3.17, 3.18, 3.19 Uniform Circular Motion

As the Earth moves in its orbit around the sun, the Earth and the sun are separated by a distance of 1.5 E 11 m. As we all know, the Earth makes one trip around the sun every 365.25 days. Even though the orbital path is slightly elliptical, consider the path to be circular. 1) Convert the orbital time from days to seconds. 2) Calculate the centripetal acceleration of the Earth in m/s^2. 3) If the Earth has a mass of 6.0 E 24 kg, calculate the centripetal force necessary to keep the Earth in orbit. 4) What is the source of the centripetal force that holds the Earth in orbit around the sun?

1) (365.25 days)(24 hrs/1 day)(3600 s/1 hr) = 31557600 s or 3.1557 E 7 s 2) T = 3.16 E 7 s r = 1.5 E 11 m v = ? v = [(2)(π)(r)]/T v = [(2)(3.14)( 1.5 X 1011 m)]/31557600 s v = 29850 m/s v = 3.0 E 4 m/s ac = v^2/r ac = (29850 m/s)2/1.5E 11 m ac = 0.005940 m/s^2, toward the center of the orbit ac = 0.0059 m/s^2, toward the center of the orbit 3) m = 6.0 E 24 kg ac = 0.005940 m/s^2, toward the center of the orbit Fc = ? Fc = mac Fc = (6.0 E 24 kg)(0.005940 m/s^2) Fc = 3.8017 E 22 N, toward the center of the orbit Fc = 3.564 E 22 N, toward the center of the orbit 4) The source of the centripetal force is the weight of the Earth.

Get a small, clear plastic container and fill it about half full of water. Take the container of water to an open area. You could go outside or just go to a large room in your home. Hold the container in your outstretched arm. The container will be positioned straight out in front of you. Keeping the container in front of you, slowly spin in a circle. The pattern is similar to a figure skater executing a slow "spin" maneuver. As you are slowly spinning, observe the level of the water in the container. Stop the motion. 1) What did you observe about the level of the water in the container as you were slowly spinning? 2) Was the level of the water higher on the side of the container away from you or the side of the container nearest you?

1) It was raised on one side. 2) It was the side away from you.

This is a "thought" experiment. Just think about the activity. Suppose you have a ball attached to the end of a strong string. You are whirling the ball in a circular path. 1) Identify the forces exerted on the ball. 2) Identify the force necessary to keep the ball moving in the circular path. 3) What is the direction of the force necessary to keep the ball moving in the circular path?

1) Perhaps you see the gravitational force and the tension force on the ball. 2) The force along the string, which could be called the tension force. 3) If you think about trying to keep a ball rolling in a circle, you would say the force needs to be "toward the center" or "inward."

A ball is attached to a string 0.50 m in length and has a mass of 1.5 kg. It is moving in a clockwise direction horizontally above the ground 2.0 meters. The speed of the ball is 25 meters. 1) What direction would the centripetal acceleration vector point when the ball is at the above location? 2) What direction would the velocity vector point when the ball is at the above location? 3) What direction would the centripetal force vector point when the ball is at the above location? 4) What is the centripetal acceleration of the ball? 5) What is the centripetal force causing this centripetal acceleration?

1) d—centripetal "center-seeking" 2) c—tangential—perpendicular to the radius 3) d—centripetal "center-seeking" 4) ) m = 1.5 kg v = 25 m/s r = 0.50 m ac = v^2/r ac = (25 m/s)^2/0.50m ac = 1250 m/s^2 ac = 1300 m/s^2 5) Fc = mac Fc = (1.5 kg)(1250 m/s2) Fc = 1875 N Fc = 1900 N

Jenna whirls a can attached to a string in a horizontal circle with a .64 meter radius. 1) If the period of the swing is 0.80 s, what is the speed of the can in the circular path? 2) Is the force exerted on the can by the string an unbalanced force or a balanced force? Explain your answer. 3) Is the can accelerating? Explain your answer.

1) r = 0.64 m T = 0.80 s v = ? v = [(2)(π)(r)]/T v = [(2)(3.14)(.64 m)]/0.80 s v = 5.02 m/s v = 5.0 m/s 2) It is an unbalanced force because the can is accelerating. 3) The direction of the can is constantly changing; therefore, it is accelerating.

