Purpose: To explore the concepts and equations associated with uniform circular motion

Introduction: This tutorial will provide you with the information essential to the understanding of the concepts associated with uniform circular motion. In addition, the equations used to solve problems concerning uniform circular motion will be presented. The equations will involve three vector quantities: velocity, acceleration, and force. Remember that vector quantities require a direction for a complete description.

Materials: none

Procedure:

1. Read the tutorial carefully. Pay special attention to the new vocabulary words and the sample problems.

2. You may want to print this tutorial for your notebook.

Tutorial:

1. Uniform Circular Motion
   
   You can probably think of many objects that travel in the seemingly circular path: a merry-go-round, the "death spiral" in pairs figure skating, a satellite in orbit, or a model airplane on a tether. How do these examples relate to uniform circular motion? Think about the phrase "uniform circular motion."
   
   Uniform means "constant."
   
   Circular refers to a "circular path."
   
   Motion implies "movement."
   
   Uniform Circular Motion refers to the motion of an object traveling at a constant speed in a circular path.

2. The Speed of an Object in Uniform Circular Motion
   
   Since the path of an object undergoing uniform circular motion is a circle, the distance that the object travels one time around the circle is the circumference of the circle. Remember that we can calculate the speed of a moving object if we know the distance traveled and the time for the travel. The time required for the object to travel one time around the circle is called the period. The symbol for the period is T. Study the following comparison and example problem:

Example Problem: Yvonne and Eddy are riding a ferris wheel at the Dade County Fair. If the radius of the ferris wheel is 10.0 m, and it rotates once every 35.0 seconds, what is the linear speed of Yvonne and Eddy?
d = 2r
T = 35.0 s
v = (2r)/T
v = [(2 )(3.14)(10.0 m)]/35.0 s
v = 1.79 m/s

3. Why Does an Object Travel in Uniform Circular Motion?
   
   An object travels in uniform circular motion because a force acts to pull the object out of a straight path and into a circular path. This force is called the centripetal force. The term "centripetal" means "center-seeking" or "toward the center." In other words, a centripetal force acting toward the center of the circular path, constantly pulls the object to maintain the circular path. If the centripetal force is removed, then the object no longer travels in a circular path. At this point, some students may confuse "centripetal" with "centrifugal." Always remember that we are considering the inward, centripetal force. Here's another very important fact. The centripetal force always acts in a direction perpendicular to the path of the moving object. It is an unbalanced force that accelerates the object by constantly changing the direction of motion, not the magnitude.
   
   Now, think back to the mini-experiment you did in the previous activity. You used a small, clear container (glass) half-ful of water. Holding it in your outstretched hand, you rotated in a circular path. You observed that the level of the water was higher on the side of the glass away from you. In other words, the water level "pointed" to you or toward the center. This revealed that there was a force, a centripetal force, acting to move the glass in a circular path.
4. Centripetal Acceleration
   
   As you just learned, an object moving in uniform circular motion does accelerate even though the speed is constant. Since the object is continually changing direction, the object accelerates. The acceleration experienced by an object undergoing uniform circular motion is called centripetal acceleration. The direction for centripetal is always "toward the center." The equation for centripetal acceleration is as follows:
   
   a_c = v²/r
   a_c = centripetal acceleration
   v = linear speed
   r = radius of the circular path
   
   Consider this sample problem: Matt is driving his car around a curve that has a radius of 45 m. If his speed is 25 m/s as he negotiates the curve, find the centripetal acceleration for the curve.
   v = 25 m/s
   r = 45 m
   a_c = ?
   a_c = v²/r
   a_c = (25 m/s)²/45 m
   a_c = 13.9 m/s², toward the center of the curve
5. Centripetal Force
   
   When you studied Newton's Second Law, you learned that an object accelerates when the forces on it become unbalanced. A "net" force is acting on the object to cause the acceleration. As mentioned above, a centripetal force acts on a object to cause it to move in a circular path. This is not really a "new" force, but could be due to string tension, gravitational force, electrical force, frictional force, or whatever force is directed at a right angle to the path of the moving object and produces circular motion. The equation for centripetal force follows the Newton's Second Law equation F = ma. The equation for centripetal force is as follows:
   
