Presentation is loading. Please wait.

Presentation is loading. Please wait.

Circular Motion and Gravitation.  Objectives:  1. Solve problems involving centripetal acceleration.  2. Solve problems involving centripetal force.

Similar presentations

Presentation on theme: "Circular Motion and Gravitation.  Objectives:  1. Solve problems involving centripetal acceleration.  2. Solve problems involving centripetal force."— Presentation transcript:

1 Circular Motion and Gravitation

2  Objectives:  1. Solve problems involving centripetal acceleration.  2. Solve problems involving centripetal force.  3. Explain how the apparent existence of an outward force in circular motion can be explained as inertia resisting the centripetal force.

3  Any object that revolves around a single axis experiences circular motion.  The axis of rotation is the line about which the object rotates.  Tangential speed ( v t ) is the speed of an object undergoing circular motion.  It is the speed of an object along a tangent line to the circular path that it follows.  When the tangential speed is constant, the object experiences uniform circular motion

4  If an object is undergoing uniform circular motion, how can it accelerate?  The acceleration is due to the change in direction of the velocity.  Centripetal acceleration is acceleration directed toward the center of a circular path.  a c = v t 2  r  Centripetal means “center seeking”

5  A rope attaches a tire to an overhanging tree limb. A girl swinging on the tire has a centripetal acceleration of 3.0 m/s 2. If the length of the rope is 2.1 m, what is the girl's tangential speed?  2.5 m/s

6  As a young boy swings a yo-yo parallel to the ground and above his head, the yo-yo has a centripetal acceleration of 250 m/s 2. If the yo-yo's string is 0.50 m long, what is the yo-yo's tangential speed?  11 m/s

7  A dog sits 1.5 m from the center of a merry-go-round. The merry-go-round is set in motion, and the dog's tangential speed is 1.5 m/s. What is the dog's centripetal acceleration?  1.5 m/s 2

8  A race car moving along a circular track has a centripetal acceleration of 15.4 m/s 2. If the car has a tangential speed of 30.0 m/s, what is the distance between the car and the center of the track?  58.4 m

9  In circular motion, any acceleration due to a change in speed is dubbed tangential acceleration.

10  Centripetal force is the net force directed towards the center of an object's circular path.  F c = mv t 2  r  Any force or combination of forces can provide the net force.


12  A 2.10 m rope attaches a tire to an overhanging tree limb. A girl swinging on the tire has a tangential speed of 2.50 m/s. If the magnitude of the centripetal force is 88.0 N, what is the girl's mass?  29.6 kg

13  A bicyclist is riding at a tangential speed of 13.2 m/s around a circular track. The magnitude of the centripetal force is 377 N, and the combination of the mass of the bicyclist and rider is 86.5 kg. What is the track's radius?  40.0 m

14  A dog sits 1.50 m from the center of a merry-go-round and revolves at a tangential speed of 1.80 m/s. If the dog's mass is 18.5 kg, what is the magnitude of the centripetal force on the dog?  40.0 N 

15  A 905 kg car travels around a circular track with a circumference of 3.25 km If the magnitude of the centripetal force is 2140 N, what is the car's tangential speed?  35.0 m/s

16  Centripetal force causes an object to maintain circular motion.  If the centripetal force is removed, the object will move in a straight line along its inertial path at the moment it was released.  If a ball is whirled in a circular path parallel to the ground and the string that it is attached to breaks, the ball will fly off horizontally in a direction tangent to the path.  The ball will then follow the parabolic path of a projectile.

17  The phenomena that occurs when an objects initializes circular motion and moves out from the center of the axis of rotation until it is stopped by an outside force is sometimes referred to as centrifugal (center fleeing) force.  This term is incorrect as the phenomena is actually caused by inertia, or the tendency of an object to continue moving in the same direction at the same speed.  NO CENTRIFUGAL FORCE!!!!!!

18  Objectives:  1. Explain how Newton's law of universal gravitation accounts for various phenomena, including satellite and planetary orbits, falling objects, and the tides.  2. Apply Newton's law of universal gravitation to solve problems.

19  Gravitational force is the mutual force of attraction between particles of matter.  The force that holds the planets in orbit around the sun and caused an apple to fall on Newton's head are one and the same.  Newton discovered that if an object were projected at just the right speed, the object would fall to Earth exactly as the Earth was curving away from it, ie, it would orbit the Earth.  The gravitational force would be just great enough to keep the object from following its inertial path.

