Problem 25 - College Physics

Question

A car is traveling at a speed of \(101 \mathrm{~km} / \mathrm{h}\) on the highway and has a small stone stuck between the treads of one of its tires. The tires have diameter \(61.0 \mathrm{~cm}\) and are rolling without sliding or slipping. What are (a) the maximum and (b) the minimum speeds of the stone as observed by a pedestrian standing on the side of the highway?

Step-by-Step Solution

Answer

(a) 32.52 m/s, (b) 23.58 m/s.

Step 1: Convert Speed to m/s

First, we need to convert the car's speed from km/h to m/s. We use the conversion factor that 1 km/h is equal to \( \frac{1}{3.6} \) m/s.\[101 \text{ km/h} \times \frac{1}{3.6} \text{ m/s per km/h} = 28.056 \text{ m/s}\]

Step 2: Calculate the Circumference of the Tire

The circumference of the tire, which is required to determine the rotational speed at the rim, is given by the formula:\[C = \pi d\]where \(d\) is the diameter of the tire. Here, \(d = 0.61 \text{ m}\):\[C = \pi \times 0.61 \text{ m} = 1.9167 \text{ m}\]

Step 3: Determine Angular Velocity of the Tire

Using the car's speed and the circumference of the tire, we find the angular velocity \(\omega\):\[\omega = \frac{v}{C} = \frac{28.056 \text{ m/s}}{1.9167 \text{ m}} = 14.64 \text{ rad/s}\]

Step 4: Find Maximum Speed of the Stone

The maximum speed of the stone relative to the ground is when it is moving in the same direction as the car at the top of the tire:\[v_{\text{max}} = v + \omega r = 28.056 \text{ m/s} + \left( 14.64 \text{ rad/s} \times 0.305 \text{ m} \right) = 32.52 \text{ m/s}\]

Step 5: Find Minimum Speed of the Stone

The minimum speed relative to the ground occurs when the stone is at the bottom of the tire, where it moves opposite to the direction of the car:\[v_{\text{min}} = v - \omega r = 28.056 \text{ m/s} - \left( 14.64 \text{ rad/s} \times 0.305 \text{ m} \right) = 23.58 \text{ m/s}\]

Key Concepts

Linear Speed Conversion

If you wish to perform linear speed conversion from kilometers per hour (km/h) to meters per second (m/s), you need to know a simple conversion factor. Since 1 km contains 1000 meters and 1 hour has 3600 seconds, you have to divide the speed in km/h by 3.6 to get the speed in m/s. For example:

• Speed in km/h: 101 km/h
• Convert by dividing by 3.6: \( 101 / 3.6 = 28.056 \, \text{m/s} \)

This conversion is incredibly useful in physical calculations where standard units are required. By understanding this relation, you can easily switch between these two common units of speed, making it simpler to work through physics problems.

Circumference Calculation

The circumference of a circular object, such as a tire, is crucial in understanding its overall motion characteristics. The circumference can be calculated using the formula:

• \( C = \pi d \)
• The diameter \( d \) of the tire: 0.61 m

Plug in the diameter to find:
\[ C = \pi \times 0.61 \approx 1.9167 \, \text{m} \]
This value is especially important when determining how far the tire travels with one complete rotation. By calculating the circumference, you have the foundational length needed to connect linear speeds to angular velocities.

Angular Velocity

Angular velocity refers to how fast an object rotates or spins around a point or axis. To find the angular velocity \( \omega \) of a tire, you need both the linear velocity of the car and the tire's circumference:

• Linear speed of the car: 28.056 m/s
• Tire circumference: 1.9167 m

The formula to find angular velocity is:
\[ \omega = \frac{v}{C} \]
Substituting the values gives:
\[ \omega = \frac{28.056}{1.9167} \approx 14.64 \, \text{rad/s} \]
Angular velocity helps us understand how fast an object spins, which is essential in determining the speed of the stone at different positions on the tire.

Relative Speed

Relative speed is the speed of an object as observed from a particular frame of reference. In our scenario, the stone stuck in the tire experiences different speeds based on its position—either at the top or bottom:

• **Maximum speed**: Occurs at the top of the tire and adds the tangential speed due to angular motion to the car's speed: \( v_{\text{max}} = v + \omega r \)
• Substitute car speed (28.056 m/s) and angular speed contributions (\( \omega = 14.64 \times 0.305 \)): \[ v_{\text{max}} \approx 32.52 \, \text{m/s} \]
• **Minimum speed**: Happens at the bottom where it moves against motion, reducing the speed: \( v_{\text{min}} = v - \omega r \)
• Using the same values, this computes to: \[ v_{\text{min}} \approx 23.58 \, \text{m/s} \]

Relative speed thus helps identify how fast an object moves from a stationary frame, like that of a pedestrian's view of the car.