns of the wheels, we can use \(\alpha = \frac{-\omega}{\theta}\) to find the angular acceleration \(\alpha\), where \(\theta\) is the total angle in radians (which is \(30\) turns \(* 2\pi\)). Hence, \(\theta = 30 * 2\pi = 60\pi \, \text{rad}\). Substituting the values, we have: \(\alpha = \frac{-709.06 \, \text{rad/s}}{60\pi \, \text{rad}} = -3.76 \, \text{rad/s}^2\).
03

Find the distance covered during braking

Finally, the distance \(d\) covered during this period can be calculated from the relation: \(d = r \times \theta\). Substituting the given radius and calculated theta: \(d = 38 \, \text{cm} * 60\pi \, \text{rad} = 7111.8 \, \text{cm or} \, 71.12 \, \text{m}\).

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Key Concepts

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

Angular Speed
In circular motion, angular speed, denoted by \(\omega\), describes how fast an object rotates around a fixed point. It's analogous to linear speed, but instead of covering distance over time, it measures how much of an angle is covered per unit of time. This is expressed in radians per second (rad/s). To find angular speed, we utilize the relationship between linear speed (\(v\)) and radius (\(r\)):
\[ \omega = \frac{v}{r} \]
In our exercise, the car travels at 97 km/h. We convert this to 26,944.44 cm/s to make calculations consistent with the wheel's diameter of 76 cm. With a radius of 38 cm (half the diameter), we calculate:
\[ \omega = \frac{26944.44 \, \text{cm/s}}{38 \, \text{cm}} = 709.06 \, \text{rad/s} \]
This means the wheels complete approximately 709 radians in one second as the car travels.
Angular Acceleration
Angular acceleration (\(\alpha\)) measures how the angular speed changes with time. It's critical for understanding how quickly an object speeds up or slows down its rotation. The unit of angular acceleration is radians per second squared (\(\text{rad/s}^2\)). When a vehicle slows down uniformly, as in our exercise, it can be captured by the formula:
\[ \alpha = \frac{-\omega}{\theta} \]
where \(\theta\) is the total angular displacement in radians. The car stops in 30 rotations, or \(30 \times 2\pi\), which is equal to \(60\pi\) radians. With an initial angular speed of 709.06 rad/s, we calculate:
\[ \alpha = \frac{-709.06 \, \text{rad/s}}{60\pi \, \text{rad}} = -3.76 \, \text{rad/s}^2 \]
The negative sign indicates deceleration, as expected when braking.
Linear and Angular Relationship
The relationship between linear and angular quantities is foundational in rotational kinematics. Linear distance, speed, and acceleration have their angular counterparts (angle, angular speed, angular acceleration). This allows us to translate motion descriptions in linear terms into rotational terms and vice versa.
  • Linear distance \(d\) is related to angular displacement \(\theta\) by the formula:
    \[ d = r \times \theta \]
  • For our problem, the car's wheels cover a distance during the braking period. Using the calculated \(\theta = 60\pi\) radians, and \(r = 38\) cm, we have:
    \[ d = 38 \, \text{cm} \times 60\pi \, \text{rad} = 7111.8 \, \text{cm} \]
  • Converting to meters: 71.12 m.
    This showcases how angular motion translates into linear advance, helping to understand the car's travel during wheel rotations.

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Most popular questions from this chapter

The flywheel of an engine is rotating at \(25.2 \mathrm{rad} / \mathrm{s}\). When the engine is turned off, the flywheel decelerates at a constant rate and comes to rest after \(19.7 \mathrm{~s}\). Calculate \((a)\) the angular acceleration (in \(\mathrm{rad} / \mathrm{s}^{2}\) ) of the flywheel, (b) the angle (in rad) through which the flywheel rotates in coming to rest, and \((c)\) the number of revolutions made by the flywheel in coming to rest.

While waiting to board a helicopter, you notice that the rotor's motion changed from 315 rev/min to 225 rev/min in \(1.00\) min. ( \(a\) ) Find the average angular acceleration during the interval. ( \(b\) ) Assuming that this acceleration remains constant, calculate how long it will take for the rotor to stop. (c) How many revolutions will the rotor make after your second observation?

If an airplane propeller of radius \(5.0 \mathrm{ft}(=1.5 \mathrm{~m})\) rotates at 2000 rev/min and the airplane is propelled at a ground speed of \(300 \mathrm{mi} / \mathrm{h}(=480 \mathrm{~km} / \mathrm{h})\), what is the speed of a point on the tip of the propeller, as seen by \((a)\) the pilot and \((b)\) an observer on the ground? Assume that the plane's velocity is parallel to the propeller's axis of rotation.

Starting from rest at \(t=0\), a wheel undergoes a constant angular acceleration. When \(t=2.33 \mathrm{~s}\), the angular velocity of the wheel is \(4.96 \mathrm{rad} / \mathrm{s}\). The acceleration continues until \(t=23.0 \mathrm{~s}\), when it abruptly ceases. Through what angle does the wheel rotate in the interval \(t=0\) to \(t=46.0 \mathrm{~s}\) ?

A speedometer on the front wheel of a bicycle gives a reading that is directly proportional to the angular speed of the wheel. Suppose that such a speedometer is calibrated for a wheel of diameter \(72 \mathrm{~cm}\) but is mistakenly used on a wheel of diameter \(62 \mathrm{~cm} .\) Would the linear speed reading be wrong? If so, in what sense and by what fraction of the true speed?
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