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 first need to determine the linear velocity and angular velocity of the car.
Given:
- Car speed (v) = 80 km/h
- Tire diameter (d) = 66 cm
Step 1: Convert the car speed to meters per second (m/s)
To convert the speed from km/h to m/s, we need to divide it by 3.6 since 1 km/h = 0.2778 m/s.
So, the car speed is 80 km/h × 0.2778 m/s = 22.22 m/s.
Step 2: Calculate the angular velocity (ω) of the wheel
The angular velocity (ω) is defined as the ratio of the linear velocity (v) to the radius (r) of the wheel.
The radius (r) is half the diameter, so r = 66 cm / 2 = 33 cm = 0.33 m.
Angular velocity (ω) = Linear velocity (v) / Radius (r)
ω = 22.22 m/s / 0.33 m = 67.33 rad/s
Step 3: Calculate the velocities at different points of the tire
(a) Center of the tire:
The velocity at the center of the tire is the same as the velocity of the car.
Velocity at the center = (22.22 m/s) i (in unit-vector notation)
(b) Top of the tire:
The top of the tire is moving forward and rotating due to the angular velocity.
Velocity at the top = (22.22 m/s + 0.33 m × 67.33 rad/s) i
(c) Bottom of the tire:
The bottom of the tire is moving forward but rotating in the opposite direction.
Velocity at the bottom = (22.22 m/s - 0.33 m × 67.33 rad/s) i
Step 4: Calculate the accelerations at different points of the tire
To calculate the accelerations, we need to consider the radial acceleration (ar) and tangential or linear acceleration (at).
(d) Center of the tire:
At the center of the tire, there is only tangential acceleration due to the change in linear velocity.
Acceleration at the center = 0 (no change in direction)
(e) Top of the tire:
At the top of the tire, there is both tangential and radial acceleration.
Radial acceleration (ar) at the top = ω²r
Tangential acceleration (at) at the top = αr (where α is angular acceleration, but here α = 0 since the angular velocity is constant)
Acceleration at the top = -ω²r j
(f) Bottom of the tire:
At the bottom of the tire, like the top, there is both tangential and radial acceleration.
Radial acceleration (ar) at the bottom = ω²r
Tangential acceleration (at) at the bottom = αr (where α is angular acceleration, but here α = 0 since the angular velocity is constant)
Acceleration at the bottom = ω²r j
Step 5: Repeat steps 3 and 4 considering the hitchhiker sitting next to the road
Relative to the hitchhiker, the linear velocity of the car is the same, but the angular velocity and accelerations differ.
(g) Center of the tire:
Velocity at the center = (22.22 m/s) (-i) (in unit-vector notation)
(h) Top of the tire:
Velocity at the top = (22.22 m/s + 0.33 m × 67.33 rad/s) (-i)
(i) Bottom of the tire:
Velocity at the bottom = (22.22 m/s - 0.33 m × 67.33 rad/s) (-i)
(j) Center of the tire:
Acceleration at the center = 0 (no change in direction)
(k) Top of the tire:
Acceleration at the top = -ω²r (-j)
(l) Bottom of the tire:
Acceleration at the bottom = ω²r (-j)
Note: Unit-vector notation refers to using unit vectors (i, j, k) to represent the directions along the x, y, and z axes, respectively. In this case, only the x-component (i) is considered since the car is moving along a level road.
To solve this problem, we need to consider both the linear velocity and the rotational velocity of the tires.
Let's start by finding the linear velocity of the car. We are given that the car travels at 80 km/h, which we can convert to meters per second. To do that, we divide 80 km/h by 3.6 since there are 3.6 seconds in an hour.
Linear velocity of the car: v_car = 80 km/h ÷ 3.6 = 22.22 m/s
Now, let's find the rotational velocity of the tires. The circumference of a circle is given by C = πd, where d is the diameter of the circle. In this case, each tire has a diameter of 66 cm, so the circumference of each tire is C = π × 66 cm.
Rotational velocity of the tire: v_rot = linear velocity / (circumference of the tire) = 22.22 m/s / (π × 66 cm)
To find the velocity at different points on the tire relative to a person riding in the car:
(a) The velocity at the center of the tire will be the same as the velocity of the car since the center of the tire is in contact with the road.
Velocity at the center of the tire: v_center = 22.22 m/s in the positive x-direction
(b) The velocity at the top of the tire will be the sum of the linear velocity of the car and the rotational velocity of the tire.
Velocity at the top of the tire: v_top = v_car + v_rot in the positive x-direction
(c) The velocity at the bottom of the tire will be the linear velocity of the car minus the rotational velocity of the tire.
Velocity at the bottom of the tire: v_bottom = v_car - v_rot in the positive x-direction
To find the magnitude of acceleration at different points on the tire, we need to consider the centripetal acceleration. Centripetal acceleration is given by a = v_rot^2 / R, where R is the radius of the tire.
(d) For the center of the tire, the radius is equal to half of the diameter.
Magnitude of acceleration at the center of the tire: a_center = v_rot^2 / (R_center) = v_rot^2 / (0.5d) in the negative x-direction
(e) For the top of the tire, the radius is the sum of half the diameter and the distance from the center to the top.
Magnitude of acceleration at the top of the tire: a_top = v_rot^2 / (R_top) = v_rot^2 / (0.5d + r) in the negative x-direction
(f) For the bottom of the tire, the radius is the difference between half the diameter and the distance from the center to the bottom.
Magnitude of acceleration at the bottom of the tire: a_bottom = v_rot^2 / (R_bottom) = v_rot^2 / (0.5d - r) in the negative x-direction
To find the velocity at different points on the tire relative to a hitchhiker sitting next to the road, we need to consider the velocity of the car and the rotational velocity of the tire.
(g) The velocity at the center of the tire will be equal to the velocity of the car since the tire is moving with the car.
Velocity at the center of the tire: v_center = 22.22 m/s in the positive x-direction
(h) The velocity at the top of the tire will be the sum of the linear velocity of the car and the rotational velocity of the tire.
Velocity at the top of the tire: v_top = v_car + v_rot in the positive x-direction
(i) The velocity at the bottom of the tire will be equal to the linear velocity of the car since the rotational velocity of the tire and the linear velocity of the car cancel each other out.
Velocity at the bottom of the tire: v_bottom = 22.22 m/s in the positive x-direction
To find the magnitude of acceleration at different points on the tire relative to a hitchhiker sitting next to the road, we only need to consider the rotational velocity of the tire.
(j) The magnitude of acceleration at the center of the tire will be zero since the tire is moving with a constant linear velocity.
Magnitude of acceleration at the center of the tire: a_center = 0
(k) The magnitude of acceleration at the top of the tire will be equal to the centripetal acceleration.
Magnitude of acceleration at the top of the tire: a_top = v_rot^2 / R = v_rot^2 / (0.5d + r) in the positive x-direction
(l) The magnitude of acceleration at the bottom of the tire will be equal to the centripetal acceleration.
Magnitude of acceleration at the bottom of the tire: a_bottom = v_rot^2 / R = v_rot^2 / (0.5d - r) in the positive x-direction
This is how you can find the velocities and magnitudes of acceleration at different points on the tire relative to a person riding in the car and a hitchhiker sitting next to the road.
Just use imagination
From car:
center (the axle) does not move
bottom moves back at 80 with the road
so top has to move forward at 80
from roadside:
center moves forward at 80
top forward at 160
bottom still where it touches stationary ground
the axle is not accelerating
the top and bottom have centripetal acceleration toward the axle.