Applications of Circular Motion
Newton's Second Law - Revisited
Newton's second law states that the acceleration of an object is directly proportional to the net force acting upon the object and inversely proportional to the mass of the object. The law is often expressed in the form of the following two equations.
In Unit 2 of The Physics Classroom, Newton's second law was used to analyze a variety of physical situations. The idea was that if any given physical situation is analyzed in terms of the individual forces which are acting upon an object, then those individual forces must add up as vectors to the net force. Furthermore, the net force must be equal to the mass times the acceleration. Subsequently, the acceleration of an object can be found if the mass of the object and the magnitudes and directions of each individual force are known. And the magnitude of any individual force can be determined if the mass of the object, the acceleration of the object, and the magnitude of the other individual forces are known. The process of analyzing such physical situations in order to determine unknown information is dependent upon the ability to represent the physical situation by means of a free-body diagram. A free-body diagram is a vector diagram which depicts the relative magnitude and direction of all the individual forces which are acting upon the object.
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In this Lesson, we will use Unit 2 principles (free-body diagrams, Newton's second law equation, etc.) and circular motion concepts in order to analyze a variety of physical situations involving the motion of objects in circles or along curved paths. The mathematical equations discussed in Lesson 1 and the concept of a centripetal force requirement will be applied in order to analyze roller coasters and other amusement park rides and various athletic movements.
To illustrate how circular motion principles can be combined with Newton's second law to analyze a physical situation, consider a car moving in a horizontal circle on a level surface. The diagram below depicts the car on the left side of the circle.
Applying the concept of a centripetal
force requirement, we know that the net force acting upon
the object is directed inwards. Since the car is positioned
on the left side of the circle, the net force is directed
rightward.
An analysis of the situation would reveal that there are
three forces acting upon the object - the force of gravity
(acting downwards), the normal force of the pavement (acting
upwards), and the force of friction (acting inwards or
rightwards). It is the friction force which supplies the
centripetal force requirement for the car to move in a
horizontal circle. Without friction, the car would turn its
wheels but would not move in a circle (as is the case on an
icy surface). This analysis leads to the free-body diagram
shown at the right. Observe that each force is represented
by a vector arrow which points in the specific direction
which the force acts; also notice that each force is labeled
according to type (Ffrict, Fnorm, and
Fgrav). Such an analysis is the first step of any
problem involving Newton's second law and a circular
motion.
Now consider the following two problems pertaining to this physical scenario of the car making a turn on a horizontal surface.
Sample problem #1 provides kinematic information (v and R) and requests the value of an individual force. As such the solution of the problem will demand that the acceleration and the net force first be determined; then the individual force value can be found by use of the free-body diagram. Sample problem #2 provides information about the individual force values (or at least information which allows for the determination of the individual force values) and requests the value of the maximum speed of the car. As such, its solution will demand that individual force values be used to determine the net force and acceleration; then the acceleration can be used to determine the maximum speed of the car. The two problems will be solved using the same general principles. Yet because the given and requested information is different in each, the solution method will be slightly different.
Solution
to Sample Problem #1
The known information and requested information in sample problem #1 is:
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Known Information: m = 945 kg |
Requested Information: Ffrict = ??? |
The mass of the object can be used to determine the force of gravity acting in the downward direction. Use the equation
Fgrav = m *
g
where g is 9.8 m/s/s. Knowing that there is no vertical acceleration of the car, it can be concluded that the vertical forces balance each other. Thus, Fgrav = Fnorm= 9261 N. This allows us to determine two of the three forces identified in the free-body diagram. Only the friction force remains unknown.
Since the force of friction is the only horizontal force, it must be equal to the net force acting upon the object. So if the net force can be determined, then the friction force is known. To determine the net force, the mass and the kinematic information (speed and radius) must be substituted into the following equation:
Substituting the given values yields a net force of 3600 Newtons. Thus, the force of friction is 3600 N.
Finally the coefficient of friction ("mu") can be determined using the equation which relates the coefficient of friction to the force of friction and the normal force.
Substituting 3600 N for Ffrict and 9261 N for Fnorm yields a coefficient of friction of 0.389.
Solution
to Sample Problem #2
Once again, the problem begins by identifying the known and requested information. The known information and requested information in the sample problem #2 is:
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Known Information: m = 945 kg |
Requested Information: v = ??? |
The mass of the car can be used to determine the force of gravity acting in the downward direction. Use the equation
Fgrav = m *
g
where g is 9.8 m/s/s. Knowing that there is no vertical acceleration of the car, it can be concluded that the vertical forces balance each other. Thus, Fgrav = Fnorm= 9261 N. Since the coefficient of friction ("mu") is given, the force of friction can be determined using the following equation:
This allows us to determine all three forces identified in the free-body diagram.
The net force acting upon any object is the vector sum of all individual forces acting upon that object. So if all individual force values are known (as is the case here), the net force can be calculated. The vertical forces add to 0 N. Since the force of friction is the only horizontal force, it must be equal to the net force acting upon the object. Thus, Fnet = 7872 N.
Once the net force is determined, the acceleration can be quickly calculated using the following equation.
Fnet =
m*a
Substituting the given values yields an acceleration of 8.33 m/s/s. Finally, the speed at which the car could travel around the turn can be calculated using the equation for centripetal acceleration:
Substituting the known values for a and R into this equation and solving algebraically yields a maximum speed of 17.1 m/s.
Each of the two sample problems above are solved using the same basic problem-solving approach. The approach can be summarized as follows.
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The method prescribed above will serve you well as you approach circular motion problems. However, one caution is in order. Every physics problem differs from the previous problem. As such, there is no magic formula for solving every one. Using an appropriate approach to solving such problems (which involves constructing a FBD, identifying known information, identifying the requested information, and using available equations) will never eliminate the need to think, analyze and problem-solve. For this reason, make an effort to develop an appropriate approach to every problem; yet always engage your critical analysis skills in the process of the solution. If physics problems were a mere matter of following a foolproof, 5-step formula or using some memorized algorithm, then we wouldn't call them "problems."
Check
Your Understanding
Use your understanding of Newton's second law and circular motion principles to determine the unknown values in the following practice problems. Click the button to check your answers.
1. A 1.50-kg bucket of water is tied by a rope and whirled in a circle with a radius of 1.00 m. At the top of the circular loop, the speed of the bucket is 4.00 m/s. Determine the acceleration, the net force and the individual force values when the bucket is at the top of the circular loop.
m = 1.5 kg
a = ________ m/s/s
Fnet = _________ N
2. A 1.50-kg bucket of water is tied by a rope and whirled in a circle with a radius of 1.00 m. At the bottom of the circular loop, the speed of the bucket is 6.00 m/s. Determine the acceleration, the net force and the individual force values when the bucket is at the bottom of the circular loop.
m = 1.5 kg
a = ________ m/s/s
Fnet = _________ N