Transcript PPT
Gravity Unanswered questions Galileo describes falling objects by rolling objects down a ramp. But why does everything accelerate the same rate regardless of its mass? Kepler describes planetary motions. What force can account for the elliptical paths of the planets? Two seemingly unrelated observations, but… Newton unites the two in one master stroke at the age of 24. Inverse Square Law 2 R r ac T2 T 2 kR3 2 Circular motion result Kepler’s observational law 2 R 1 r ac ~ 2 3 kR R Centripetal acceleration behaves like inverse square FG maC Newton’s 2nd Law for centripetal direction 2 1 ~m 2 R 1 R2 GMm FG R2 ~ Mm Newton’s 3rd Law suggests both masses are important Constant of proportionality determined by Cavendish G 6.67 10 11 Nm 2 kg2 The Force Law GMm FG R2 M m Direction? Along a line connecting the center of the two masses. Action at a distance. How does the force GET from the Earth to the moon? “I feign no hypothesis regarding action at a distance.” Potential Energy Revisited Remember how we used to write the force of gravity. FG mg From this, we derived an expression for gravitational potential energy: UG mgh This only applies near surface of earth! (When GME/RE2=9.8 m/s2) More generally, we have UG GMm R As usual, we have a choice for where we set UG = 0. When using this equation, the choice is made for us. Two Applications Terrestrial (free fall near the surface of a planet or star) FG mg Celestial (circular orbit around a planet or star) FG GMm R2 This most general expression is always true, but sometimes the first expression is simpler to implement (it has limited application, however, so be careful!). Terrestrial Application Example: The radius of the Earth is 6.4 x 106 m and the value of g is 9.8 m/sec2. What is the mass of the Earth? GM E m R2 FG gm GM E m gm 2 R gR 2 ME G FG 9.86.4 10 6 2 6.67 10 11 6.0 10 24 kg Celestial Application The ISS orbits the earth with a speed of approximately 7680 m/s. What is the orbital radius of the station, and what is its altitude? Fc FG m s me v2 ms G 2 R R me me 2 v G or, R G 2 R v 24 11 6 x10 R 6.67 x10 7680 2 6.79 x10 6 m Radius of the earth is about 6.38 x 106 m at the equator. That gives the altitude above the surface to be: 6.79x106 – 6.38x106 = 0.41x 106 m, or 410 km. “g” ~ 8.68 m/s2 Torque Dynamics r F1 r F2 r F3 Which applied force results in the largest angular acceleration of the bolt? Dynamics r F2 r F1 r F3 Which applied force results in the largest angular acceleration of the bolt? Dynamics r F1 r F2 Which applied force results in a clockwise angular acceleration? A counter-clockwise angular acceleration? Dynamics quantified Consider a force acting on a rigid body, some distance away from a fixed pivot point. r FPerp r FP ar Can split the force into components. Only the perpendicular part contributes to rotation! Where does the parallel part go? r r r FPerp r r r O Dynamics quantified r F1 r F2 r F3 Which applied force results in the largest angular acceleration of the bolt? Dynamics quantified r F2 r F1 r F3 Which applied force results in the largest angular acceleration of the bolt? Direction of torque r r r r r F1 r F2 Direction of torque is the direction the applied force tends to cause the object to rotate. F1 provides a clockwise torque. F2 provides a counter-clockwise torque. But torque is a vector, and vectors only point in a single direction. Direction of torquer is given by right hand rule. r Draw r and F tip to tail r Point fingers of right hand in the direction of r r Curl fingers in direction of F Thumb points in direction of torque: either into (clockwise, negative) or out of positive) page (counter-clockwise, Newton’s Laws Newton’s 1st Law – If there is no net torque on an object, then the object rotates at a constant angular velocity (could be zero angular velocity). Newton’s 2nd Law – If the net torque on an object about a point is not zero, then the net torque produces an angular acceleration about that point. r F rF r r rma rmr mr 2 O The quantity mr2 is the rotational equivalent of mass, and is called moment of inertia. Newton’s 3rd Law – For every action, there is an equal and opposite reaction. Example r FA r FB A B r FT 8000 N 20m 40m A truck crosses a massless bridge supported by two piers. What force much each pier exert when the truck is at the indicated position? r r r r r r net, A T, A B, A net, B T, B A, B 0 ()FT rT ()FB rB 0 FT rT FB rB Fr 8000(20) FB T T 2667 N rB (60) 0 ()FT rT ()FA rA 0 FT rT FA rA Fr 8000(40) FA T T 5333 N rA (60) 2667 5333 8000 N Sample Torque Problems X = __________ X = __________