0 - 100 real quick!

A car is driving at 100 mph 100\text{ mph} down a straight road. What is the speed at the point on the tire where the wheel touches the ground?

Note: Assume that the tires are not slipping or skidding.

0 mph 0\text{ mph} 25 mph 25\text{ mph} 50 mph 50\text{ mph} 100 mph 100\text{ mph}

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2 solutions

Zach Abueg
Jan 15, 2017

When a wheel rolls, it is moving in two ways: rotationally, around its center, and horizontally, in the direction of the travel. At its point of contact with the ground, both of these motions cancel each other out, leaving a net speed of 0 0 .

Moderator note:

This is a perfect circumstance where thinking with vectors is helpful. Visualize the rotational vector as it spins around the wheel.

The point when the rotational vector is exactly opposed to the forward motion of the car is when the point is touching the ground.

If the point is on the top of the wheel, both vectors are facing the same direction and the speed is doubled!

Yes, on a fixed ground the two motions will cancel at the bottom most point if v = R ω v= R \omega , here v v is the speed of the car, R R is the radius of the tire and ω \omega is the angular velocity of the tire.

Rohit Gupta - 4 years, 4 months ago

So is the point on the top of the wheel moving 200 mph?

Max Tepper - 4 years, 3 months ago

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Yes, it is moving at 200 mph.

Rohit Gupta - 4 years, 3 months ago

But that's velocity Speed= Distance/Time and Velocity = DISPLACEMENT/Time Speed is a scalar, "Velocity" is a VECTOR quantity!

Mihir Amdekar - 4 years, 1 month ago

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I think you are confused between speed and average speed.

Speed or instantaneous speed is the magnitude of the velocity. Whereas, average speed is total distance traveled by total time taken.

Rohit Gupta - 4 years, 1 month ago

You talk about velocity without talking about the referential system according to which you are calculating this velocity!. Let's put the referential (o,x,y,z) on the car wheel axis, the whole car will be at rest relatively to this system except the wheel that is spining around it. In this case, the point will not be at rest because you have just one vector which is tangential to the wheel and cannot be canceled. Isn't it ?.

Tunisna Haut - 3 years, 1 month ago

But the question didn't mention which reference are we calculating this speed in.

Malika Oubilla - 2 years, 9 months ago
Rohit Gupta
Feb 3, 2017

It is very surprising at first that how can a moving rigid object have a point on it that is at rest. However, it's possible in a way that the point does not remain at rest forever rather it's instantaneous speed that is zero. In the next instant, the bottom most point is changed and some other point of the wheel takes this position.

Unless the tire is skidding/slipping, the contact points must have equal velocities. Hence, the bottom most point remains at rest.

Moreover, if the bottom most point has a velocity in the direction of motion of the car. The car is said to be skidding forward. This case generally occurs when hard brakes are applied on a moving vehicle.

Similarly, if the bottom most point has a velocity opposite to the direction of motion of the car. The car is said to be going under backward skidding. This situation generally occurs when tires are stuck in the mud and we try to accelerate the car. In this case, the tire rotates more than moving forward.

Nice explanation! It seems counter intuitive that point is at rest because it's at rest only for an instant.

We can often use the fact that the point is instantaneous rest to simplify calculations. We can assume the wheel to performing purely rotational motion and no translation about this point. This is known as the instantaneous axis of rotation.

Pranshu Gaba - 4 years, 4 months ago

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Yes, I agree with you. The concept of the instantaneous axis of rotation is helpful in many cases. Like in this case, we can write the kinetic energy of the wheel directly as 1 2 I b o t t o m ω 2 \frac{1}{2} I_{bottom} \omega^2 .

Rohit Gupta - 4 years, 4 months ago

The explanation which says the that point on the wheel in contact with the ground is at rest is off the mark. The wheel is a rigid object. All the points on its circumference are moving at the same rotational speed and linear speed which depends on its radius and circumference. The car is moving at 100 mph so any point in contact with the ground, absent skidding, must be moving backwards with respect to the car at 100 mph. True its position, x, is a particular point at time t. But the first derivative of x with respect to time, dx/dt is -100 mph with respect to the car and wrt to the ground at that moment. The moment passes instantaneously as that point on the tire lifts off the ground. This is reminiscent of Zeno's paradox about the flight of an arrow which purposely confuses the position of an arrow with its speed. At each moment, the arrow is at a particular location but it is not at rest at that location. Just because the ground is at rest does not mean the wheel is at rest. That would be true if they were in contact for any finite time, but they are not. Instantaneity is not finite.

Paul Palmer - 4 years, 3 months ago

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If we throw a ball vertically upwards, what would be its speed when it is at its topmost point? It would be zero, even if it is just for an instant. It would have a non-zero speed before and after it is at the topmost point.

We can use a similar argument for the bottommost point of the wheel.

Pranshu Gaba - 4 years, 3 months ago

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Asked value should be velocity not speed If the answer is 0

gaurav modak - 4 years, 3 months ago

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@Gaurav Modak Speed is the magnitude of the velocity. Hence, if the velocity is zero so does the speed.

Rohit Gupta - 4 years, 3 months ago

@Pranshu Gaba Sorry, no connection between the two examples. The acceleration remains constant in both cases. The speed of every point on the circumference of the wheel is constant though the velocity keeps changing as its direction changes. In the ball example, the acceleration is the constant acceleration of gravity. Yes, the speed passes through the value zero at the top but zero is just some number it goes through, just like it goes through other numbers. You can't put your mind on an instant of zero duration and then pretend that it is the same as an instant of finite duration because our minds can't easily encompass zero duration any more than they can infinite duration. The only thing we can do is to view a single point as a limit. Before it touches the ground (at every point actually) a point on the wheel is moving at a finite speed. After the contact, the same point is moving at a finite speed. Condense that down to the limit as it touches the ground and the limit is still the same speed.

Paul Palmer - 4 years, 2 months ago

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@Paul Palmer In the frame of the car, the bottom most point of the tyre moves backwards with velocity ω r \omega r . The car itself is moving forward with velocity ω r \omega r . Adding vectorially, the velocity at bottom most point in the ground frame is ω r ω r = 0 \omega r - \omega r = 0 .

Any point on the wheel moves in the shape of a cycloid. Note that its speed approaches zero when it touches the ground.

Pranshu Gaba - 4 years, 2 months ago

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@Pranshu Gaba If you replace the wheel with caterpillar tracks, it is becoming absolutly obvious: Each plate of the tracks is placed on the ground once per turn, lying motionlessly on the ground for a short time, until it is lifted off again (and transported forward on the upper side with double speed in relation to the ground).

Motion is always relative, and only makes sense in regard to a reference system. The wheel turns continuesly und all its surface points move—relative to the axis(!)—with constant angular speed, resulting in this case to 100 mph constant linear speed (seen relative to the axis!). But in this problem the ground is meant to be the reference system.

Martin Ramsch - 3 years, 4 months ago

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@Martin Ramsch I agree, caterpillar tracks is a great way to visualize this!

Pranshu Gaba - 3 years, 4 months ago

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