20 F g = G m1 m2 r 2 gravitational force = constant x mass 1 x mass 2 (distance b/t masses) 2

21  G is the constant of universal gravitation.  Newton never knew the value of G, but it has been found to be  G = x N▪ m 2  kg 2  Newton also discovered that the distance between an object exerting a gravitational force and the object experiencing a gravitational force is measured from the center of gravity of each object.


23  Gravitational forces act between all objects.  The acceleration of the larger object towards the smaller object is less apparent as acceleration is inversely proportional to mass.  The gravitation force between two orbiting bodies is a centripetal force and the bodies orbit the center of mass of the two-body system.

24  What must be the distance between two kg balls if the magnitude of the gravitational force between them is 8.92 x N?  m

25  Mars has a mass of about 6.4 x kg, and its moon Phobos has a mass of about 9.6 x kg. If the magnitude of the gravitational force between the two bodies is 4.6 x N, how far apart are Mars and Phobos?  9.4 x 10 6 m

26  Find the magnitude of the gravitational force a 66.5 kg person would experience while standing on the surface of each of the following planets:  a. Earth m=5.97 x kg; r = 6.38 x 10 6 m  b. Mars m = 6.42 x kg; r = 3.40 x 10 6 m  c. Pluto m = 1.25 x kg; r = 1.20 x 10 6 m  a. 651 N  b. 246 N  c N

27  Tides occur as a result of the difference in the gravitational force of the moon at the Earth's center and at points on the surface.  Two high tides occur at points on the Earth in line with the moon.  On the side closer to the moon, the gravitational force is greater than at the Earth's center, so the water bulges outward towards the moon creating a high tide.  On the side farthest from the moon, the gravitational force is less, so the water bulges outward as well.

28  1798-Henry Cavendish conducted an experiment that determined the value of G.  Once the value of G was known, Newton's law of gravitation was used to determine Earth's mass.

29  Masses create a gravitational field in space.  Gravitational force is an interaction between a mass and the gravitational field created by other masses.  According to field theory, gravitational potential energy is stored in the gravitational field.  Earth's gravitational field can be described by the gravitational field strength (g).  g = F g  m

30  The value for g is equal to the acceleration due to gravity.  They are not, however, the same thing.  Consider a mass hanging on a spring scale.  You are measuring the gravitational field strength.  The mass is at rest, so there is no acceleration.  Gravitational field strength rapidly decreases as distance from the Earth increases.

31  Weight is equal to mass times gravitational field strength by F g = Gm1 m2  r 2  Thus, g = F g = Gm 1 m 2 = G m 2  m 1 m 1 r 2 r 2  Gravitational field strength depends on mass and distance from the object creating the gravitational field.  Thus, your weight depends on location.

32  As you get farther from Earth, your weight decreases.  Your weight would be different on different planets because their masses and radii are different from the mass of Earth.

33  Newton's 2 nd Law (F=ma) uses inertial mass  An object's tendency to resist acceleration  Newton's Law of Gravitation uses gravitational mass  relates how objects attract each other  It was not always accepted that these are the same mass.  We now know these are the same because free fall acceleration on Earth's surface is always the same.

34  Objectives:  1. Describe Kepler's laws of planetary motion.  2. Relate Newton's mathematical analysis of gravitational force to the elliptical planetary orbits proposed by Kepler.  3. Solve problems involving orbital speed and period.

35  First Law: Each planet travels in an elliptical orbit around the sun, and the sun is at one of the focal points.  Second Law: An imaginary line drawn from the sun to any planet sweeps out equal areas in equal time intervals.  Third Law: The square of a planet's orbital period ( T 2 ) is proportional to the cube of the average distance (r 3 ) between the planet and the sun.  T is the time it takes a planet to complete one full revolution.  r is the distance between the planet and the sun.

36  Kepler's second law basically states that when a planet is closer to the sun, it travels faster.  Kepler's third law relates the orbital periods and distances of one planet to those of another planet.  The laws apply to all orbiting bodies (satellites, moons, etc).

37  Newton verified his law of gravitation using Kepler's laws.  He demonstrated that Kepler's third law could be derived using his law of gravitation.

38  You can mathematically show how Kepler's third law can be derived from Newton's law of universal gravitation (assuming circular orbits.)  To begin, recall that the centripetal force is provided by the gravitational force.  Set the equations for gravitational and centripetal force equal to one another and solve for v t 2.  Because speed equals distance divided by time and because the distance for one period is the circumference (2  r), v t = 2  r/ T.

39  Square this value, subtract the squared value into your previous equation, and then isolate T 2.  How does your result relate to Kepler's third law?

