v lim a t = d v dt a n = v2 R curvature

Transcription

1 PHY 02 K. Solutions for Problem set # 6. Textbook problem 5.27: The acceleration vector a of the particle has two components, the tangential acceleration a t d v dt v lim t 0 t (1) parallel to the velocity vector and the normal acceleration a n v 2 R curvature (2) perpendicular to the velocity. Given the magnitude a and the angle θ between the a and v vectors, we have a t a cosθ 1.05 m/s 2 cos m/s 2, a n a sinθ 1.05 m/s 2 sin m/s 2. () (a) The particle is traveling in a circle, so the curvature radius of its path is simply the circle s radius R 2.90 m. Given this radius and the normal acceleration, we can find the particle s speed as a n v2 R v R a n 2.90 m 0.56 m/s m/s. (4) (b) If the tangential acceleration of the particle is constant, then its speed increases linearly with time as v(t) v 0 + a t t. (5) Hence, two seconds after the data in part (a), the particle has speed v 1.27 m/s m/s s.05 m/s. (6) 1

2 Textbook problem 5.7: (a) Given the plane s speed v 100 km/h 810 MPH 60 m/s and the upper limit on its centripetal acceleration there is a lower limit on the radius of the circle it can fly: a c v2 R < amax c, (7) R > R min v2 a max c (60 m/s) m. (8) m/s2 (b c) The apparent weight of a body is W app m( g a). (9) At the bottom of the circle, the centripetal acceleration of the plane points up, so the artificial gravity m a points down and adds to the real gravity m g. Hence, the apparent weight of the pilot is W bot app m(g +a c) m(g +6.0 g) 7.0 mg kg 9.8 m/s N 1200 lb. (10) At the top of the circle, the centripetal acceleration points down so the artificial gravity m a points up and subtracts from the real gravity. Hence, W top app m g a c m g 6.0 g 5.0 mg kg 9.8 m/s N 860 lb. (11) Note that since a c > g, this apparent weight is directed up rather than down! But that s OK since at the top of the circle, the plane is flying upside down. 2

3 Non-textbook problem #1: According to Newton s Law of Gravity, your weight on Earth is W E m g E G M E m R 2 E (12) where m is your mass, M E and R E are Earth s mass and radius, and G is the universal gravitational constant (also known as Newton s constant, even though Newton didn t known its value). Likewise, your weight on Titan would be W T m g T G M T m R 2 T. (1) Note that G and m in eqs. (12) and (1) are exactly the same G is universal, and your mass m does not depend on where you happen to be. Consequently, these variables cancel out of the ratio W T W E G M T m M T R 2 T RT 2 / ME R 2 E M T/M E (R T /R E ) (0.468) / G ME m In other words, your weight on Titan would be 0.10 your weight on Earth, whatever it happens to be. For example, if you weight 150 pounds on Earth, on Titan you would weigh only 15.5 pounds. R 2 E (14) Non-textbook problem #2: (a) According to Kepler s First Law, planets go around the Sun in elliptic orbits, with the Sun being in one of the focal points of the ellipse. This is illustrated at the hyperphysics web page at see the Kepler Laws link in the supplementary notes section of the homework page. The top picture at

4 that page shows the Sun is located on the major axis of the ellipse (whose length is 2a) at the distance e a from the middle of the axis, where e is the eccentricity of the ellipse. The closest point of the orbit to the Sun the perihelion is at one end of the major axis, and the most distant point the aphelion it at the other end of the major axis. Thus, at the perihelion, the distance between the planet and the Sun is R p a e a (1 e) a, (15) and at the aphelion, the distance is R a a + e a (1+e) a. (16) Mercury s orbit has semi-major axis a 0.87 au and eccentricity e Hence, at the perihelion, Mercury is only R p (1 0.20) 0.87 au 0.1 au km away from the Sun, while at the aphelion the distance increases to R a (1+0.20) 0.87 au 0.46 au km. (b) According to Kepler s Third Law, planet s year the time it takes to complete one orbit around its star is related to the semi-major axis a of its orbit at a T 2π. (17) GM star Comparing two planets orbiting the same star such as Mercury and Earth and taking the ratio of their orbital periods, we have T M 2π T E a M GM Sun a M GM Sun a M a E / a E 2π GM Sun / a E GM Sun (18) ( am a E ) /2. Note the G and the Sun s mass cancel out from this ratio. 4

