5-29-98
When an object is experiencing uniform circular motion, it is traveling in a circular path at a constant speed. If r is the radius of the path, and we define the period, T, as the time it takes to make a complete circle, then the speed is given by the circumference over the period. A similar equation relates the magnitude of the acceleration to the speed:
These two equations can be combined to give the equation:
This is known as the centripetal acceleration; v2 / r is the special form the acceleration takes when we're dealing with objects experiencing uniform circular motion.
In circular motion many people use the term centripetal force, and say that the centripetal force is given by:
I personally think that "centripetal force" is misleading, and I will use the phrase centripetal acceleration rather than centripetal force whenever possible. Centripetal force is a misleading term because, unlike the other forces we've dealt with like tension, the gravitational force, the normal force, and the force of friction, the centripetal force should not appear on a free-body diagram. You do NOT put a centripetal force on a free-body diagram for the same reason that ma does not appear on a free body diagram; F = ma is the net force, and the net force happens to have the special form when we're dealing with uniform circular motion.
The centripetal force is not something that mysteriously appears whenever an object is traveling in a circle; it is simply the special form of the net force.
Whenever an object experiences uniform circular motion there will always be a net force acting on the object pointing towards the center of the circular path. This net force has the special form
, and because it points in to the center of the circle, at right angles to the velocity, the force will change the direction of the velocity but not the magnitude.
It's useful to look at some examples to see how we deal with situations involving uniform circular motion.
Example 1 - The Earth orbiting the Sun. What is the magnitude of the acceleration experienced by the Earth in its orbit around the Sun? We can work out the answer two different ways.
The free-body diagram of the Earth in this situation involves only one force, the gravitational force exerted by the Sun on the Earth, given by Newton's law of universal gravitation, which we've seen before:
The two masses are the mass of the Sun and the mass of the Earth, and r is the distance between them. Because this is the only force acting on the Earth, we can set this equal to ma, remembering that a has the special form a = v2 / r:
This gives us two ways to get the acceleration,
Method 1 - To get the acceleration using the first equation, we need to know the mass of the Sun and the distance from the Earth to the Sun. Roughly, the Sun has a mass of 2 x 1030 kg, and it's 93 million miles = 150 million km = 1.5 x 1011 m away. Plugging these values in gives an acceleration of :
Method 2 - the second method should give us the same answer, but we need to know the speed of the Earth as it orbits the Sun. The period, T, the time it takes the Earth to orbit the Sun, is one year, which, very conveniently, happens to work out to very close to 
. The speed of the Earth is:
This gives an acceleration of:
That's pretty good agreement between the two methods, given the approximating we're doing here...which means the approximations are rather good. The acceleration is relatively small...but sufficient to make the Earth keep orbiting the Sun.
Example 2 - Twirling an object tied to a rope in a horizontal circle. (Note that the object travels in a horizontal circle, but the rope itself is not horizontal). If the tension in the rope is 100 N, the object's mass is 3.7 kg, and the rope is 1.4 m long, what is the angle of the rope with respect to the horizontal, and what is the speed of the object?
As always, the place to start is with a free-body diagram, which just has two forces, the tension and the weight. It's simplest to choose a coordinate system that is horizontal and vertical, because the centripetal acceleration will be horizontal, and there is no vertical acceleration.
The tension, T, gets split into horizontal and vertical components. We don't know the angle, but that's OK because we can solve for it. Adding forces in the y direction gives:
This can be solved to get the angle:
In the x direction there's just the one force, the horizontal component of the tension, which we'll set equal to the mass times the centripetal acceleration:
We know mass and tension and the angle, but we have to be careful with r, because it is not simply the length of the rope. It is the horizontal component of the 1.4 m (let's call this L, for length), so there's a factor of the cosine coming in to the r as well.
Re-arranging this to solve for the speed gives:
which gives a speed of v = 5.73 m/s.
Example 3 - Identical objects on a turntable, different distances from the center. Let's not worry about doing a full analysis with numbers; instead, let's draw the free-body diagram, and then see if we can understand why the outer objects get thrown off the turntable at a lower rotational speed than objects closer to the center.
In this case, the free-body diagram has three forces, the weight, the normal force, and a frictional force. The friction here is static friction, because even though the objects are moving, they are not moving relative to the turntable. If there is no relative motion, you have static friction. The frictional force also points towards the center; the frictional force acts to oppose any relative motion, and the object has a tendency to go in a straight line which, relative to the turntable, would carry it away from the center. So, a static frictional force points in towards the center.
Summing forces in the y-direction tells us that the normal force is equal in magnitude to the weight. In the x-direction, the only force there is is the frictional force.
