Circular Motion
Article objectives
Conceptual framework
Circular motion does not produce an outward force
The easiest way to understand circular motion is to bring back the parable of the bowling ball in the pickup truck. As the truck makes a left turn, the driver looks in the rearview mirror and thinks that some mysterious force is pulling the ball outward, but the truck is accelerating, so the driver's frame of reference is not an inertial frame. Newton's laws are violated in a noninertial frame, so the ball appears to accelerate without any actual force acting on it. Because we are used to inertial frames, in which accelerations are caused by forces, the ball's acceleration creates a vivid illusion that there must be an outward force.
Figure a: In the turning truck's frame of reference, the ball appears to violate Newton's laws, displaying a sideways acceleration that is not the result of a force-interaction with any other object. 2. In an inertial frame of reference, such as the frame fixed to the earth's surface, the ball obeys Newton's first law. No forces are acting on it, and it continues moving in a straight line. It is the truck that is participating in an interaction with the asphalt, the truck that accelerates as it should according to Newton's second law.
Figure b: This crane fly's halteres help it to maintain its orientation in flight.
In an inertial frame everything makes more sense. The ball has no force on it, and goes straight as required by Newton's first law. The truck has a force on it from the asphalt, and responds to it by accelerating (changing the direction of its velocity vector) as Newton's second law says it should.
Example 1: The halteres
Another interesting example is an insect organ called the halteres, a pair of small knobbed limbs behind the wings, which vibrate up and down and help the insect to maintain its orientation in flight. The halteres evolved from a second pair of wings possessed by earlier insects. Suppose, for example, that the halteres are on their upward stroke, and at that moment an air current causes the fly to pitch its nose down. The halteres follow Newton's first law, continuing to rise vertically, but in the fly's rotating frame of reference, it seems as though they have been subjected to a backward force. The fly has special sensory organs that perceive this twist, and help it to correct itself by raising its nose.
Circular motion does not persist without a force To make curve around with the car, you need some force, friction from the seat, or a normal force from the side of the car. (In fact, all forces were probably adding together.) One of the reasons why Galileo failed to refine the principle of inertia into a quantitative statement like Newton's first law is that he was not sure whether motion without a force would naturally be circular or linear. In fact, the most impressive examples he knew of the persistence of motion were mostly circular: the spinning of a top or the rotation of the earth, for example. Newton realized that in examples such as these, there really were forces at work. Atoms on the surface of the top are prevented from flying off straight by the ordinary force that keeps atoms stuck together in solid matter. The earth is nearly all liquid, but gravitational forces pull all its parts inward.
Uniform and nonuniform circular motion
Circular motion always involves a change in the direction of the velocity vector, but it is also possible for the magnitude of the velocity to change at the same time. Circular motion is referred to as uniform if |v| is constant, and nonuniform if it is changing.
Your speedometer tells you the magnitude of your car's velocity vector, so when you go around a curve while keeping your speedometer needle steady, you are executing uniform circular motion. If your speedometer reading is changing as you turn, your circular motion is nonuniform. Uniform circular motion is simpler to analyze mathematically, so we will attack it first and then pass to the nonuniform case.
Figure c: 1. An overhead view of a person swinging a rock on a rope. A force from the string is required to make the rock's velocity vector keep changing direction. 2. If the string breaks, the rock will follow Newton's first law and go straight instead of continuing around the circle.
Figure d: Sparks fly away along tangents to a grinding wheel.
Only an inward force is required for uniform circular motion.
Figure c showed the string pulling in straight along a radius of the circle, but many people believe that when they are doing this they must be “leading” the rock a little to keep it moving along. That is, they believe that the force required to produce uniform circular motion is not directly inward but at a slight angle to the radius of the circle. This intuition is incorrect, which you can easily verify for yourself now if you have some string handy. It is only while you are getting the object going that your force needs to be at an angle to the radius. During this initial period of speeding up, the motion is not uniform. Once you settle down into uniform circular motion, you only apply an inward force.
If you have not done the experiment for yourself, here is a theoretical argument to convince you of this fact. We have discussed the principle that forces have no perpendicular effects. To keep the rock from speeding up or slowing down, we only need to make sure that our force is perpendicular to its direction of motion. We are then guaranteed that its forward motion will remain unaffected: our force can have no perpendicular effect, and there is no other force acting on the rock which could slow it down. The rock requires no forward force to maintain its forward motion, any more than a projectile needs a horizontal force to “help it over the top” of its arc.
Figure e: To make the brick go in a circle, an inward force on the rope had to be exerted.
