AP Physics 1 Chapter 7 Circular Motion and Gravitation

Chapter 7: Circular Motion and Angular Measure Gravitation Angular Speed and Velocity Uniform Circular Motion and Centripetal Acceleration Angular Acceleration Newton s Law of Gravitation Kepler s Laws and Earth Satellites

Homework for Chapter 7 Read Chapter 7 Chapter 7: #s 15,16,17,21,23,25,26,27,29,51 #18,19,22,30,31,32,54,55,60,65,66,68,70, # 62,63,66,68,70, 71,75 # 33,35,36,37,38,40 #34,39,56,57,59#41-47,53

Angular Measure Angular Speed and Velocity

Angular Measure rotation axis of rotation lies within the body (example: Earth rotates on its axis) revolution axis of rotation lies outside the body (example: Earth revolves around the Sun) The relationship between rectangular coordinates and polar coordinates are: x = r cos Ө y = r sin Ө Circular motion is conveniently described using polar coordinates (r,ө) because r is a constant and only Ө varies. Ө is measured counter-clockwise from the +x axis.

Angular Measure Angular distance ( Ө = Ө Ө 0 ) may be measured in either degrees or radians (rad). 1 rad 57.3 or 2π rad = 360

Angular Measure

Angular Measure Example 7.1: When you are watching the NASCAR Daytona 500, the 5.5 m long race car subtends and angle of 0.31. What is the distance from the race car to you?

Angular Speed and Velocity Linear analogy: v = x t Linear analogy: a = v t

Angular Speed and Velocity The units of angular acceleration are rad/s 2. The way to remember this is the right-hand rule: When the fingers of the right hand are curled in the direction of rotation, the extended thumb points in the direction of the angular velocity or angular acceleration vector.

Angular Speed and Velocity a) Tangential and angular speeds are related by v = rω, with ω in radians per second. Note, all of the particles rotating about a fixed axis travel in circles. All of the particles have the same angular speed (ω). Particles at different distances from the axis of rotation have different tangential speeds. b) Sparks from a grinding wheel illustrate instantaneous tangential velocity.

Angular Speed and Velocity For every linear quantity or equation there is an analogous angular quantity or equation. (Assume x 0 = 0, θ 0 = 0, t 0 = 0). Substitute θ x, ω v, α a. Quantity Linear / Tangential Angular distance (arc length) s rθ tangential speed v rω tangential acceleration a rα displacement x = v t θ = ω t average velocity v = v + v 0 2 ω = ω + ω 0 2 kinematics eqn. #1 v = v 0 + at ω = ω 0 + αt kinematics eqn. #2 x = v 0 t + ½ at 2 θ = ω 0 t + ½ αt 2 kinematics eqn. #3 v 2 = v 0 2 + 2ax ω 2 = ω 0 2 + 2αθ

Angular Speed and Velocity When angular speed and velocity are given in units of rpm (revolutions per minute) you should first convert them to rad/s before trying to solve the problem. Example 7.2a: Convert 33 rpm to rad/s.

Angular Speed and Velocity f = frequency T = period ω = angular speed

Angular Speed and Velocity The SI unit of frequency is 1/sec or hertz (Hz).

Angular Speed and Velocity Example 7.2b: A bicycle wheel rotates uniformly through 2.0 revolutions in 4.0 s. a) What is the average angular speed of the wheel? b) What is the tangential speed of a point 0.10 m from the center of the wheel? c) What is the period? d) What is the frequency?

