A car travels at 80 km/h on a level road in the positive direction of an x axis. Each tire has a diameter of 66 cm. Relative to a woman riding in the car and in unit-vector notation, what are the velocity 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 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 (l) bottom of each tire?
To solve this problem, we'll need to use some concepts from physics and kinematics. We'll start by finding the velocity at the center, top, and bottom of the tire, and then calculate the magnitude of acceleration using the given information.
First, we need to convert the car's speed from km/h to m/s since the SI units are used in physics. Speed can be converted by dividing by 3.6, so 80 km/h is equal to (80/3.6) m/s ≈ 22.22 m/s.
Now, let's calculate the velocity at the center, top, and bottom of the tire relative to the woman inside the car.
(a) Velocity at the center of the tire:
Since the car is traveling in the positive direction of the x-axis, the velocity at the center of the tire will also be in the positive x-direction. Therefore, the velocity at the center is v = (22.22 m/s) * (i-hat) = 22.22i, where "i" represents the unit vector in the x-direction.
(b) Velocity at the top of the tire:
The top of the tire is moving forward and rotating about the center. Since the tire is rolling without slipping, the velocity at the top of the tire is the sum of the translational velocity of the car and the tangential velocity due to rotation. The tangential velocity can be calculated using the formula v = ω * r, where ω is the angular velocity (in radians per second) and r is the radius of the tire.
The radius of the tire is half the diameter, so r = 66/2 cm = 0.33 m.
Since the tire is rolling without slipping, the rotational speed can be related to the translational speed as v = ω * r. So, ω = v / r = (22.22 m/s) / (0.33 m) ≈ 67.33 rad/s.
The velocity at the top of the tire will be the sum of the translational and tangential velocities. The translational velocity is in the positive x-direction, and the tangential velocity is in the clockwise direction due to rotation. Therefore, the velocity at the top of the tire is v = (22.22 m/s) * (i-hat) + (67.33 m/s) * (j-hat) = 22.22i + 67.33j.
(c) Velocity at the bottom of the tire:
The bottom of the tire is moving forward as the tire is rolling and rotating about the center. Again, the velocity at the bottom of the tire is the sum of the translational velocity and the tangential velocity.
Using the same calculations, the velocity at the bottom of the tire is v = (22.22 m/s) * (i-hat) + (-67.33 m/s) * (j-hat) = 22.22i - 67.33j.
Now, let's find the magnitude of the acceleration at the center, top, and bottom of each tire.
(d) Magnitude of acceleration at the center of the tire:
Since the car is moving at a constant speed on a level road, there is no acceleration at the center of the tire. Therefore, the magnitude of acceleration at the center is 0.
(e) Magnitude of acceleration at the top of the tire:
The only acceleration present at the top of the tire is due to the rotation. The magnitude of the tangential acceleration can be calculated using the formula a = α * r, where α is the angular acceleration and r is the radius of the tire. Since the car is moving at a constant speed, the angular acceleration is 0.
Therefore, the magnitude of acceleration at the top of the tire is 0.
(f) Magnitude of acceleration at the bottom of the tire:
Similar to the top of the tire, the only acceleration present at the bottom of the tire is due to rotation, and the magnitude is 0.
Now, let's calculate the velocity and acceleration relative to the hitchhiker sitting next to the road.
(g), (h), and (i) Velocity at the center, top, and bottom of the tire relative to the hitchhiker:
Since the hitchhiker is next to the road and observing the car, the velocity relative to the hitchhiker will be the same as the velocity of the car. Therefore, the velocities at the center, top, and bottom of the tire relative to the hitchhiker are the same as calculated earlier: v = 22.22i, v = 22.22i + 67.33j, and v = 22.22i - 67.33j respectively.
(j), (k), and (l) Magnitude of acceleration at the center, top, and bottom of the tire relative to the hitchhiker:
Again, since the car is moving at a constant speed, there is no acceleration observed by the hitchhiker. Therefore, the magnitude of acceleration at the center, top, and bottom of the tire relative to the hitchhiker is 0.
So, in unit-vector notation, the answers are:
(a) Velocity at the center: 22.22i
(b) Velocity at the top: 22.22i + 67.33j
(c) Velocity at the bottom: 22.22i - 67.33j
(d) Magnitude of acceleration at the center: 0
(e) Magnitude of acceleration at the top: 0
(f) Magnitude of acceleration at the bottom: 0
(g) Velocity at the center relative to the hitchhiker: 22.22i
(h) Velocity at the top relative to the hitchhiker: 22.22i + 67.33j
(i) Velocity at the bottom relative to the hitchhiker: 22.22i - 67.33j
(j) Magnitude of acceleration at the center relative to the hitchhiker: 0
(k) Magnitude of acceleration at the top relative to the hitchhiker: 0
(l) Magnitude of acceleration at the bottom relative to the hitchhiker: 0