Figure f: A series of three hammer taps makes the rolling ball trace a triangle, seven hammers a heptagon. If the number of hammers was large enough, the ball would essentially be experiencing a steady inward force, and it would go in a circle. In no case is any forward force necessary.
Figure g: When a car is going straight at constant speed, the forward and backward forces on it are canceling out, producing a total force of zero. When it moves in a circle at constant speed, there are three forces on it, but the forward and backward forces cancel out, so the vector sum is an inward force.
Why, then, does a car driving in circles in a parking lot stop executing uniform circular motion if you take your foot off the gas? The source of confusion here is that Newton's laws predict an object's motion based on the total force acting on it. A car driving in circles has three forces on it
(1) an inward force from the asphalt, controlled with the steering wheel;
(2) a forward force from the asphalt, controlled with the gas pedal; and
(3) backward forces from air resistance and rolling resistance.
You need to make sure there is a forward force on the car so that the backward forces will be exactly canceled out, creating a vector sum that points directly inward.
Example 2: A motorcycle making a turn
The motorcyclist in figure h is moving along an arc of a circle. It looks like he's chosen to ride the slanted surface of the dirt at a place where it makes just the angle he wants, allowing him to get the force he needs on the tires as a normal force, without needing any frictional force. The dirt's normal force on the tires points up and to our left. The vertical component of that force is canceled by gravity, while its horizontal component causes him to curve.
In uniform circular motion, the acceleration vector is inward Since experiments show that the force vector points directly inward, Newton's second law implies that the acceleration vector points inward as well. This fact can also be proven on purely kinematical grounds.
Uniform circular motion
A simple and very useful equation for the magnitude of the acceleration of an object undergoing constant acceleration is derived using Law of Sines shown in figure i.
Figure i: The law of sines
Figure j: Deriving |a|=\(|v|^{2}\)/r for uniform circular motion.
The derivation is brief, but the method requires some explanation and justification. The idea is to calculate a Δv vector describing the change in the velocity vector as the object passes through an angle θ. We then calculate the acceleration, a=Δv/Δt . The astute reader will recall, however, that this equation is only valid for motion with constant acceleration. Although the magnitude of the acceleration is constant for uniform circular motion, the acceleration vector changes its direction, so it is not a constant vector, and the equation a=Δv/Δt does not apply. The justification for using it is that we will then examine its behavior when we make the time interval very short, which means making the angle θ very small. For smaller and smaller time intervals, the Δv/Δt expression becomes a better and better approximation, so that the final result of the derivation is exact.
In figure j/1, the object sweeps out an angle θ. Its direction of motion also twists around by an angle θ, from the vertical dashed line to the tilted one. Figure j/2 shows the initial and final velocity vectors, which have equal magnitude, but directions differing by θ. In j/3, the vectors are reassembled in the proper positions for vector subtraction. They form an isosceles triangle with interior angles θ, η, and η. The law of sines gives
$$\frac{|\Delta{} v|}{sinθ} = \frac{v}{sin η}$$
This tells us the magnitude of Δv , which is one of the two ingredients we need for calculating the magnitude of a=Δv/Δt . The other ingredient is Δ t. The time required for the object to move through the angle θ is
$$Δt = \frac{\text{length of arc}}{v}$$
Now if we measure our angles in radians we can use the definition of radian measure, which is (angle)=(length of arc)/(radius) , giving Δt=θr/|v| . Combining this with the first expression involving |Δ v| gives
$$|a| = |Δv|/Δt$$ $$=\frac{|v^2|}{r} ⋅ \frac{sinθ}{θ} ⋅ \frac{1}{sin η}$$
When θ becomes very small, the small-angle approximation sin θ≈ 0 applies, and also η becomes close to 90°, so sin η ≈ 1, and we have an equation for |a| :
$$|a| = \frac{|v|^2}{r} \; \; \; \; \text{[uniform circular motion]}$$
Example 3: Force required to turn on a bike
◊ A bicyclist is making a turn along an arc of a circle with radius 20 m, at a speed of 5 m/s. If the combined mass of the cyclist plus the bike is 60 kg, how great a static friction force must the road be able to exert on the tires?
◊ Taking the magnitudes of both sides of Newton's second law gives
$$|F| =|ma| =m|a|$$
$$\text{Substitutiing} \; |a|=|v|^2/r \; \text{gives}$$
$$|F| =m|v|^2/r ≈80 N$$
(rounded off to one sig fig).
Example 4: Don't hug the center line on a curve!