The maximum possible value of the static force of friction is
As the velocity increases, the frictional force has to increase to provide the necessary force required to keep the object spinning in a circle. If we continue to increase the rotation rate of the turntable, thereby increasing the speed of an object sitting on it, at some point the frictional force won't be large enough to keep the object traveling in a circle, and the object will move towards the outside of the turntable and fall off.
Why does this happen to the outer objects first? Because the speed they're going is proportional to the radius (v = circumference / period), so the frictional force necessary to keep an object spinning on the turntable ends up also being proportional to the radius. More force is needed for the outer objects at a given rotation rate, and they'll reach the maximum frictional force limit before the inner objects will.
A good example of uniform circular motion is a car going around a banked turn, such as on a highway off-ramp. These off-ramps often have the recommended speed posted; even if there was no friction between your car tires and the road, if you went around the curve at the "design speed" of the curve you would be fine. With no friction, if you went faster than the design speed you would veer towards the outside of the curve, and if you went slower than the design speed you would veer towards the inside of the curve.
In theory, then, accounting for friction, there is a range of speeds at which you can negotiate a curve. Practically speaking, however, in most cases the coefficient of friction is sufficiently high, and the angle of the curve sufficiently small, that there is no danger of going too slow around the curve. Going too fast is another story, however.
The textbook does a good analysis of a car on a banked curve without friction, arriving at a connection between the angle of the curve, the radius, and the speed. The speed is known as the design speed of the curve (the speed at which you're safest negotiating the curve) and is given by:
Consider now the role that friction plays, and think about how to determine the maximum speed at which you can negotiate the curve without skidding. The first thing to realize is that the frictional force is static friction; even though the car is moving, the car tires are not slipping on the road surface, so the part of tire in contact with the road is instantaneously at rest with respect to the road. Also, if we're worried about the maximum speed at which we can go around the banked turn, if there was no friction the car would tend to slide towards the outside of the curve, so the friction opposes this tendency and points down the slope.
The diagram, and a free-body diagram, of the situation is shown here. Note that the diagram looks similar to that of a box on an inclined plane. There is a critical difference, however; for the box on an inclined plane, the coordinate system was parallel and perpendicular to the slope, because the box was either moving, and/or accelerating, up or down the slope. In this case the coordinate system is horizontal and vertical, because the centripetal acceleration points horizontally in towards the center of the circle and there is no vertical component of the acceleration.
Moving from the free-body diagram to the force equations gives:
This can be re-arranged to solve for the normal force:
Note that we're solving for the maximum speed at which the car can go around the curve, which will correspond to the static force of friction being a maximum, which is why it's valid to say that 
.
In the x-direction, the force equation is:
Substituting , and plugging in the equation for the normal force, gives:
There is an m in every term, so the mass cancels out. The equation can then be re-arranged to solve for the maximum speed:
Note that when the coefficient of friction is zero (i.e., the road is very slippery), the maximum speed reduces to the design speed,
and for certain combinations of theta and the coefficient of friction (both large, in general) the denominator turns out to be negative, implying that there is no maximum speed; in those cases, you could drive as fast as you wanted without worrying about skidding. Note that this does not apply to standard highway off-ramps! The appropriate conditions for no maximum safe speed (or at least a very high maximum) would be found at racetracks like the Indianapolis Speedway, for example.
Some roller-coasters have loop-the-loop sections, where you travel in a vertical circle, so that you're upside-down at the top. To do this without falling off the track, you have to be traveling at least a particular minimum speed at the top. The critical factor in determining whether you make it completely around is the normal force; if the track has to exert a downward normal force at the top of the track, you're fine, but if the normal force drops to zero you're in trouble.
The normal force changes as you travel around a vertical loop; it changes because your speed changes and because your weight has a different effect at each part of the circle. To keep going in a circular path, you must always have a net force equal to mv2 / r pointing towards the center of the circle. If the net force drops below the required value, you will veer off the circular path away from the center, and if the net force is more than the required value you will veer off towards the center.
Consider what happens at the bottom of the loop, and compare it to what happens at the top. At the bottom, mg points down and the normal force points up, towards the center of the circle. The normal force is then not simply equal to the weight, but is larger because it must also supply the required centripetal acceleration:
This is why you actually feel heavier at the bottom of a loop like this, because your apparent weight is equal to the normal force you feel.
At the top of the loop, on the other hand, the normal force and the weight both point towards the center of the circle, so the normal force is less than the weight:
If you're going at just the right speed so mv2 / r = mg, the normal force drops to zero, and you would actually feel weightless for an instant. Faster than this speed and there is a normal force helping to keep you on the circular path; slower and the normal force would go up, which means you'll fall out of the coaster and/or the coaster will fall off the track unless you're strapped in and the coaster is held down to the track.