 |
 |
Tire Speed, Top vs. Bottom
Name: Dwight
Status: other
Age: N/A
Location: N/A
Country: N/A
Date: N/A
Question:
This is a question about rotation and
acceleration. We heard a puzzler on a radio program where they
stated that on the tire of a moving vehicle, the point on the
tire in contact with the ground is not moving and the point
opposite that is moving at twice the speed of the vehicle. It is
perfectly clear on the face of it how this can be. But it is
difficult to wrap one's brain around. One student noted that
every point on the outside surface of the tire would have to
accelerate from, say, 0 to 120MPH in the time and distance it
takes for the wheel to make half a turn: like less than three
feet distance in a few hundredths of a second. This is almost
like being shot out of a gun. Why doesn't that kind of
acceleration cause the tire to fly apart? Is this constant
acceleration the same as centripetal force? Can you give us a
formula for calculating the acceleration on a 24 inch diameter
wheel going 60 miles an hour?
Replies:
The wheel does not increase its speed at all. In this case, the outer
surface of the tire is moving at 60 mph relative to the axle at all
times; only its direction is changing. The acceleration is v*v/r,
where v is the speed and r is the radius of the tire.
Tim Mooney
The velocity of a point on the tire relative to the ground is the sum of
the velocity of the point on the tire relative to the car plus the
velocity of the car relative to the ground. That is
V(tire,ground) = V(tire,car) + V(car,ground)
Since velocities are vectors this could be relatively complicated but
for the case of points on top or bottom of the tire the addition is
simple since the movement of those points on the tire is parallel to the
movement of the car.
So, if the car is moving at 60 mph, V(car,ground)=60 mph
The top of the tire is moving 60 mph relative to the car in the same
direction as the car. You should be able to convince yourself that this
is true.
V(tire top,car)= 60 mph
The bottom of the tire is moving 60 mph relative to the car in the
direction opposite to the car's motion so
V(tire bottom,car)= -60 mph.
Thus the top of the tire is moving 60+60=120 mph relative to the ground
and the bottom of the tire is moving 60-60=0 mph relative to the ground.
This has to be true otherwise the tire would be skidding.
David Kupperman
The problem becomes simpler if you use a different frame of
reference. Try using the car axle as the frame of reference and you
see that you can view this situation as one of stable linear motion
of a center of rotation. From the point of view of the forces on
the tire the only motion is a rotation about the axle, and that is
at a steady rate. As long as the tire can hold up to the rotational
acceleration there is no danger.
Of course, I have ignored many things here, such as deformation of
the tire where it contacts the ground, etc, but these are not
significant to the question.
Greg Bradburn
Click here to return to the Physics Archives
| |
Update: June 2012
|
|
