Problem 1 - Fundamentals of Physics

Question

A car travels at \(80 \mathrm{~km} / \mathrm{h}\) on a level road in the positive direction of an \(x\) axis. Each tire has a diameter of \(66 \mathrm{~cm} .\) Relative to a woman riding in the car and in unit-vector notation, what are the velocity \(\vec{v}\) at the (a) center, (b) top, and (c) bottom of the tire and the magnitude \(a\) of the acceleration at the (d) center, (e) top, and (f) bottom of each tire? Relative to a hitchhiker sitting next to the road and in unit-vector notation, what are the velocity \(\vec{v}\) at the (g) center, (h) top, and (i) bottom of the tire and the magnitude \(a\) of the acceleration at the (j) center, (k) top, and (I) bottom of each tire?

Step-by-Step Solution

Answer

Velocity and acceleration are zero at the bottom and non-zero elsewhere, relative to the car. The centers move with the car and match its velocity. Acceleration remains unchanged as it only depends on rotation.

Step 1: Understanding the Given Information

The car travels at a speed of \(80 \mathrm{~km/h}\), and each tire has a diameter of \(66 \mathrm{~cm}\). We need to find out the velocity and acceleration at the center, top, and bottom of the tires from the perspective of both a woman in the car and a hitchhiker on the road.

Step 2: Calculating the Radius and Angular Velocity

Convert the tire's diameter to meters: \(66 \mathrm{~cm} = 0.66 \mathrm{~m}\). Thus, the radius \(r = 0.33 \mathrm{~m}\). Convert the car's speed into m/s: \(80 \mathrm{~km/h} = \frac{80 \times 1000}{3600} = 22.22 \mathrm{~m/s}\). The angular velocity \(\omega\) is \(\omega = \frac{v}{r} = \frac{22.22}{0.33} \approx 67.33\, \mathrm{rad/s}\).

Step 3: Velocity Relative to the Woman in the Car

Relative to the woman in the car, the speed at the center of the tire \(\vec{v}_{\text{center}}\) is \(0\) because the center moves with the car, so \(\vec{v}_{\text{center}} = \mathbf{0}\). At the top of the tire \(\vec{v}_{\text{top}}\): it moves at \(2v,\) so \(\vec{v}_{\text{top}} = 2 \cdot 22.22 \hat{i} = 44.44 \hat{i} \mathrm{~m/s}\). At the bottom, \(\vec{v}_{\text{bottom}} = \mathbf{0}\) as it contacts the ground.

Step 4: Acceleration Relative to the Woman in the Car

For uniform circular motion, the centripetal acceleration \(a_c\) is \(\omega^2 r\). The acceleration is \(a_c = (67.33)^2 \times 0.33 \approx 1488.4 \mathrm{~m/s^2}\). The magnitudes for the top and bottom positions remain the same, as they also involve centripetal acceleration.

Step 5: Velocity Relative to the Hitchhiker

For the hitchhiker, the velocity \(\vec{v}_{\text{center}}\) is the car's speed \(22.22 \hat{i} \mathrm{~m/s}\). The top \(\vec{v}_{\text{top}}\) moves at \(2v\), so \(\vec{v}_{\text{top}} = 44.44 \hat{i} \mathrm{~m/s}\). The bottom \(\vec{v}_{\text{bottom}} = 0 \) since it's instantaneously at rest with respect to the ground.

Step 6: Acceleration Relative to Hitchhiker

Centripetal acceleration is the same for the center, top, and bottom relative to the hitchhiker as it is relative to the woman in the car, because it only depends on the wheel's rotation, not its linear motion.

Key Concepts

Angular Velocity

In circular motion, the angular velocity is a crucial concept. It represents how quickly an object rotates about a circular path. Angular velocity is denoted by \(\omega\) and measured in radians per second (rad/s). For a car's tire rolling down the road, every point on the tire follows a circular path. This means understanding the angular velocity helps determine how fast each part of the tire moves.

To calculate angular velocity, use the formula:

• \(\omega = \frac{v}{r}\)

Where:

• \(v\) is the linear velocity (how fast the car is moving forward)
• \(r\) is the radius of the tire

In our example, the car travels at \(22.22\, \text{m/s}\) with a tire radius of \(0.33 \text{ m}\), so

• \(\omega = \frac{22.22}{0.33} \approx 67.33 \, \text{rad/s}\).

This figure shows how quickly the tire is turning to maintain the car's speed.

Centripetal Acceleration

Centripetal acceleration is a type of acceleration in circular motion that keeps an object moving along its circular path. It points towards the center of the circle and ensures the object doesn't fly off in a straight line due to inertia.

In the case of the car tire, both the top and bottom parts of the tire experience this rotational force, so it’s a key factor in calculating their accelerations.
Use the formula:

• \(a_c = \omega^2 r\)

Where:

• \(a_c\) is the centripetal acceleration
• \(\omega\) is the angular velocity
• \(r\) is the radius of the circular path

Plugging in our example values:

• \(a_c = (67.33)^2 \times 0.33 \approx 1488.4 \, \text{m/s}^2\).

This is the centripetal acceleration felt at all points on the tire, constant as it rolls.

Relative Velocity

Relative velocity describes how the velocity of one object seems from another object's perspective. In this exercise, we calculate relative velocities from two points of view: a woman inside the car and a hitchhiker by the road.

• **From the Woman in the Car:**
  • Center of tire: Appears stationary (\(\vec{v}_{\text{center}} = 0\)).
  • Top of tire: Moves at double the car's speed because of rotation (\(\vec{v}_{\text{top}} = 44.44 \, \text{m/s}\)).
  • Bottom of tire: Instantaneously stationary as it meets the ground.
• **From the Hitchhiker:**
  • Center of tire: Matches the car's forward speed (\(\vec{v}_{\text{center}} = 22.22 \, \text{m/s}\)).
  • Top of tire: Rotates to appear at double the car speed (\(\vec{v}_{\text{top}} = 44.44 \, \text{m/s}\)).
  • Bottom of tire: Appears stationary, momentarily, due to contact with ground.

Understanding these perspectives helps grasp how different observers experience the same motion.

Uniform Circular Motion

Uniform circular motion occurs when an object travels in a circular path at a constant speed. It's essential because it means the basic rotational properties do not change over time.

In a rolling tire scenario, uniform circular motion keeps everything consistent:

• **Constant Speed:** Despite the differential velocities at various points, the tire maintains the same overall speed in forward motion.
• **Unchanging Acceleration:** Centripetal acceleration remains constant, acting perpendicular to linear velocity, guiding the rotating tire.

This uniformity simplifies calculations, offering predictable movement patterns for both internal and external observers. It shows that regardless of observational frame (inside the car or stationary), the underlying mechanics maintain the same rules and help streamline complex motion into manageable concepts.