Check for Understanding

Uniform Circular Motion and Centripetal Acceleration Angular Acceleration

Uniform Circular Motion and Centripetal Acceleration Physics Warmup # 35

Uniform Circular Motion and Centripetal Acceleration uniform circular motion An object moves at a constant speed in a circular path. The speed of an object in uniform circular motion is constant, but the object s velocity changes in the direction of motion. Therefore, there is an acceleration. Fig. 7.8 p.218

Uniform Circular Motion and Centripetal Acceleration centripetal acceleration center-seeking For and object in uniform circular motion, the centripetal acceleration is directed towards the center. There is no acceleration component in the tangential direction. If there were, the magnitude of the velocity (tangential speed) would change. a c = v 2 = rω 2 r Fig. 7.10, p.219

Uniform Circular Motion and Centripetal Acceleration From Newton s second law, F net = ma. Therefore, there must be a net force associated with centripetal acceleration. In the case of uniform circular motion, this force is called centripetal force. It is always directed toward the center of the circle since we know the net force on an object is in the same direction as acceleration. F c = ma c = mv 2 = mrω 2 r Centripetal force is not a separate or extra force. It is a net force toward the center of the circle. A centripetal force is always required for objects to stay in a circular path. Without it, an object will fly out along a tangent line due to inertia.

Uniform Circular Motion and Centripetal Acceleration The time period T, the frequency of rotation f, the radius of the circular path, and the speed of the particle undergoing uniform circular motion are related by: T = 2 π r = 1 = 2 π v f ω centrifugal force center-fleeing force; a fictitious force; something made up by nonphysicists; the vector equivalent of a unicorn Hint: Do not label a force as centripetal force on your free-body diagram even if that force does act toward the center of the circle. Rather, label the actual source of the force; i.e., tension, friction, weight, electric force, etc. Question 1: What provides the centripetal force when clothes move around a dryer? (the inside of the dryer) Question 2: What provides the centripetal force upon a satellite orbiting the Earth? (Earth s gravity)

Uniform Circular Motion and Centripetal Acceleration Example 7.a:

Uniform Circular Motion and Centripetal Acceleration Example 7.3: A car of mass 1500 kg is negotiating a flat circular curve of radius 50 meters with a speed of 20 m/s. a) What is the source of centripetal force on the car? b) What is the magnitude of the centripetal acceleration of the car? c) What is the magnitude of the centripetal force on the car?

Uniform Circular Motion and Centripetal Acceleration Example 7.3a: A car approaches a level, circular curve with a radius of 45.0 m. If the concrete pavement is dry, what is the maximum speed at which the car can negotiate the curve at a constant speed?

Uniform Circular Motion and Centripetal Acceleration Check for Understanding: 1. In uniform circular motion, there is a a. constant velocity b. constant angular velocity c. zero acceleration d. net tangential acceleration Answer: b

Uniform Circular Motion and Centripetal Acceleration Check for Understanding: 2. If the centripetal force on a particle in uniform circular motion is increased, a. the tangential speed will remain constant b. the tangential speed will decrease c. the radius of the circular path will increase d. the tangential speed will increase and/or the radius will decrease Answer: d; F c = mv 2 r

Uniform Circular Motion and Centripetal Acceleration Check for Understanding: 3. Explain why mud flies off a fast-spinning wheel. Answer: Centripetal force is proportional to the square of the speed. When there is insufficient centripetal force (provided by friction and adhesive forces), the mud cannot maintain the circular path and it flies off along a tangent.

Angular Acceleration Average angular acceleration ( ) is = t The SI unit of angular acceleration is rad/s 2. The relationship between tangential and angular acceleration is a t = r (This is not to be confused with centripetal acceleration, a c ).

Angular Acceleration

In uniform circular motion, there is centripetal acceleration but no angular acceleration (α = 0) or tangential acceleration (a t = r α = 0). In nonuniform circular motion, there are angular and tangential accelerations. a t = v = (rω) = r ω = rα t t t Fig. 7.16, p.226

7.4 Angular Acceleration There is always centripetal acceleration no matter whether the circular motion is uniform or nonuniform. It is the tangential acceleration that is zero in uniform circular motion. Example 7.4: A wheel is rotating wit a constant angular acceleration of 3.5 rad/s 2. If the initial angular velocity is 2.0 rad/s and is speeding up, find a) the angle the wheel rotates through in 2.0 s b) the angular speed at t = 2.0 s

Angular Acceleration Example 7.5: The power on a medical centrifuge rotating at 12,000 rpm is cut off. If the magnitude of the maximum deceleration of the centrifuge is 50 rad/s 2, how many revolutions does it rotate before coming to rest?