Elizabeth is riding the merry-go-round at the Hamilton County Fair. Elizabeth's horse is located 7.5 m from the center of the merry-go-round. Ginger has a stopwatch, and she discovered that the merry-go-round has a period of 34 seconds. 1) Calculate the speed of Elizabeth's horse in the circular path. 2) Calculate the centripetal acceleration of Elizabeth and her horse. 3) If Elizabeth has a mass of 55 kg, calculate the centripetal force necessary to keep her in this circular path.

1) r = 7.5 m T = 34 s v = ? v = [(2)(π)(r)]/T v = [(2)(3.14)(7.5 m)]/34 s v = 1.39 m/s v = 1.4 m/s 2) v = 1.39 m/s r = 7.5 m ac = ? ac = v^2/r ac = (1.39 m/s)^2/7.5 m ac = 0.26 m/s^2, toward the center of the merry-go-round 3) ac = 0.26 m/s^2, toward the center of the merry-go-round m = 55 kg Fc = ? Fc = ma Fc = (55 kg)(0.26 m/s^2) Fc = 14.3 N, toward the center of the merry-go-round Fc = 14 N, toward the center of the merry-go-round

Centripetal Acceleration

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: ac = v^2/r ac = centripetal acceleration v = linear speed r = radius of the circular path

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-full 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.

Consider whirling the stone on a shorter string—that is, of smaller radius.

For a given speed, the rate that the stone changes direction is more. This indicates that as the radius decreases, acceleration increases.

The whirling stone's direction of motion keeps changing.

If it moves faster, its direction changes faster. This indicates that as speed increases, acceleration increases.

Uniform Circular Motion

In the simulation, the ladybug is moving in uniform circular motion. This motion is the movement of an object in a circle with a constant or uniform speed. Notice, the length of the velocity vector does not change, only its direction. The magnitude of the velocity vector is the instantaneous speed of the object. The direction of the vector is always directed tangent to the circle. Because the ladybug is changing velocities, it must have acceleration. From the simulation, you can see that the acceleration is constant in magnitude but always pointing toward the center of the circular path. Centripetal force is the force that causes this acceleration. If the centripetal force is removed from an object moving in uniform circular motion, then that object will continue to move along a straight-line path according to the velocity vector. You can see this by selecting "Linear" at anytime the ladybug is moving in a circular path.

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*m^2/kg^2 mEarth = 5.98 E24 kg mmoon = 7.35 E22 kg rEarth - moon = 3.85 E8 m We can set the centripetal force [Fc = (mv^2)/r] equal to the gravitation force [Fg = (Gm1m2)/d^2] 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. (mmoonv^2)/r = (GmEarthmmoon)/d^2 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. (v^2) = (GmEarth)/d Solving for v: v = sqrt[(GmEarth)/d] v = sqrt[(6.67 E-11N*m^2/kg^2)(5.98 E24 kg)/(3.85 E8 m)] v = 1017.84 m/s v = 1020 m/s

Consider the phrase "uniform circular motion." In your opinion, what is the meaning of this phrase?

Looking at the terms, you might say uniform "not changing"; circular "going in a circle"; motion "moving"; and give the definition as moving in a circle with a speed that does not change.

Paul Hewitt's Concept Development Practice Page 9-2: Acceleration and Circular Motion

Newton's Second Law, a = F/m, tells us that net force and its corresponding acceleration are always in the same direction. (Both force and acceleration are vector quantities.) Remember, however, that force and acceleration are not always in the direction of velocity (another vector).

Forces for Circular Motion

Pictures of loops in roller coasters..What force allows the cart to make the loop safely? By now, you know of many forces that can act on an object at a given time. Forces on a coaster going through a loop are the normal force, the force of gravity, and air resistance, to name a few. All objects that move in a circular motion experience a unique force called the centripetal force. The coasters go up and through the loop because of the centripetal force provided by the track. Centripetal force is responsible for keeping these coasters moving in a circle. Remember, all objects require an unbalanced net force to speed up, slow down, or change direction. So objects moving with a constant speed around a circle require a centripetal force to change their direction.

