CAPA due today. Today will finish up with the hinge problem I started on Wednesday. Will start on Gravity. Universal gravitation

Transcription

1 CAPA due today. Today will finish up with the hinge problem I started on Wednesday. Will start on Gravity. Universal gravitation

2 Hinge Problem from Wednesday

3 Hinge Problem cont. F x = 0 = F Nx T cosθ F y = 0 = T sinθ W 1 W 2 F Ny = 0 Guessed wrong on direction τ = 0 = T sinθ L W 2 L W 1 L 2 T = (W 2 + W 1 2 ) /sinθ F Nx = (W 2 + W 1 2 )cotθ F Ny = W 1 /2

4 Things to keep in mind Force of gravity acts at the center of mass Tension force must be in the same direction as the rope There are often multiple choices for a reasonable axis about which to calculate torques. All of them are OK and you should still be able to solve the problem. A good choice just makes the problem easier.

5 Clicker question 1 Set frequency to BA A mass M is placed on a very light board supported at the ends, as shown. The free-body diagram shows directions of the forces, but not their correct relative sizes. M F L F R (2/3)L L/3 Mg A: 2/3 B: 1/3 C: 1/2 D: 2 E: some other answer What is the ratio F R /F L?

6 Clicker question 1 Set frequency to BA A mass M is placed on a very light board supported at the ends, as shown. The free-body diagram shows directions of the forces, but not their correct relative sizes. M F L F R (2/3)L L/3 Mg A: 2/3 B: 1/3 C: 1/2 D: 2 E: some other answer What is the ratio F R /F L? Sum of the Torques = 0, or - F L *(2/3L ) + F R *(L/3) = 0, so F R /F L = 2/1.

7 Newton s Law of Gravity Newton and Einstein are generally thought to be the two greatest physicists ever. Not only did Newton come up with the three laws of motion and invent calculus, he was the first to realize that the force associated with things falling was also responsible for astronomical phenomena. Newton s Law of Gravitation can be written as Between any two masses (here m 1 & m 2 ) there is an attractive force proportional to the product of the masses and inversely proportional to the square of the distance between them.

8 Gravitational Force is the force of gravity which is felt by each mass and directed towards the other mass. Newton figured out the 1/r 2 dependence assuming that the celestial objects and the Earth were point particles. By inventing integral calculus he could prove that for a mass m 2, outside a spherical mass m 1, the force of gravity was as if all of the mass m 1 was in the center of the sphere. Therefore for any two spherically symmetric objects, the distance r that enters into the force of gravity is the distance between the centers of the spheres.

9 Newton s Shell Theorem A uniform spherical shell of matter attracts a particle that is outside the shell as if all of the shell s mass were concentrated at its center. Let ρ = M 4πR 2 da = (2πRsinθ)Rdθ dm = ρda = 1 2 M sinθ dθ df = GmdM s 2 cosα = 1 2 GmM cosα sinθdθ s 2 Find F = GMm /r 2

10 Force rules is the force of gravity with Newton s 2 nd law still works. The net force on an object determines the object s acceleration: Remarkably, the mass in Newton s 2 nd law (called the inertial mass) is the same as the mass in the law of gravitation (called the gravitational mass). Einstein figured out (230 years later) that this coincidence could be explained by assuming space and time were curved (in the theory of general relativity). Remember, force is still a vector and the law of superposition still works. To find the net gravitational force on an object, determine the magnitude and direction of the force from all other masses and then add these forces together. F 1,net = F 12 + F 13 + F F 1n

11 Clicker question 2 Set frequency to BA Two asteroids in inter-galactic space are a distance r = 20 km apart. Asteroid 2 has 10 times the mass of asteroid 1. The magnitudes of the accelerations of asteroids 1 and 2 are a 1 and a 2, respectively. What is the ratio a 1 /a 2? r = 20 km A. 1/100 m 1 m 2 B. 1/10 C. 1 D. 10 E. 100

12 Clicker question 2 Set frequency to BA Two asteroids in inter-galactic space are a distance r = 20 km apart. Asteroid 2 has 10 times the mass of asteroid 1. The magnitudes of the accelerations of asteroids 1 and 2 are a 1 and a 2, respectively. What is the ratio a 1 /a 2? A. 1/100 B. 1/10 C. 1 D. 10 E. 100 The force on m 1 is the same as the force on m 2 : Acceleration is force divided by mass r = 20 km m 1 m 2 so and which gives us