Angular Acceleration Check for Understanding: 1. The angular acceleration in circular motion a. is equal in magnitude to the tangential acceleration divided by the radius b. increases the angular velocity if in the same direction c. has units of rad/s 2 d. all of the above Answer: d

Angular Acceleration Check for Understanding: 2. Can you think of an example of a car having both centripetal acceleration and angular acceleration? Answer: Yes, when a car is changing its speed on a curve.

Angular Acceleration Check for Understanding: 3. Is it possible for a car in circular motion to have angular acceleration but not centripetal acceleration? Answer: No, this is not possible. Any car in circular motion always has centripetal acceleration.

Newton s Law of Gravitation Kepler s Laws and Earth Satellites

Uniform Circular Motion and Centripetal Acceleration Physics Warmup # 48 Solution: It would decrease. You would have mass below you pulling downward and mass above you pulling upward. At the center of the earth, you would weigh zero.

Newton s Law of Gravitation G is the universal gravitational constant.

Newton s Law of Gravitation F α 1 r 2 Inverse Square Law

Any two particles, or point masses, are gravitationally attracted to each other with a force that has a magnitude given by Newton s universal law of gravitation. For homogeneous spheres, the masses may be considered to be concentrated at their centers. Fig. 7.17, p.228

Newton s Law of Gravitation

Newton s Law of Gravitation We can find the acceleration due to gravity, a g, by setting Newton s 2 nd Law = the Law of Gravitation F = ma g = GmM r 2 (m cancels out) so, a g = GM This is the acceleration due to gravity at a r 2 distance r from a planet s center. At the Earth s surface: a ge = g = GM E R E 2 M E = 6.0 x 10 24 kg R E = 6.4 x 10 6 m where M E and R E are the mass and radius of the Earth. At an altitude h above the Earth s surface: a g = GM E (R E + h) 2

Newton s Law of Gravitation Example 7.7: Calculate the acceleration due to gravity at the surface of the moon. The radius of the moon is 1750 km and the mass of the moon is 7.4 x 10 22 kg.

Newton s Law of Gravitation Note: it is just r, not r 2, in the denominator.

Newton s Law of Gravitation

Gravitational potential energy U = - Gm 1 m 2 r Fig. 7.20, p. 231 Note: U = mgh only applies to objects near the surface of the earth. On Earth, we are in a negative gravitational potential energy well. Work must be done against gravity to get higher in the well: in other words, U becomes less negative. The top of the well is at infinity, where the gravitational potential energy is chosen to be zero.

Newton s Law of Gravitation Example 7.6: The hydrogen atom consists of a proton of mass 1.67 x 10-27 kg and an orbiting electron of mass 9.11 x 10-31 kg. In one of its orbits, the electron is 5.4 x 10-11 m from the proton and in another orbit, it is 10.6 x 10-11 m from the proton. a) What are the mutual attractive forces when the electron is in these orbits, respectively? a) If the electron jumps from the large orbit to the small one, what is the change in potential energy?

Newton s Law of Gravitation: Check for Understanding 1. The gravitational force is a. a linear function of distance b. an infinite-range force c. applicable only to our solar system d. sometimes repulsive Answer: b

Newton s Law of Gravitation: Check for Understanding 2. The acceleration due to gravity on the Earth s surface a. is a universal constant like G b. does not depend on the Earth s mass c. is directly proportional to the Earth s radius d. does not depend on the object s mass Answer: d

Newton s Law of Gravitation: Check for Understanding 3. Astronauts in a spacecraft orbiting the Earth or out for a spacewalk are seen to float in midair. This is sometimes referred to as weightlessness or zero gravity (zero g). Are these terms correct? Explain why an astronaut appears to float in or near an orbiting spacecraft. Answer: No. Gravity acts on the astronauts and the spacecraft, providing the necessary centripetal force for the orbit, so g is not zero and there is weight by definition (w=mg). The floating occurs because the spacecraft and astronauts are falling ( accelerating toward Earth at the same rate).

