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 = F_c = ma_c, it is supplying the centripetal force necessary to keep the block moving in its circular path.

 

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 = F_c = ma_c.

 

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 = F_c = ma_c.

 

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.

 

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 = F_c = ma_c. 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

v_critical = √ (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 v_critical then the car would drift up the incline. If the speed of the car is less than v_critical then the car would slip down the incline.

 

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.