◊ You're driving on a mountain road with a steep drop on your right. When making a left turn, is it safer to hug the center line or to stay closer to the outside of the road?
◊ You want whichever choice involves the least acceleration, because that will require the least force and entail the least risk of exceeding the maximum force of static friction. Assuming the curve is an arc of a circle and your speed is constant, your car is performing uniform circular motion, with |a|=\(|v|^{2}\)/r . The dependence on the square of the speed shows that driving slowly is the main safety measure you can take, but for any given speed you also want to have the largest possible value of r. Even though your instinct is to keep away from that scary precipice, you are actually less likely to skid if you keep toward the outside, because then you are describing a larger circle.
Example 5: Acceleration related to radius and period of rotation
◊ How can the equation for the acceleration in uniform circular motion be rewritten in terms of the radius of the circle and the period, T, of the motion, i.e., the time required to go around once?
◊ The period can be related to the speed as follows:
$$|v| =\frac{circumference}{T} =2πr/T$$
$$\text{Substituting into the equation} \; |a|=|v|^2/r \; \text{gives}$$
$$|a|=\frac{4π^2r}{T^2}$$
Example 6: A clothes dryer
◊ My clothes dryer has a drum with an inside radius of 35 cm, and it spins at 48 revolutions per minute. What is the acceleration of the clothes inside?
◊ We can solve this by finding the period and plugging in to the result of the previous example. If it makes 48 revolutions in one minute, then the period is 1/48 of a minute, or 1.25 s. To get an acceleration in mks units, we must convert the radius to 0.35 m. Plugging in, the result is 8.8 m/\(s^2\) .
Example 7: More about clothes dryers!
◊ Let's make the assumption that the clothes remain against the inside of the drum as they go over the top. In light of the previous example, is this a correct assumption?
◊ No. We know that there must be some minimum speed at which the motor can run that will result in the clothes just barely staying against the inside of the drum as they go over the top. If the clothes dryer ran at just this minimum speed, then there would be no normal force on the clothes at the top: they would be on the verge of losing contact. The only force acting on them at the top would be the force of gravity, which would give them an acceleration of g=9.8 m/\(s^2\) . The actual dryer must be running slower than this minimum speed, because it produces an acceleration of only 8.8 m/\(s^2\) .
Figure l: An artist's conception of a rotating space colony in the form of a giant wheel. A person living in this noninertial frame of reference has an illusion of a force pulling her outward, toward the deck, for the same reason that a person in the pickup truck has the illusion of a force pulling the bowling ball. By adjusting the speed of rotation, the designers can make an acceleration \(|v|^{2}\)/r equal to the usual acceleration of gravity on earth. On earth, your acceleration standing on the ground is zero, and a falling rock heads for your feet with an acceleration of 9.8 m/\(s^{2}\) . A person standing on the deck of the space colony has an upward acceleration of 9.8 m/\(s^{2}\), and when she lets go of a rock, her feet head up at the nonaccelerating rock. To her, it seems the same as true gravity.
Nonuniform circular motion
Figure m: 1. Moving in a circle while speeding up. 2. Uniform circular motion. 3. Slowing down.
What about nonuniform circular motion? Although so far we have been discussing components of vectors along fixed x and y axes, it now becomes convenient to discuss components of the acceleration vector along the radial line (in-out) and the tangential line (along the direction of motion). For nonuniform circular motion, the radial component of the acceleration obeys the same equation as for uniform circular motion,
$$a_r = v^2 / r$$
where v=|v| , but the acceleration vector also has a tangential component,
$$a_t =\text{slope of the graph of} \; v \; \text{versus} \; t.$$
The latter quantity has a simple interpretation. If you are going around a curve in your car, and the speedometer needle is moving, the tangential component of the acceleration vector is simply what you would have thought the acceleration was if you saw the speedometer and didn't know you were going around a curve.
Example 8: Slow down before a turn, not during it.
◊ When you're making a turn in your car and you're afraid you may skid, isn't it a good idea to slow down?
◊ If the turn is an arc of a circle, and you've already completed part of the turn at constant speed without skidding, then the road and tires are apparently capable of enough static friction to supply an acceleration of \(|v|^2\)/r . There is no reason why you would skid out now if you haven't already. If you get nervous and brake, however, then you need to have a tangential acceleration component in addition to the radial component you were already able to produce successfully. This would require an acceleration vector with a greater magnitude, which in turn would require a larger force. Static friction might not be able to supply that much force, and you might skid out. As in the previous example on a similar topic, the safe thing to do is to approach the turn at a comfortably low speed.