7.3 Uniform Circular Motion and Centripetal Acceleration Physics Warmup # 33 Boeing 747 Freedom 7 Space Shuttle ISS Hubble

Satellites Johannes Kepler (1571-1630) German astronomer and mathematician formulated three law of planetary motion The laws apply not only to planets, but to any system of a body revolving about a more massive body (such as the Moon, satellites, some comets)

Satellites

Satellites When a planet is nearer to the sun, the radius of orbit is shorter, and so its linear momentum must be greater in magnitude (it orbits with greater speed) for angular momentum to be conserved.

Satellites

Satellites We can find the tangential velocity of a planet or satellite where m is orbiting M. Set: Centripetal Force = Force of Gravity F = mv 2 = GmM r r 2 Solve for v: v = GM tangential velocity r of an orbiting body Kepler s third law can be derived from this expression. Since v = 2 π r / T (circumference / period), and M is the mass of the Sun, 2 π r = GM T r Squaring both sides and solving for T 2 gives For our solar system, K = 2.97 x 10-19 s 2 /m 3 T 2 = 4 π 2 r 3 or T 2 = Kr 3 Kepler s 3 rd Law or GM Kepler s Law of Periods

Satellites Example 7.8: The planet Saturn is 1.43 x 10 12 m from the Sun. How long does it take for Jupiter to orbit once about the Sun?

Satellites Example 11: A satellite is placed into a circular orbit 1000 km above the surface of the earth (r = 1000 km + 6400 km = 7400 km). Determine a) the time period (T) of the satellite b) the speed (v) of the satellite

Satellites escape speed the initial speed needed to escape from the surface of a planet or moon. At the top of a planet s potential energy well, U = 0. An object projected to the top of the well would have an initial velocity of v esc. At the top of the well, its velocity would be close to zero. From the conservation of energy, final equals initial: K 0 + U 0 = K + U ½ mv esc 2 GmM = 0 + 0 r v esc = 2GM r escape speed On Earth, since g = GM E / R E2, v esc = 2gR E A tangential speed less than the escape speed is required for a satellite to orbit. Notice, escape speed does not depend on the mass of the satellite.

Satellites Example 7.9: If a satellite were launched from the surface of the Moon, at what initial speed would it need to begin in order for it to escape the gravitational attraction of the Moon?

Satellites

7.6 Satellites Note: a geosynchronous satellite orbits the earth with a period of 24 hours so its motion is synchronized with the earth s rotation. Viewed by an observer on earth, a geosynchronous satellite appears to be stationary. All geosynchronous satellites with circular orbits have the same orbital radius (36,000 km above sea level for Earth).

Check for Understanding A Space Shuttle orbits Earth 300 km above the surface. Why can t the Shuttle orbit 10 km above Earth? a) The Space Shuttle cannot go fast enough to maintain such an orbit. b) Because r appears in the denominator of Newton s law of gravitation, the force of gravity is much larger closer to the Earth; this force is too strong to allow such an orbit. c) The closer orbit would likely crash into a large mountain such as Everest because of its elliptical nature. d) Much of the Shuttle s kinetic energy would be dissipated as heat in the atmosphere, degrading the orbit.

Check for Understanding Answer: d. A circular orbit is allowed at any distance from a planet, as long as the satellite moves fast enough. At 300 km above the surface Earth s atmosphere is practically nonexistent. At 10 km, though, the atmospheric friction would quickly cause the shuttle to slow down.

7.6 Check Satellites for Understanding

7.6 Check Satellites for Understanding The period of a satellite is given by the formula: T 2 = K r 3. This means a specific period maps onto a specific orbital radius. Therefore, there is only one orbital radius for a geosynchronous satellite with a circular orbit.

Check for Understanding

Internet Activity Put a satellite in orbit: http://www.lon-capa.org/~mmp/kap7/orbiter/orbit.htm