   F_c = (mv²)/r
   F_c = centripetal force
   m = mass of the object
   v = linear speed
   r = radius of the circular path
   
   Consider this sample problem: Ryan is swinging a 0.011 kg ball tied to a string around his head in a flat, horizontal circle. The radius of the circle is 0.40 m and linear speed is 1.3 m/s. Find the centripetal force acting on the ball to keep it in the circular path.
   m = 0.011 kg
   r = 0.40 m
   v = 1.3 m/s
   Fc = ?
   Fc = (mv²)/r
   Fc = [(0.011 kg)(1.3 m/s)²]/0.40 m
   Fc = 0.046 (kg*m/s²)
   Fc = 0.046 N, toward the center of the circular path
6. Centripetal Force due to friction
   
   When you studied Force Diagrams, one of the forces in the diagrams was the frictional force. Friction is the force that opposes the motion of one surface over another. The direction of the frictional force is always opposite motion or opposite pending motion.
   
   Friction is the force that keeps your car moving in a circular manner. The Normal Force is the weight of the car if the car is on a level surface, and the coefficient of friction is the determined by the material of the road and the material of the road and tires.
   
   The equation for friction is as follows:
   
   F_f = F_N
   
   F_f is the Frictional Force
   F_N is the Normal Force
   is the coefficient of friction
   
   Use the following site to determine which is greater: the static friction between the block and table or the kinetic friction.
   
   Frictional force
   
   Consider this sample problem: Friction provides the centripetal force necessary for a car to travel around a flat circular race track. What is the maximum speed at which a car can safely travel around a circular track of radius 50.0m if the coefficient of friction between the tire and road is 0.25?
   
   The equation for Centripetal force is
   F_c = (mv²)/r
   
   The equation for calculating Frictional force is
   F_f = F_N
   
   The Centripetal force is unbalanced Frictional force.
   F_c = F_f
   
   Setting them equal to each other we have
   (mv²)/r = F_N
   
   The Normal force is the equal to the weight of the car because the car is neither accelerating up nor down.
   F_N = mg
   
   Now the equation becomes
   (mv²)/r = mg
   
   Because mass is on both sides of the equation we can write
   v²/r = g
   
   Solving for the velocity, you can now calculate the safe velocity for this road. Notice that the mass of the car did not enter into the equation only the radius of the road, gravitational acceleration at that location, and the coefficient of friction between the car and the road.
   
   v = sqrt(rg)
   v = sqrt [(0.25)(50.0 m)(9.81 m/s/s)]
   v = 11.0736 m/s
   v = 11 m/s
7. Centripetal Force due to weight
   
   When you studied Newton's Law of Universal Gravitation, you discovered that the gravitational force between two masses is directly proportional to the product of the two masses and inversely proportional to the distance between the masses.
   
   The equation for Newton's Law of Universal Gravitation is as follows:
   F_g = (Gm₁m₂)/d²
   
   Consider this sample problem: The gravitational force between the Earth and moon provides the centripetal force necessary to keep the moon in orbit. What is the average speed of the moon as it revolves around the Earth?
   
   In this problem, the weight is the Centripetal Force.
   Remember the constant "G" = 6.67 E -11 N*m²/kg²
   m_Earth = 5.98 E24 kg
   m_moon = 7.35 E22 kg
   r_{Earth - moon} = 3.85 E8 m
   
   We can set the Centripetal Force [F_c = (mv²)/r] equal to the Gravitation Force [F_g = (Gm₁m₂)/d²]
   
   As we look at the Centripetal Force equation, which mass should we use? Since we are looking for the force on the moon caused by the Earth, and it is the moon that is doing the revolving, we use the moon.
   
   (m_moonv²)/r = (Gm_Earthm_moon)/d²
   
   The mass of the moon cancels out of the equation. The r in the Centripetal Force equation is equal to the d in the Universal Law of Gravitation equation, so the r falls out of the left side of the equation. The d² becomes d.
   
   (v²) = (Gm_Earth)/d
   
   Solving for v:
   v = sqrt[(Gm_Earth)/d]
   v = sqrt[(6.67 E-11N*m²/kg²)(5.98 E24 kg)/(3.85 E8 m)]
   v = 1027.84 m/s
   v = 1030 m/s

Copyright 1997 Paul Hewitt, modified by FLVS.