PHY 302 K. Solutions for problem set #9. Non-textbook problem #1: (a) Rotation frequency of 1 Hz means one revolution per second, or 60 revolutions per minute (RPM). The pre-lp vinyl disks rotated at 78 RPM or 78 60 the frequency is f 1.3 Hz. 1.3 revolutions per second, i.e. (b) The period T and the angular frequency ω of uniform rotation are related to the cyclic frequency f as T 1 f, ω 2πf 2π rad T. (S.1) For the disk in question T 1/(1.3 Hz) 0.77 s and ω 2π rad 1.3 Hz 8.2 rad/s. (c) The coin lies 10 cm from the axis of rotation, so it moves along a circle of radius r 10 cm. The coin s speed is v r ω 0.1 m 8.2 s 1 8.2 m/s. (S.2) Note: in formulae relating rotation to linear motion, we can treat angles in radians as dimensionless. Thus, ω 8.2 rad/s is the same as ω 8.2 s 1. (d) The centripetal acceleration of the coin is a c v2 r r ω 2 0.1 m (8.2 s 1 ) 2 6.7 m/s 2. (S.3) (e) We assume that the coin is pushed so slowly that we may neglect its acceleration due to changing radius r compared to the centripetal acceleration a c ; this allows us to approximate a a c. The force providing for this acceleration (and hence for keeping the coin moving in a circle) is the static friction force f between the coin and the disk. This force cannot be stronger than µ s N µ s mg, hence ma c µ s mg a c µ s g. (S.4) As the coin is pushed away from the axis of rotation, the radius r or its circular motion increases, while the angular velocity ω remains constant. Consequently, the linear velocity 1
v of the coin increases, and the centripetal acceleration also increases as a c ω 2 r. When this acceleration exceeds the limit (S.4), the static friction force can no longer keep it fixed relative to the disk, and the coin slips away; this happens when a c ω 2 r a max c µ s g. (S.5) Given the angular velocity ω 8.2 s 1 and the radius r 12 cm where the coin slips off the disk, we can solve eq. (S.5) for the static friction coefficient: µ s ω2 r g (8.2 s 1 ) 2 0.12 m 9.8 m/s 2 0.82. (S.6) Textbook problem 7.14: In a rotating space habitat, the inertial force F in m a c acts as artificial gravity. The direction of this force is opposite to the centripetal acceleration, i.e. away from the axis of rotation, so the outer wall of the habitat acts as a floor. The magnitude of the inertial force is F m a c m ω 2 r. If we want it to have the same value as the weight W mg on Earth, we need a c ω 2 r g. (S.7) For the radius r 5 miles 8 km, this requires ω g r 9.8 m/s 2 8000 m 0.035 s 1 (S.8) i.e. 0.035 rad/s. This angular velocity corresponds to rotation period T 2π/ω 180 s or 3 minutes. 2
Textbook problem 7.24: (a) The forces acting on the car are the gravity force m g, the normal force N, and the static friction force f. Together, they provide for the centripetal acceleration of the car a c, thus m g + N + f m a c. (S.9) The directions of the friction and normal forces depend on the banking on the road: f is parallel to the road s surface while N is perpendicular to it. Consequently, the magnitudes f and N also depend on the banking, so that eq. (S.9) holds true for a range of banking angles θ. We want to find θ for which f 0. This means that eq. (S.9) reduces to m g + N m a c, (S.10) or in components 0 + N sin θ F net x ma c, mg + N cos θ F net y 0. (S.11) Consequently, N sin θ N cos θ ma c mg sin θ cos θ a c g (S.12) and therefore tan θ a c g v2 rg. (S.13) (b) The road curve in question has radius r 50.0 m and is designed for cars moving at speed v 30.0 miles per hour or 13.4 m/s. Consequently, the centripetal acceleration is a c v 2 /r (13.4 m/s) 2 /50.0 m 3.60 m/s 2, and we need to bank the road at θ arctan a c g 3.60m/s2 9.8 m/s 2 20. (S.14) 3
Textbook problem 7.28: (a) At point A, the roller-coaster track has curvature radius r A 10 m, so the centripetal acceleration of the car is a c (A) v2 A r A (20 m/s)2 10 m 40 m/s 2. (S.15) The direction of this acceleration is up note the center of the circle above point A so a y +a c and therefore N mg +ma c. (S.16) Solving this equation for the normal force N we find N mg + ma 500 kg 40 m/s 2 + 500 kg 9.8 m/s 2 24.9 kn 25 kn. (S.17) (b) At point B the centripetal acceleration a c (B) is pointing down rather then up, hence a y a c and N mg ma c N m(g a c ). (S.18) The normal force can push the car up but cannot pull it down, so we must have N 0, or else the car will fly off the track. Hence, the centripetal acceleration at point B should not exceed g, a c (B) g v2 B r B g v 2 B r B g. (S.19) Which means that the speed at point B should not exceed v max (B) r B g 15 m 9.8 m/s 2 12 m/s, (S.20) or about 27 MPH. 4