40  T 2 = (4  2 ) r3  (Gm)

41  T = 2  ( r 3 /Gm)  v =  [G ( m/r)]  Note that m is the mass of the object being orbited.  r is the distance between the centers of the two bodies.

42  Using Table 1 on p250 of the text, find the orbital speed and period of a satellite traveling at a mean altitude of 361 km above Earth, Jupiter, and Earth's moon.  Earth: 7.69 x 10 3 m/s; 5.51 x 103 s  Jupiter: 4.20 x 10 4 m/s; 1.08 x 10 4 s  moon: 1.53 x 10 3 m/s; 8.63 x 10 3 s

43  At what distance above the Earth would a satellite have a period of 125 min?  1.90 x 10 6 m

44  Weight is the magnitude of the force due to gravity.  When you step on a scale, it measures the downward force applied to it.  From Newton's third law, we know the normal force of the scale equals the downward force applied to it.  The scale reading is equal to the normal force acting on you.

45  When you are standing on a scale in an elevator and the elevator is at rest, the scale reads your weight.  When the elevator accelerates downward, the normal force will be smaller and the scale will read less than your weight.  When the elevator's acceleration is equal to free-fall acceleration, the scale will read zero and you will have no normal force acting on you.  Both you and the elevator are falling with free-fall acceleration.  Apparent weightlessness.

46  Astronauts in orbit experience apparent weightlessness.  The shuttle is accelerating at the same rate as the astronauts.  Force due to gravity keeps the shuttle and astronauts in orbit, but the astronauts feel weightless because there is no normal force acting on them.  Actual weightlessness only occurs in deep space.

47  Objectives:  1. Distinguish between torque and force.  2. Calculate the magnitude of a torque on an object.  3. Identify the six types of simple machines.  4. Calculate the mechanical advantage of a simple machine.

48  Rotational motion is the motion of a rotating rigid object that spins about its center of mass.  Objects can exhibit multiple types of motion at the same time, but they can be separated and analyzed separately.

49  Torque is a quantity that measures the ability of a force to rotate an object about some axis.  The farther the force is from the axis of rotation, the easier it is to rotate the object and the more torque is produced.  The perpendicular distance from the axis of rotation to a line drawn along the direction of the force is called the lever arm.  Forces do not have to be perpendicular to an object to cause the object to rotate.


51  When the angle is less than 90 ํํํํํํ, the object will not rotate as easily.  T = Fd sin Ө  Torque = force x lever arm!!  Ө is the angle between the applied force and the lever.  Units of torque are N▪m



54  Torque is a vector and, as thus, has a magnitude and a direction.  Torque is positive if the rotation is counterclockwise and negative if it is clockwise.  Net torque can be found by finding the sum of the individual torques acting on an object.

55  Find the magnitude of the torque produced by a 3.0 N force applied to a door at a perpendicular distance of 0.25 m from the hinge.  0.75 N▪m

56  A simple pendulum consists of a 3.0 kg point mass hanging at the end of a 2.0 m long light string that is connected to a pivot point.  a. Calculate the magnitude of the torque (due to gravitational force) around this pivot point when the string makes a 5.0˚ angle with the vertical.  b. Repeat this calculation for an angle of 15.0˚.  a. 5.1 N▪m  b. 15 N▪m

57  If the torque required to loosen a nut on the wheel of a car has a magnitude of 40.0 N▪m, what minimum force must be exerted by a mechanic at the end of a 30.0 cm wrench to loosen the nut?  133 N

58  A machine is any device that transmits or modifies force, usually by changing the force applied to an object.  All machines are combinations or modifications of six simple machines  lever  pulley  inclined plane  wheel and axle  wedge  screw


60  Mechanical advantage is the comparison of the output force to the input force.  MA = output force = F out  input force F in  When using a lever (friction disregarded)  T in = T out  F in d in = F out d out  MA = F out = d in  F in d out


62  Mechanical energy is conserved (in the absence of friction) by machines as well.  Work will remain constant.  When using an inclined plane to move an object, the force required is decreased but the distance moved is increased.

63  Real machines are not frictionless and some mechanical energy is dissipated.  Efficiency relates the useful work output to the work input.  eff = W out  W in  If the machine is frictionless, the efficiency is 1.  All real machines have an efficiency of less than 1.

Download ppt "Circular Motion and Gravitation.  Objectives:  1. Solve problems involving centripetal acceleration.  2. Solve problems involving centripetal force."

Similar presentations

Ads by Google