5 By definition of the astronomical unit, Earth orbit around the Sun has a E 1 au. The problem gives us Mercury orbit s semi-major axis in astronomical units, a M 0.87 au, which means that the ratio a M /a E is Consequently, the ratio of orbital periods of Mercury and Earth is T M T E ( am a E ) /2 (0.87) / (19) In other words, the year on Mercury is Earth s years, or about 88 Earth s days. Non-textbook problem #: The year on the planet in question is given by eq. (17) in terms of the semi-major axis (or the radius) of its orbit and the mass of its star. Comparing it to the year on Earth and taking the ratio, we have T P 2π T E a P GM star a P GM star (a P /a E ) M star /M Sun (2 au/1 au) 1.6. /1 / a E 2π GM Sun / a E GM Sun 8 (20) In other words, the year on the planet in question is about 1.6 Earth s years, or 596 Earth s days. Non-textbook problem #4: A satellite appears to hand stationary relative to the planetary surface if its orbital motion matches the planetary spin. This requires a circular orbit (so that ω const) in the planet s equatorial plane, and the satellite s period should be equal to the planet s sidereal day T p. 5

6 In terms of the orbital radius r and the planet s mass M p, the satellite s period is r T sat 2π, (21) GM p cf. eq. (17), so we need r 2π T p, GM p r GM p ( ) 2 Tp, 2π r GM p (T p /2π) 2. (22) The Earth has mass M E kg and sidereal day T s or 2 hours, 56 minutes, and 4 seconds. Therefore, the geostationary orbit of a satellite which appears to hang stationary has radius r(e) GM E (T E /2π) 2 ( N m 2 /kg 2 ) ( kg) (86164 s/2π) 2 (2) 42,200 km. More accurately, using G M E m /s 2, we get r 42,164 km. Note that this is the radius of the orbit, i.e. the distance between the satellite and the Earth s center. The altitude of the satellite above the Earth s surface is h r R E 5,786 km or about 22,240 miles. Mars has a smaller mass than Earth M M kg and a slightly longer The sidereal day during which the Earth spins through exactly 60 is slightly shorter than the mean solar day of exactly 24 hours, because during the mean solar day the Earth spins through to catch up with Earth s motion around the Sun. Over a year, the difference amounts to an extra day: there are mean solar days in 1 year but sidereal days. 6

7 sidereal day T M s. Hence, a synchronous satellite of Mars has orbit of radius r(m) GM E (T E /2π) 2 ( N m 2 /kg 2 ) ( kg) (88642 s/2π) 2 (24) 20,400 km. Non-textbook problem #5: To maintain a constant speed of the bucket, the man should pull on the rope with force T equal to the full bucket s weight mg 10.0 kg 9.8 m/s 2 98 N. This force acts on the bucket over a distance L equal to the well s depth D, and the direction of the force is the same as the direction of the bucket s displacement. Hence, the mechanical work is W TL mgd. (25) Given the force and the work, we can use this equation to find the well s depth as D W mg 6.00 kj 6000 J 98 N 61 m, (26) or about 200 feet. Non-textbook problem #6: The kinetic energy of a body depends on its mass and speed as K 1 2 mv2 (27) Therefore, two bodies of respective masses m 1 and m 2 and speeds v 1 and v 2 have same kinetic energies whenever 1 2 m 1 v m 2v 2 2 (28) Given the masses of the two balls in question and the speed of the bowling ball, we can sole 7

8 this equation for the speed of the ping-pong ball as v2 2 m 1 v m 1, 2 2 v 2 m1 v 1 m g.00 m/s 2.45 g 160 m/s, (29) or almost 60 miles per hour. I have seen some fast ping-pong games, but never this fast! Non-textbook problem #7: The bullet hits the tree with a kinetic energy K mv2 1 2 (0.009 kg)(250 m/s)2 280 J. (0) As the bullet penetrates the wood, there is a resisting force F opposing its motion. The work of this force over a distance x is W Fx (1) where the minus sign comes from the opposite directions of the force F and the bullet s displacement x. By the work energy theorem, the bullet s kinetic energy becomes K K 0 +W K 0 Fx, (2) and when this kinetic energy drops to zero, the bullet stops. Hence, given K J and the stopping distance x 0.08 m, we can find the resisting force F as F K 0 x 280 J 0.08 m 500 N 800 lb. () 8