Textbook problem 7.52: Looking at the figure P7.52 we see that the ball moves in a circle of radius r L sin θ where L 1.5 m is the length of the string on which it is suspended. acceleration of the ball is a c v2 r v 2 L sin θ The centripetal (S.21) in the horizontal direction. This acceleration is provided by the vector sum of two forces: the string tension T and the weight mg, hence in components 0 + T sin θ ma x ma c, mg + T cos θ ma y 0. (S.22) Combining the two equations and proceeding as in problem 7.24, we have T sin θ T cos θ ma c mg sin θ cos θ a c g. (S.23) Combining this formula with eq. (S.21), we arrive at and therefore sin θ cos θ v2 /(L sin θ) g sin 2 θ cos θ v2 Lg. (S.24) (S.25) (a) Given the angle θ and the string length L, we find the ball s speed v by solving eq. (S.25): v 2 Lg sin2 θ cos θ 1.5 m 9.8 m/s 2 sin2 30 14.7 m 2 /s 2 1/4 3/2 4.24 m 2 /s 2, cos 30 (S.26) v 2.06 m/s. 5
(b) Now we know the ball s speed v and need to find the angle θ. Evaluating the right hand side of eq. (S.25) we have and therefore v 2 Lg (4.0 m/s) 2 1.088 (S.27) 1.5 m 9.8 m/s2 sin 2 θ cos θ 1.088. (S.28) To solve this trigonometric equation, we use sin 2 θ 1 cos 2 θ. This turns eq.(s.28) into 1 cos 2 θ cos θ 1.088 (S.29) and hence cos 2 θ + 1.088 cos θ 1 0 (S.30) a quadratic equation for the cos θ. Solving this equation we get two roots: cos θ 1.088 ± 1.088 2 + 4 2 +0.594 and 1.683. (S.31) Since we cannot have cos θ < 1, we reject the second root as unphysical. Thus, cos θ +0.594 and therefore θ arccos(+0.594) 53.5. (c) According to the second equation (S.22), the string tension T follows from the angle θ as T cos θ mg T mg cos θ. (S.32) Thus, to assure that T does not exceed the maximum tension T max 9.8 N and the string does not tear itself apart, we must make sure that cos θ mg T max and the angle θ does not exceed θ max 0.5 kg 9.8 m/s2 9.8 N 0.5 (S.33) arccos mg T max arccos(0.5) 60. (S.34) The ball s speed v is related to the string angle θ via eq. (S.25), and it s clear from that equation that the speed increases with the angle. Thus, the maximum speed corresponds to 6
the maximum angle: sin 2 θ max cos θ max v2 max Lg (S.35) and therefore v 2 max v max Lg sin2 θ max cos θ max 1.5 m 19.8 m/s 2 sin2 60 14.7 m 2 /s 2 3/4 1/2 22 m 2 /s 2 4.7 m/s. cos 60 (S.36) Non-textbook problem #2: According to Newton s Law of Gravity, your weight on Earth is W E m g E G M E m R 2 E (S.37) 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. (S.38) Note that G and m in eqs. (S.37) and (S.38) are exactly the same G is universal, and your mass m does not depend on where you happen to be. Consequently, these variables 7
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 ) 2 0.0226 (0.468) 2 0.103. / G ME m R 2 E (S.39) In other words, your weight on Titan would be 0.103 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. Non-textbook problem #3: To escape from the gravitational potential well of a planet of mass M and radius R, a body needs to increase its potential energy by U U( ) U(surface) GMm R m g surface R. (S.40) In a typical space flight, a rocket gives a spacecraft high speed v 0 before its lifts very high above the surface. Then the rocket runs out of fuel and shuts off, and the net mechanical energy of the spacecraft stays constant until it reaches its destination, K + U const K U. (S.41) Therefore, to escape the planet, spacecraft s initial kinetic energy must be greater than the potential energy rise (S.40), K 0 1 2 mv2 0 > U GMm R v 0 > 2GM R. (S.42) In other words, spacecraft s initial speed (when the rocket shots off) must be greater than 8
the escape speed v e 2GM R. (S.43) To find the escape speed from Titan, we need to know its mass M T and radius R T. The previous problem gives us these data in units of Earth s mass M E and radius R E, which gives us two ways to evaluate eq. (S.43). We can look up (on the back inside cover of the textbook) the values of M E 5.98 10 24 kg, R E 6.38 10 6 m, and G 6.67 10 11 N m 2 /s 2 /kg 2, then evaluate Titan s mass and radius in metric units as M T 0.0226 M E 1.35 10 23 kg and R T 0.468 R E 2.99 10 6 m, and then directly evaluate eq. (S.43) for Titan: v e (T ) 2GM T R T 2 6.67 10 11 N m 2 /kg 2 1.35 10 23 kg 2.99 10 6 m 2460 m/s (S.44) Alternatively, we may calculate the ratio of escape speeds from Titan and from Earth as v e (T ) v e (E) 2GM T R T 2GM T R T / 2GM E R E / 2GME R E 0.0226 0.468 0.220. M T /M E R T /R E (S.45) This way, we don t need to know the values of G, M E and R E, as long as we know the escape speed from Earth which is indeed given in this problem as 11.2 km/s. Consequently, the escape speed from Titan is v e (T ) 0.220 v e (E) 0.220 11.2 km/s 2.46 km/s. (S.46) 9