13 Comments about Earth The density of the Earth is higher than most of the other planets in our solar system. Sources vary when it comes to the density of the Earth. ~5.5 g/cm 3 Inner Core: solid, ~13 g/cm 3 mainly Fe + Ni Outer Core: Thought to be mainly responsible for earth s magnetic field.~11g/cm 3 Lower Mantle: Silicon, magnesium, Oxygen ~3.5g/cm 3

14 Earth is not a sphere + rotating! The equator radius is larger than the polar radius by 21 km! Means gravitational acceleration is larger at poles than at equator! R = m F = F N ma g = m( v 2 /R) F N = ma g mω 2 R g = a g ω 2 R ω = dθ dt = 2π radians 24 hr g = 9.8m /s m /s 2

15 Force of gravity on Earth How does correspond to our new force? If we consider mass 2 to be the Earth (M E ) and r to be the radius of the Earth (R E ) then we can write Using known values we can find that So, on the surface of the Earth, the force of gravity between the Earth and an object m 1 is We can only use if the distance above the surface is very small compared to the radius.

16 Clicker question 3 Set frequency to BA Used to find near the Earth s surface Planet X has the same mass as the Earth, but ½ the radius due to its higher density. What is the acceleration of gravity on Planet X? A. ¼ g B. ½ g For Earth C. g D. 2 g E. 4 g

17 Clicker question 3 Set frequency to BA Used to find near the Earth s surface Planet X has the same mass as the Earth, but ½ the radius due to its higher density. What is the acceleration of gravity on Planet X? A. ¼ g B. ½ g For Earth C. g D. 2 g For Planet X E. 4 g The higher density of Krypton (being made of Kryptonite) makes the force of gravity at the surface stronger, meaning Superman must be stronger to do any old normal thing.

18 Clicker question 4 A rock is released from rest in space beyond the orbit of the Moon. The rock falls toward the Earth and crosses the orbit of the Moon. At this point, the acceleration of the rock is A. greater B. smaller C. the same as Set frequency to BA Earth Moon rock the acceleration of the Moon.

19 Clicker question 4 Set frequency to BA A rock is released from rest in space beyond the orbit of the Moon. The rock falls toward the Earth and crosses the orbit of the Moon. At this point, the acceleration of the rock is A. greater B. smaller C. the same as Earth Moon rock the acceleration of the Moon. If the Moon and the rock are a distance r from the center of the Earth then the acceleration of either mass can be determined by independent of whether it is the Moon or a rock Note, the speeds are probably not the same but the accelerations are!

20 Gravitational potential energy When we used we found a potential energy of. What is the potential energy associated with the force? To make sense, potential energy should increase as the distance increases and be smallest when the objects are closest together. A while ago we learned force is the derivative of potential energy. The potential energy gives the force when you take the derivative with respect to r.

21 Gravitational potential energy Potential energy increases (less negative) as the separation increases. This is what we wanted. Maximum potential energy is 0 when r approaches infinity. Since two objects cannot share the same space, r > 0. The minimum potential energy is when the objects are touching.

22 Earth s gravitational potential energy Potential energy due to Earth s gravity is where r is the distance from the center and h is the height above the surface of the Earth. R E = radius of Earth = 6380 km Distance from center of the Earth (km) Suppose a rock is released from rest at r = km. Initially it only has potential energy. It will start falling, converting potential energy to kinetic energy. The total energy remains the same. The rock cannot go past r = km because it would have negative kinetic energy at that point (which is impossible).

23 Effect of total energy on trajectory If the total energy were 0 then it is possible for the object to make it to r =. We can identify three basic scenarios for a total energy which is positive, negative, or 0. R E = radius of Earth = 6380 km Distance from center of the Earth (km) For total energy < 0 the object is bound by the gravitational field (and orbits are ellipses). Examples are planets around the sun. For total energy of 0 the object is barely unbound (parabolic orbit). For total energy > 0, object is unbound with a hyperbolic orbit.