Problem 29 - Applied Physics

Question

An automobile is traveling at \(60.0 \mathrm{~km} / \mathrm{h}\). Its tires have a radius of \(33.0 \mathrm{~cm}\). (a) Find the angular speed of the tires (in \(\mathrm{rad} / \mathrm{s}\) ). (b) Find the angular displacement of the tires in \(30.0 \mathrm{~s}\). (c) Find the linear distance traveled by a point on the tread in \(30.0 \mathrm{~s}\). (d) Find the linear distance traveled by the automobile in \(30.0 \mathrm{~s}\).

Step-by-Step Solution

Answer

Angular speed is 50.52 rad/s, angular displacement is 1515.6 rad, both the tread and car travel 500.148 m.

Step 1: Convert linear speed to meters per second

The automobile speed is given as \(60.0 \; \text{km/h}\). First, convert this to meters per second by using the conversion factor \(1 \; \text{km/h} = \frac{1000}{3600} \; \text{m/s}\). Therefore, the speed in \(\text{m/s}\) is:\[v = 60.0 \times \frac{1000}{3600} = 16.67 \; \text{m/s}\]

Step 2: Calculate the angular speed of the tires

Use the formula for angular speed \(\omega\), which is \(\omega = \frac{v}{r}\), where \(v\) is the linear speed and \(r\) is the radius of the tires. Here, \(r = 33.0\,\text{cm} = 0.33\,\text{m}\):\[\omega = \frac{16.67}{0.33} \approx 50.52 \; \text{rad/s}\]

Step 3: Find the angular displacement in 30 seconds

The angular displacement \(\theta\) over a time \(t\) is given by \(\theta = \omega \times t\). We found \(\omega = 50.52 \, \text{rad/s}\), and the time \(t = 30.0\,\text{s}\):\[\theta = 50.52 \times 30 = 1515.6 \, \text{rad}\]

Step 4: Calculate the linear distance traveled by a point on the tread

The linear distance \(s\) traveled by a point on the tread is related to the angular displacement \(\theta\) by \(s = r \times \theta\). Using \(r = 0.33\,\text{m}\) and \(\theta = 1515.6 \, \text{rad}\):\[s = 0.33 \times 1515.6 = 500.148 \, \text{m}\]

Step 5: Find the linear distance traveled by the automobile

Assuming no slippage, the linear distance traveled by the automobile is the same as the distance traveled by the point on the tread, which is:\[500.148 \, \text{m}\]

Key Concepts

Linear Speed Conversion

When dealing with problems involving motion, it's often necessary to convert speeds to the same units to simplify calculations. In this exercise, the automobile's speed is given in kilometers per hour (km/h). However, for calculations involving angular motion, meters per second (m/s) is more suitable. This conversion helps because the formula for angular speed involves meters and seconds. Use the conversion factor: - 1 km/h is equivalent to \(\frac{1000}{3600} \text{ m/s}\) as there are 1000 meters in a kilometer and 3600 seconds in an hour. By applying this conversion, the speed of 60 km/h becomes \(16.67 \text{ m/s} \). This conversion is crucial for further calculations of angular speed and displacement.

Angular Displacement

Angular displacement represents the angle through which a point or line has been rotated in a specified sense about a specified axis. It provides a measure of how much rotation has occurred. In the problem, we calculate angular displacement \( \theta \), using the formula: - \( \theta = \omega \times t \), where \( \omega \) is the angular speed and \( t \) is time. Angular speed \( \omega \) is previously found as \( 50.52 \text{ rad/s} \), and time \( t \) is given as 30 seconds. So, the angular displacement is \( 1515.6 \text{ rad} \). This measure indicates how much the tires have rotated in radians over the 30 seconds.

Tire Radius

The radius of the tire is an important factor in determining both the angular speed and displacement. The radius is given as 33 cm, which converts to 0.33 meters for consistency with other units (meters and seconds). This dimension feeds into the calculations for several aspects:

• In the formula for converting linear speed to angular speed: \( \omega = \frac{v}{r} \)
• In calculating the linear distance traveled from angular displacement: \( s = r \times \theta \)

A larger radius would mean slower rotation but cover more distance per rotation, while a smaller radius results in faster rotation. Understanding the role of the tire's radius helps in visualizing how different sizes impact tire behavior in motion.

Linear Distance Calculation

Calculating the linear distance traveled by a point on a tire's tread involves understanding its relation to angular displacement. The linear distance \( s \) is tied to how far any given point travels as the tire rotates.This is given by: - \( s = r \times \theta \)- Here \( r \) is the radius (0.33 m) and \( \theta \) is the angular displacement (1515.6 rad).Performing this multiplication yields the distance: \( 500.148 \text{ m} \). Thus, over the span of 30 seconds, the automobile, by virtue of its tires, traverses this distance, assuming no slipping occurs. Understanding this relationship allows you to analyze how effectively the tires convert rotational motion into linear travel.