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: Simple Speed Calculation: d = distance t = time v = d/t Uniform Circular Motion Speed Calculation d = distance = circumference = 2πr t = time = period = T v = (2πr)/T

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: Fc = (mv2)/r The equation for calculating frictional force is: Ff = μFN The centripetal force is unbalanced frictional force. Fc = Ff Setting them equal to each other we have (mv2)/r = μFN The normal force is equal to the weight of the car because the car is neither accelerating up or down. FN = mg Now the equation becomes (mv2)/r = μmg Because mass is on both sides of the equation, we can write v2/r = μg 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(μrg) v = sqrt [(0.25)(50.0 m)(9.81 m/s/s)] v = 11.0736 m/s v = 11 m/s

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 determined by the material of the road and the material of the road and tires. The equation for friction is as follows: Ff = μFN Ff = frictional force FN = normal force μ = coefficient of friction These cars remain on the track because of the static friction between the tires and the road. If the friction is removed, the cars would continue in a straight path rather than follow the circular track.

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: Fg = (Gm1m2)/d^2

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: Fc = (mv^2)/r Fc = centripetal force m = mass of the object v = linear speed r = radius of the circular path

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. This athlete demonstrates uniform circular motion by keeping the rope under tension while swinging a hammer in a circle.

You continue driving and round a sharp curve to the left at constant speed.

Your body leans outward. The direction of the car's acceleration is inward. The force on the car acts inward.

You're in a car at a traffic light. The light turns green and the driver "steps on the gas."

Your body lurches backward. The car accelerates forward. The force on the car acts forward.

You're driving along and approach a stop sign. The driver steps on the brakes.

Your body lurches forward. The car accelerates backward. The force on the car acts backward.

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 = 2πr T = 35.0 s v = (2πr)/T v = [(2 )(3.14)(10.0 m)]/35.0 s v = 1.79 m/s

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 = (mv^2)/r Fc = [(0.011 kg)(1.3 m/s)^2]/0.40 m Fc = 0.046 (kg*m/s^2) Fc = 0.046 N, toward the center of the circular path

Shanise is playing a pinball machine. She notices that the 0.014 kg steel ball is shot from the plunger and moves into a circular path with a radius of 0.25 m. If the magnitude of the force necessary to keep the ball in the circular path is 0.026 N, what is the speed of the steel ball?

m = 0.014 kg r = 0.25 m Fc = 0.026 N v = ? Fc = (mv^2)/r v = sqrt (Fcr/m) v = sqrt [(0.026 N )(0.25 m)]/0.014 kg v = 0.68 m/s

Determine the speed of the Hubble space telescope orbiting above the Earth at a height of 5.98 E 5 meters above the Earth's surface. The radius of the Earth is 6.38 E 6 meters, and the mass of the Hubble is 11600 kg.

r = rEarth + hheight above Earth = 6.38 E6 m + 5.98 E5 m = 6978000 m G = 6.67 E -11 N*m^2/kg^2 mEarth = 5.98 E24 kg We can set the centripetal force [Fc = (mv^2)/r] equal to the gravitation force [Fg = (Gm1m2)/d^2] As we look at the centripetal force equation, which mass should we use? Since we are looking for the force on the Hubble caused by the Earth and it is the Hubble that is doing the revolving, we use the Hubble. (mHubblev^2)/r = (GmEarthmHubble)/d^2 The mass of the Hubble cancels out of the equation and we are left with the "r" in the Centripetal equation, which is equal to the "d" in the Universal Gravitation equation: (v^2)/r = (GmEarth)/d^2 The r in the denominator on one side of the equation cancels out one of the d on the other side of the equation: (v^2) = (GmEarth)/d Solving for v: v = sqrt[(GmEarth)/d] v = sqrt[(6.67 E-11N*m^2/kg^2)(5.98 E24 kg)/(6978000 m)] v = sqrt [52634732.12 m^2/s^2] v = 7254.9798 m/s v = 7.25 E3 m/s

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 ac = ? ac = v^2/r ac = (25 m/s)^2/45 m ac = 13.9 m/s^2, toward the center of the curve