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PHYSICS 231 Lecture 19: We have lift-off! Comet Kohoutek Remco Zegers Walk-in hour: Monday 9:15-10:15 am Helproom PHY 231 1 Previously… v ac r Centripetal acceleration: ac=v2/r=2r Caused by force like: •Gravity •Tension •Friction •F=mac for rotating object The centripetal acceleration is caused by a change in the direction of the linear velocity vector, not a change in magnitude PHY 231 2 quiz (extra credit) 3 children are sitting on a rotating disc in a playground. The disc starts to spin faster and faster. Which of the three is most likely to start sliding first? top view a) child A A b) child B c) child C d) the same for all three B C ac= 2r. A= B= c rA<rB<rc acA<acB<acC PHY 231 demo 3 The gravitational force, revisited Newton: m1m2 F G 2 r G=6.673·10-11 Nm2/kg2 The gravitational force works between every two massive particles in the universe, yet is the least well understood force known. PHY 231 4 Gravitation between two objects A B The gravitational force exerted by the spherical object A on B can be calculated by assuming that all of A’s mass would be concentrated in its center and likewise for object B. Conditions: B must be outside of A A and B must be ‘homogeneous’ PHY 231 5 Gravitational acceleration m1mEARTH F G 2 r F=mg g=GmEARTH/r2 On earth surface: g=9.81 m/s2 r=6366 km On top of mount Everest: r=6366+8.850km g=9.78 m/s2 Low-orbit satellite: r=6366+1600km g=6.27 m/s2 Geo-stationary satellite: r=6366+36000km g=0.22 m/s2 PHY 231 6 Losing weight easily? You are standing on a scale in a stationary space ship in low-orbit (g=6.5 m/s2). If your mass is 70 kg, what is your weight? F=mg=70*6.5=455 N And what is your weight if the space ship would be orbiting the earth? Weightless! PHY 231 7 launch speed 4 km/s 6 km/s PHY 231 8 km/s 8 Gravitational potential energy So far, we used: PEgravity=mgh Only valid for h near earth’s surface. More general: PEgravity=-GMEarthm/r PE=0 at infinity distance from the center of the earth See example 7.12 for consistency between these two. Example: escape speed: what should the minimum initial velocity of a rocket be if we want to make sure it will not fall back to earth? KEi+PEi=0.5mv2-GMEarthm/REarth KEf+PEf=0 v=(2GMearth/REarth)=11.2 km/s PHY 231 9 Kepler’s laws Johannes Kepler (1571-1630) PHY 231 10 Kepler’s First law Ellipticity e(0-1) p+q=constant An object A bound to another object B by a force that goes with 1/r2 moves in an elliptical orbit around B, with B being in one of the focus point of the ellipse; planets around the sun. PHY 231 11 launch speed is 10 km/s PHY 231 12 Kepler’s second law Area(D-C-SUN)=Area(B-A-SUN) A line drawn from the sun to the elliptical orbit of a planet sweeps out equal areas in equal time intervals. PHY 231 13 Kepler’s third law r3 T2 r3=T2/Ks Consider a planet in circular motion around the sun: M sun M planet M planetv 2planet G 2 r r s 2r v t T 2 3 4 2 r K s r 3 T GM sun K s 2.97 10 19 s 2 / m 2 r3=constant*T2 T: period-time it takes to make one revolution PHY 231 14 Chapter 8. Torque Top view It is much easier to swing the door if the force F is applied as far away as possible (d) from the rotation axis (O). Torque: The capability of a force to rotate an object about an axis. Torque =F·d (Nm) Torque is positive if the motion is counterclockwise Torque is negative if the motion is clockwise demo: opening a cellar door demo: turning a screw PHY 231 15 Decompositions F= FL F// Top view What is the torque applied to the door? Force parallel to the rotating door: F//=Fcos600=150 N Force perpendicular to rotating door: FL=Fsin600=260 N Only FL is effective for opening the door: =FL·d=260*2.0=520 Nm PHY 231 16 Multiple force causing torque. 100 N 0.3 m Top view Two persons try to go through a rotating door at the same time, one on the l.h.s. of the rotator and one the r.h.s. of the rotator. If the forces are applied as shown in the drawing, what will happen? 50 N 0.6 m 1=F1·d1=-100*0.3=-30 Nm 2=F2·d2=50*0.6 =30 Nm 0 Nm + Nothing will happen! The 2 torques are balanced. demo: balance with unequal masses PHY 231 17 dpull Center of gravity. Fpull 1 2 3………………………n Fgravity dgravity? =Fpulldpull+Fgravitydgravity PHY 231 Vertical direction (I.e. side view) We can assume that for the calculation of torque due to gravity, all mass is concentrated in one point: The center of gravity: the average position of the mass dcg=(m1d1+m2d2+…+mndn) (m1+m2+…+mn) 18 Center of Gravity; more general The center of gravity m x m i xCG i i i i m y m i y CG i i i i demo center of gravity PHY 231 19 Top view Fp -d d CG -Fp Object in equilibrium Newton’s 2nd law: F=ma Fp+(-Fp)=ma=0 No acceleration, no movement… But the block starts to rotate! =Fpd+(-Fp)(-d)=2Fpd There is movement! Translational equilibrium: F=ma=0 The center of gravity does not move! Rotational equilibrium: =0 The object does not rotate Mechanical equilibrium: F=ma=0 & =0 No movement! PHY 231 20