Chapter 5 Circular Motion; Gravitation

Transcription

1 Chapter 5 Circular Motion; Gravitation

2 Kinematics of Uniform Circular Motion Dynamics of Uniform Circular Motion Highway Curves, Banked and Unbanked Non-uniform Circular Motion Centrifugation Will be covered after chapter 7 (5.6) Newton s Law of Universal Gravitation (5.7) Gravity Near the Earth s Surface

3 Recalling Last Lecture

4 Remind: Midterm Date: October 16, 1pm :15pm. Location: CL There are four problems. One of them will be selected between problems 5 and 87 found in assignment 5. The midterm includes everything covered until last Thursday, including friction and inclines (chapters 1 to 4). - You ARE allowed to use a calculator. An equation sheet will be provided (you can find it at ) - Solutions of the assignments can be found on the webpage listed above. You can also find a link to old midterms from this webpage. - I will be eceptionally available for discussions, questions, etc between 14:30 and 15:00h today, and between 13:00 and 15:00h tomorrow. - I strongly recommend the Wednesday s tutorial session: an old midterm will be solved. - Remember, your marked assignments are available for you. Feel free to come anytime to my office (LB-1) if you have failed to pick it up in classroom.

5 Friction and Inclined Planes Three forces can ALWAYS be identified acting on an object moving on an inclined place: Gravity (vertical); friend Friction (along the surface); Normal (perpendicular to the surface). friend 1 you (4.5) (4.6)

6 Tension in a Fleible Cord Problem 4-41 (tetbook A 15.0-kg bo is released on a 3º incline and accelerates down the incline at 0.30 m/s. Find the friction force impeding its motion. What is the coefficient of kinetic friction? = 0.30 m/s 3 o

7 Problem 4.41 (tetbook): A free-body diagram for the bo is shown. Write Newton s nd law for each direction: F = mg sinθ F = ma F = F mg cosθ = ma = 0 y N Notice that the sum in the y direction is 0, since there is no motion (and hence no acceleration) in the y direction. Solve for the force of friction. fr y F r fr F r N θ mg r y θ mg sin θ F = ma fr ( ) ( )( ) F = mg = = fr o sinθ ma 15.0 kg 9.80 m s sin m s N 73 N Now solve for the coefficient of kinetic friction. Note that the epression for the normal force comes from the y direction force equation above. F N fr F = µ F = µ mg cos θ µ = = = 0.59 fr k N k k o mg cosθ 15.0 kg 9.80 m s cos 3 ( )( )( )

8 Tension in a Fleible Cord Problem 4-65 (tetbook A bicyclist of mass 65 kg (including the bicycle) can coast down a 6.0º hill at a steady speed of 6.0 Km/h because of air resistance. How much force must be applied to climb the hill at the same speed and same air resistance?

9 Problem 4.65 (tetbook): Consider a free-body diagram for the cyclist coasting downhill at a constant speed. Here we call F fr the friction due to air resistance (and not sliding or static friction). Since there is no acceleration, the net force in each direction must be zero. Write Newton s nd law for the direction. F = mg sinθ F = 0 F = mg sinθ fr fr F r fr F r N y This establishes the size of the air friction force at 6.0 km/h, and so can be used in the net part. Now consider a free-body diagram for the cyclist climbing the hill. F p is the force pushing the cyclist uphill. Again, write Newton s nd law for the direction, with a net force of 0. F = F + mg sinθ F = 0 fr F = F + mg sinθ = mg sinθ P fr ( )( )( ) = = o 65 kg 9.8 m s sin N P y F r P θ θ θ mg r θ mg r F r N F r fr

10 Newton s Law of Universal Gravitation If the force of gravity is being eerted on objects on Earth, what is the origin of that force? Newton s realization was that the force must come from the Earth. He further realized that this force must be what keeps the Moon in its orbit.

11 Newton s Law of Universal Gravitation More than that: the Earth eerts a downward force on you, and you eert an upward force on the Earth. This follows from Newton s third law. When there is such a difference in masses, the reaction force is undetectable, but for bodies more equal in mass it can be significant. Therefore, the gravitational force must be proportional to both masses.

12 Newton s Law of Universal Gravitation By observing planetary orbits, Newton also concluded that the gravitational force must decrease as the inverse of the square of the distance between the masses. Summary: The gravitational force is directly proportional to the masses and inversely proportional to the square of their distance. Or, as stated in Newton s law of universal gravitation: Every particle in the universe attracts every other particle with a force that is proportional to the products of their masses and inversely proportional to the square of the distance between them. This forces acts along the line joining the two particles.

13 Newton s Law of Universal Gravitation The magnitude of this force is given by: (4.7) Where:

14 Newton s Law of Universal Gravitation We know the relation between force and acceleration:. But what is a in? If m is the object eerting a force on m 1, then the gravitational acceleration felt by object 1 can be identified as: (4.8) The opposite is also true: object will feel an acceleration due to the gravitational force applied by object 1. You can then write for the force acting on m 1 and for the force acting on m.

15 Gravity Near the Earth s Surface An object of mass m ON the surface of the Earth will feel a force given by: Where, m E = mass of the Earth ; r E = m = radius of the Earth. (4.9) g has been measured and it is known to be 9.80 m/s. In fact, the value of g can be considered constant at any position near the Earth s surface (this is what we have been assuming without much discussion so far). Note that knowing G, r E and g, we can calculate the mass of the Earth. From eq. 4.9: More accurate calculations lead to m E = Kg

16 Newton s Law of Universal Gravitation Problem 5-31 (tetbook): A hypothetical planet has a radius 1.5 times that of Earth, but has the same mass. What is the acceleration due to gravity near its surface? Solution: The acceleration due to gravity at any location on or above the surface of a planet is given by g = GM r planet Planet where r is the distance from the center of the planet to the location in question. M M 1 M m s g = G = G = G = g = = planet r Planet Earth Earth Earth ( 1.5REarth) 1.5 R Earth 4.4 m s

17 Newton s Law of Universal Gravitation Problem 5-34 (tetbook): Calculate the effective value of g, the acceleration of gravity, at (a) 300 m, and (b) 300 km, above the Earth s surface. Solution: The acceleration due to gravity at any location at or above the surface of a planet is given by where r is the distance from the center of the planet to the location in question. For 4 this problem, M = M = kg P lanet E arth (a) (b) g = G M r plan et P lanet 6 r = R Earth m = m m Earth 4 ( kg) ( ) 6 ( m m) M g = G = N m kg = 9.77 m s r r = R Earth km = m m = m Earth ( ) ( kg ) 6 ( m) M g = G = N m kg = 4.34 m s r 11