Non-textbook problem #4: (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 http://hyperphysics.phy-astr.gsu.edu/hbase/kepler.html, see the Kepler Laws link in the supplementary notes section of the homework page. The top picture at 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, (S.47) and at the aphelion, the distance is R a a + e a (1 + e) a. (S.48) Mercury s orbit has semi-major axis a 0.387 au and eccentricity e 0.20. Hence, at the perihelion, Mercury is only R p (1 0.20) 0.387 au 0.31 au 46 10 6 km away from the Sun, while at the aphelion the distance increases to R a (1 + 0.20) 0.387 au 0.46 au 69 10 6 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 3. GM star (S.49) Comparing two planets orbiting the same star such as Mercury and Earth and taking 10
the ratio of their orbital periods, we have T M T E 2π a 3 M GM Sun a 3 M GM Sun a 3 M a 3 E / a 3 E 2π GM Sun / a 3 E GM Sun (S.50) ( am a E ) 3/2. Note the G and the Sun s mass cancel out from this ratio. 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 0.387 au, which means that the ratio a M /a E is 0.387. Consequently, the ratio of orbital periods of Mercury and Earth is T M T E ( am a E ) 3/2 (0.387) 3/2 0.241. (S.51) In other words, the year on Mercury is 0.241 Earth s years, or about 88 Earth s days. 11
Non-textbook problem #5: The year on the planet in question is given by eq. (S.49) 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 T E 2π a 3 P GM star a 3 P GM star (a P /a E ) 3 M star /M Sun (2 au/1 au) 3 1.63. 3/1 / a 3 E 2π GM Sun / a 3 E GM Sun 8 3 (S.52) In other words, the year on the planet in question is about 1.63 Earth s years, or 596 Earth s days. Non-textbook problem #6: 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. In terms of the orbital radius r and the planet s mass M p, the satellite s period is T sat r 2π 3, GM p (S.53) cf. eq. (S.49), so we need r 2π 3 T p, GM p r 3 GM p ( ) 2 Tp, 2π r 3 GM p (T p /2π) 2. (S.54) 12
The Earth has mass M E 5.98 10 24 kg and sidereal day T 86164 s or 23 hours, 56 minutes, and 4 seconds. Therefore, the geostationary orbit of a satellite which appears to hang stationary has radius r(e) 3 GM E (T E /2π) 2 3 (6.67 10 11 N m 2 /kg 2 ) (5.98 10 24 kg) (86164 s/2π) 2 (S.55) 42, 200 km. More accurately, using G M E 3.9757 10 14 m 3 /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 35, 786 km or about 22,240 miles. Mars has a smaller mass than Earth M M 6.43 10 23 kg and a slightly longer sidereal day T M 88642 s. Hence, a synchronous satellite of Mars has orbit of radius r(m) 3 GM E (T E /2π) 2 3 (6.67 10 11 N m 2 /kg 2 ) (6.43 10 23 kg) (88642 s/2π) 2 (S.56) 20, 400 km. Textbook problem 7.36: (a) The speed of a satellite on a circular orbit of radius r is determined by the condition that the centripetal acceleration of the satellite is equal to the gravitational free-fall acceleration g at that distance from Earth. Thus, v 2 r GM R r 2 v GME r. (S.57) A circular orbit of altitude h R E above Earth s surface has radius r h + R E 2R E. The sidereal day during which the Earth spins through exactly 360 is slightly shorter than the mean solar day of exactly 24 hours, because during the mean solar day the Earth spins through 360 + 0.985 to catch up with Earth s motion around the Sun. Over a year, the difference amounts to an extra day: there are 365.25 mean solar days in 1 year but 365.25 + 1 sidereal days. 13
Hence, a satellite on this orbit has speed v GM R 2R E (6.67 10 11 N m 2 /kg 2 ) (5.98 10 24 kg) 2 6.38 10 6 m 5580 m/s. (S.58) (b) The orbital period of a satellite in a circular orbit is T 2πr v 2πr GME /r 2π r 3, GM e (S.59) cf. eq. (S.53). For the satellite in question r 2R E, hence T 2π 2R E v 2π 2 6.38 106 m 5580 m/s 14, 400 s or about 4 hours. (c) The gravitational force on the satellite is F GM Em r 2 GM Em 4R 2 E,. (S.60) Comparing this force to the weight this satellite would have on the surface of the Earth F sur mg sur GM Em R 2 E, (S.61) we see that the gravitational force on the satellite in orbit is one quarter of the weight it would have on the surface: F F sur 4 m g sur 4 600 kg 9.8 m/s2 4 1470 N. (S.62) 14