Uniform Circular Motion: Centripetal Forces Printer Friendly Version
In order for an object to travel in a circular path, an unbalanced central force must be exerted upon it. Otherwise, the object would continue to travel along a straight-line path based on its inertia. Typical sources of centripetal forces are tensions in strings, friction, banked curves, and gravitation. Magnetism is a non-mechanical force that provides the necessary force to cause charged particles to travel in circular paths.

Example #1: Strings and Flat Surfaces

Suppose that a mass is tied to the end of a string and is being whirled in a circle along the top of a frictionless table as shown in the diagram below.

A freebody diagram of the forces on the mass would show
The tension is the unbalanced central force: T = Fc = mac, it is supplying the centripetal force necessary to keep the block moving in its circular path.

Refer to the following information for the next two questions.

Suppose a 250-gram mass is being whirled in a circular path by a string on the surface of a smooth table.
 If the maximum tensile strength in the string is 10 N, how fast would the mass slide along the table when the radius of its path equals 40 cm?

 How many revolutions would it make is 60 seconds?

Example #2: Conical Pendulums

Our next example is also an object on the end of string but this time it is a conical pendulum. Notice, that its path also tracks out a horizontal circle in which gravity is always perpendicular to the object's path.

A freebody diagram of the mass on the end of the pendulum would show the following forces.

T cos θ is balanced by the object's weight, mg. It is T sin θ that is the unbalanced central force that is supplying the centripetal force necessary to keep the block moving in its circular path: T sin θ = Fc = mac.

Refer to the following information for the next four questions.

Suppose a 75-gram ball is being whirled as a conical pendulum by a child. The ball is attached to a 50-cm string and tracks out a horizontal circle with a radius of 40 cm.
 What is the measure of angle θ?

 What is the tension in the rope?

 How fast is the ball traveling as it swings?

 How many revolutions does the stopper complete each second?

Example #3: Flat Curves

Many times, friction is the source of the centripetal force. Suppose in our initial example that a car is traveling through a curve along a flat, level road. A freebody diagram of this situation would look very much like that of the block on the end of a string, except that friction would replace tension.

Friction is the unbalanced central force that is supplying the centripetal force necessary to keep the car moving along its horizontal circular path: f = Fc = mac.

Since f = μN and N = mg on this horizontal surface, most problems usually ask you to solve for the minimum coefficient of friction required to keep the car on the road.

Refer to the following information for the next three questions.

A 1500-kg car is traveling at 24 m/sec through a flat 100-meter radius turn.
 How large is the frictional force required to keep the car moving in its circular path?

 What is the correspondingly minimal coefficient of friction between the road surface and the car's tires?

 How would this coefficient of friction be changed if a 3000-kg pickup truck were traveling through the same curve?

Example #4: Banked Curves

If instead, the curve is banked then there is a critical speed at which the coefficient of friction can equal zero and the car still travel through the curve without slipping out of its circular path.

A freebody diagram of the forces acting on the car would show weight and a normal. Since the car is not sliding down the bank of the incline, but is instead traveling across the incline, components of the normal are examined.

N sin θ is the unbalanced central force; that is, N sin θ = Fc = mac. This component of the normal is supplying the centripetal force necessary to keep the car moving through the banked curve.
Dividing the equations
N sin θ = mv2/r
N cos θ = mg
yields the equation
tan θ = v2/rg.
Solving for v produces the desired result
vcritical = √ (rg tan θ).

At this critical speed, there is no need for any friction between the car and the road's surface. If the speed of the car were to exceed vcritical then the car would drift up the incline. If the speed of the car is less than vcritical then the car would slip down the incline.

Refer to the following information for the next three questions.

At a local NASCAR racetrack, cars travel through a 316-meter radius curve that is banked at 31º.
 At what speed would the race cars be traveling if they wanted to pass through this curve frictionlessly?

 Does your answer to the previous question depend on the mass of the car?

 If the cars actually travel through this turn in excess of 195 mph, or 87 m/sec, what would supply the additional centripetal force?

Example #5: Gravitation and Satellites

Another major category of forces that produce uniform circular motion is gravitation. At a radius, r, from the center of the Earth, the gravitational field strength, g, can be calculated as:

Alternatively, we can also state that for a satellite in circular orbit that gravity supplies the unbalanced central force required to keep it in orbit.

Thus, the field strength of the Earth's gravitational field, g, is equivalent to the centripetal acceleration experienced by a satellite in circular orbit at radius r from the center of the Earth.

The formula derived above is the speed of a satellite orbiting a central mass, M, at a center-to-center distance of r. This speed can also be expressed as

where T is the period, or time for the satellite to make one complete revolution.

Substituting this expression for v into our earlier equation gives us a statement of Kepler's 3rd Law.

Refer to the following information for the next two questions.

Communication satellites orbiting the Earth have a period of 24 hours so that they remain geosynchronous. Use these statistics as well as the following constants to answer the next series of questions:

G = 6.67 x 10-11 Nm2/kg2
ME = 5.98 x 1024 kg
RE = 6.37 x 106 m

 Based on Kepler's 3rd Law, at what height above the surface of the Earth are geosynchronous satellites orbiting?

 How fast are geosynchronous satellites moving?

Standard pendulums are not included in this lesson since they are not examples of uniform circular motion. As they swing back and forth through equilibrium, their speeds increase and decrease. They are part of a larger category of behaviors known as vertical circular motion.