9 Non-textbook problem #8: Let s start with free-body diagram of forces acting on the crate: y v N T x θ 20 f mg Thepictureontherightgivesusdirectionsofalltheforces. Fortwooftheforces, wearegiven the magnitudes: The pulling force is T 125 N, and the weight mg 12.0 kg 9.8 m/s N. The normal force s magnitude follows from the crate not moving in the y direction ( to the incline): N mgcosθ ma y 0 N mgcosθ 118 N cos N. (4) Finally, the kinetic friction force is given by f µ k N N 44.2 N. (5) And now that we know all the forces, we can calculate how much fork they do. (a d) The work of a force F is a scalar product of the force vector and the displacement vector r of the body on which the force acts, W( F) F r F x x + F y y. (6) The crate in question is moving through 5.00 m along the incline, hence in our coordinates 9

10 x 5.00 m and y 0. Consequently, the work of each force acting on the box is W( F) F x x + 0 F x 5.0 m. (7) Specifically, for the four forces in question, W( T) T x x T x 125 N 5.0 m 625 J, W( f) f x x f x 44.2 N 5.00 m 221 J, W( N) N x x 0 because N x 0, (8) W(m g) mg x x mgsinθ x 118 N sin m 201 J. (e) By the work kinetic energy theorem, the change of the body s kinetic energy is equal to the net work of all the forces acting on it. K W net i W( F i ). (9) For the crate in question, the net work follows from eqs. (8): W net W( T) + W( f) + W( N) + W(m g) +625 J 221 J J (40) +20 J. Consequently, the crate s kinetic energy increases by 20 J. (f) The initial kinetic energy of the crate was K mv kg (5.00 m/s)2 150 J. (41) Once the care has been pulled through 5.00 m, its kinetic energy increases by the net work 10

11 of all the forces acting on it, K K 0 + K K 0 + W net 150 J + 20 J 5 J. (42) Consequently, the speed of the crate now satisfies 1 2 mv2 K 5 J (4) and therefore v 2K m 2 5 J 12.0 kg 7.7 m/s. (44) Non-textbook problem #9: The uphill climb could be steep and short, or long and gentle, we do not know. In any case, horse s potential energy increases by U mgh, (45) and this has to come at the expense of the horse s mechanical work. Thus, to climb the hill, the horse has to perform work and given the horse s power P 1 hp, this takes time Numerically, in metric units W mgh, (46) t W P mgh P. mgh 900 kg 9.8 m/s 2 00 m J, P 746 W, t J 746 W 550 s 1 hour, (47) 11

12 or in English units, mgh 2000 lbf 1000 ft ft lbf (feet-pounds), P 550 ft lbf/s, t ft lbf 550 ft lbf/s 640 s 1 hour. (48) Note: there are small differences between numbers calculated in different units because 2000 pounds is not exactly 900 kg and 1000 feet is not exactly 00 m. But given the low precision of the problem s data, both answers should be rounded to 1 hour. Textbook problem 6.69: The net force resisting the car s uphill motion is F r f + mg sinθ (49) where f 650 N combines the air resistance and the rolling resistance of the tires. This resistive force generates negative power P r vf r, and to keep the car moving at constant speed, the engine has to compensate for it by delivering positive power P e +F r v (50) Given the desired speed of the car v 75 km/h 47 MPH 21 m/s and the maximal power the engine can deliver P max e on the car should be no more than 120 hp 90 kw, we find that the net resisting force F max r Pmax e v 90 kw 21 m/s 400 N (51) if the resisting force is larger than this, the car would have to slow down. Consequently, 12

13 in light of eq. (49), mgsinθ F max r f 650 N (52) and hence sinθ Fmax r f mg 650 N 0.1. (5) 1200 kg 9.8 m/s2 Thus, the steepest hill this car can climb at constant 75 km/h speed is θ max arcsin(0.1) 18. (54) 1