Translations and rotations in a car's tire...?

Question

Consider a 66-cm-diameter tire on a car traveling at 80 km/h on a level road in the positive direction of an x-axis.

Relative to a woman in the car, what are (a) the translational velocity v and (b) the magnitude a of the translational acceleration of the center of the wheel?

What are (c) v and (d) a for a point at the top of the tire?

What are (e) v and (f) a for a point at the bottom of the tire?

Now repeat the questions relative to a hitchhiker sitting near the road: What are (g) v at the wheel’s center, (h) a at the wheel’s center, (i) v at the tire top, (j) a at the tire top, (k) v at the tire bottom, and (l) a at the tire bottom?

What equations would you use to solve this, and can you show your work?

Answer 1

Relative to a woman in the car:

(a) the translational velocity is ZERO since the woman is moving at the same translational velocity as the car relative to the ground.

(b) the magnitude of the translational acceleration of the center of the wheel is also ZERO because the center of the wheel is NOT moving and NOT rotating relative to the woman.

(c) v, velocity of the point at the top of the tire is = 80 km/h because that is how fast the car is moving, thus the wheel must also be rotating at the same speed.

(d) a, acceleration of the point at the top of the tire is given by:

a = v^2/r = (22.22)^2/(0.33) = 1,496.45 m/s^2

(e) v, velocity of the point at the bottom of the tire is = -80 km/h because at the bottom of the tire, the tire is moving in the opposite direction

(f) a, acceleration of the point at the bottom of the tire is the same as (d) a = v^2/r = 1,496.45 m/s^2

Now relative to a hitch hiker on the ground:

(g) v at the center of the wheel is = 80 km/h or 22.22 m/s, since the car and the center of the wheel (which is not moving relative to the car) are moving at the same speed relative to the ground.

(h) a at the center of the wheel is = v^2/r = 1,496.45 m/s^2, since the center of the wheel is rotating about the bottom point of the tire as the tire rolls on the ground.

(i) v at the top of the tire is v(top) = ωd, where ω is the angular velocity of the rotating tire and d is the diameter of the wheel. The reason we use diameter of the wheel is because the top of the tire is rotating abou the bottom of the tire, not the center.

ω = v/r = (22.22)/(0.33) = 67.33 rad/sec
v(top) = ωd = (67.33)(0.66) = 44.44 m/s

(j) a at the top of the tire is = v(top)^2/d
= (44.44)^2/(0.66) = 2,992.29 m/s^2

(k) v at the bottom of the tire is equal to ZERO because the bottom of the tire is not moving relativ to the hitch hiker or the ground.

(l) a at the bottom of the tire is also equal to ZERO since the velocity is zero.

Answer 2

Dude, that would be work... Look, for A and B, the center of the wheel is fixed relative to the passenger, so 0 for both. The bottom of the tire is sitting still on the ground, so it's moving -80kph relative to the woman, whereas the top is moving at +80kph. The bottom is being accelerated up, and the top is being accelerated down; I'm not dong the math to find the numbers but it's probably sines and cosines of time. For the hitchhiker, g) 80kph, h)0 (constant v), the tire bottom is not moving (stuck to the road), the top is moving at twice the speed, so 160kph and the accelerations should be the same as for the passenger.

Answer 3

The wheel's center is stationary with respect to (wrt) any point in or on the car.
The wheels rotate at an angular rate w. You can determine w from the tire radius r and car's velocity Vc.
r=D/2
w=Vc/r
Circular motion is resolved into X (hor., +=fwd) and Y (vert., +=up) components. For a point on the "top" (defined at time t=0) of a wheel on the right side, displacements Dx=r*sin(theta) and Dy=r*cos(theta), where theta is w*t. A point on the bottom has displacements and motions of equal magnitude and opposite sign.

Velocity magnitude = w * displacement magnitude, and velocity is the derivative of displacement. So we have, wrt the car,
top, velocity: Vx=w*r*cos(w*t), Vy=-w*r*sin(w*t)
bottom, velocity: Vx=-w*r*cos(w*t), Vy=w*r*sin(w*t)

Acceleration magnitude = w * velocity magnitude, and acceleration is the derivative of velocity. Thus, wrt the car,
top, accel: Ax=-w^2*r*sin(w*t), Ay=-w^2*r*cos(w*t)
bottom, accel: Ax=w^2*r*sin(w*t), Ay=w^2*r*cos(w*t)

For the stationary observer, just add Vc to the Ax and Vx terms above.

Answer 4

Do U think ppl here r scientists?

Answer 5

wow just reading that made my head hurt,, try math that is not so hard:)