Chap 5. Circular Motion: Gravitation

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1 Chap 5. Cicula Motion: Gavitation Sec Unifom Cicula Motion A body moves in unifom cicula motion, if the magnitude of the velocity vecto is constant and the diection changes at evey point and is diected pependicula to the adius vecto. Centipetal acceleation: The acceleation in a unifom cicula motion is along the adius vecto and is diected towads the cente of the cicle. The magnitude of the centipetal acceleation is given as a = v 2 / The acceleation vecto a is pependicula to the velocity vecto v at evey point The time taken to complete one evolution is called as the peiod (T ). The numbe of evolutions pe second is called as the fequency (f ) of evolution, such that f = 1 T

2 In one evolution the object completes an angle of 360 o 2π adians. Theeby, the object, coves a distance of 2π in time (T ) v = 2π T Sec Dynamics of Cicula motion Any motion in a cuved path epesents acceleated motion, and equies a foce diected towad the cente of cuvatue of the path. This foce is called the centipetal foce which means cente seeking foce. The foce has the magnitude: 2 F = m a = m v 2 / The centifugal and coiolis foces ae effective foces which ae invoked to explain the behavio of objects fom a fame of efeence which is otating (Non Inetial Fame). This explains why an object tied to a sting that is made to evolve, flies off tangentially instead of adially outwad when eleased. Centifugal foce is a consequence of Newton s thid law of motion. Action-Reaction foce. Gavity supplies the necessay centipetal foce to hold a satellite in obit about the eath. The cicula obit is a special case since obits ae geneally ellipses, o hypebolas in the case of objects which ae meely deflected by the planet s gavity but not captued.

3 Poblem A: Calculate the centipetal acceleation of the Eath and the centipetal foce exeted on it when in its obit aound the Sun. Assume Eath evolves in a cicula obit of adius m. Soln A: m E = Kg R = m P eiod of evolution T = s 3 The velocity of the Eath aound the Sun V E = 2πR T = The centipetal acceleation is given as = m/s V E 2 a E = R = = m/s The centipetal foce is given as F c = m E a E = = N Poblem B: What is the appaent weight of a peson of mass 50 Kg on the Feis wheel at the top of the cicle. The adius of the wheel is 7.4 m. If the velocity is 8.5m/s Soln B: Appaent weight at bottom of the Feis wheel v F N mg = m 2 F N = mg + m v2 Appaent weight at the top of the Feis wheel = = 978N mg F N = m v 2 F N = mg m v2 = = 2N

4 Poblem 7 Text: A ball at the end of a sting is evolved at a unifom ate in a veticle cicle of adius 72 cm. If its speed is 4.00 m/s and mass is Kg calculate the tension in the sting when the ball is at the top and the bottom of it path. Soln 7 Text: At the top of its path mg + F T = m v2 At the bottom of its path F T mg = m v2 F T = m v2 F T = m v2 ( ) 4 2 mg = ( ) mg = = 3.73N = 9.61N 4 Sec Highway Cuves, Banked and Unbanked Fo a cuved oad, the centipetal foce is popotional to the squae of the velocity, implying that a doubling of speed will equie fou times the centipetal foce to keep the motion in a cicle. If the centipetal foce must be povided by fiction alone on a cuve, an incease in speed could lead to an unexpected skid if fiction is insufficient. Once the ca skids the static fiction becomes kinetic fiction and the ca skids moe. Thus in such cases we sometimes need to bank (give an angle) oads in ode to pevent skidding.

5 Theshold speed on a banked highway to pevent sliding up the incline. 5

6 Poblem C: If the adius of a tun is 100 m and the oad is pefectly banked fo a ca tavelling at 20 m/s. Find the banking angle. Soln C: Fo a pefectly banked cuve thee is no need of fiction hence µ s = 0 6 tan θ = v2 g = = 22.2 Find µ s fo a ca so that it does not to skid when its velocity is 30m/s g(sinθ + µ s cosθ) v = cosθ µ s sinθ This equation can also be witten as v 2 g = (tanθ + µ s) 1 µ s tanθ We have used the identity tan(θ + α) = (tanθ + tanα) 1 tanα tanθ v 2 g = tan(θ + α) µ s = tanα (θ + α) = tan 1 ( v2 g ) = 30 2 tan 1 ( ) = Thus we get α = and µ s = tanα = 0.371

7 Sec Non Unifom Cicula motion 7 In an non unifom cicula motion the foce is not diected towads the cente of the cicle but at an angle. Thus, thee exists a tangential component of the foce that changes the magnitude velocity of the object and hence contibutes to tangential acceleation The total foce in this case is the vecto sum of the centipetal foce F c and the tangential foce F t. The same applies to acceleation: F = F t + F c Sec Centifugation a = a t + a c = v t ˆt + v2 ˆ In centifugation we use the concept of cicula motion to sediment mateials. The idea is to do the sedimantation at a faste ate by inceasing the otation speeds that povides effective gavity much lage than the nomal gavity. See example as shown in text Pg 117.

8 Sec. 5.6,7 - Newton s Laws of Gavitation 8 Evey paticle in the univese attacts evey othe paticle with a foce that is popotional to the poduct of thei masses and invesely popotional to the squae of the distance between them. The foce acts along the line joining the two paticles. The foce between mass m 1 and mass m 2 sepaated by distance is given as: F = ˆGm 1m 2 2 = m 1 a Hee G is the univesal gavitation constant G = Nm 2 /Kg 2 This foce satisfies Newton s thid law such that F 12 = F 21 This foce is the weakest foce hence pominant in heavie masses. As distance inceases this foce becomes less effective. If m 1 = M E and m 2 = m you then the acceleation you expeience due to Eath s gavitational pull on the suface of the Eath is F = ˆGM Em you R 2 E = m you a E a E = ˆGM E R 2 E = 9.8m/s 2 If m 1 = M E and m 2 = m you then the acceleation you expeience due to Eath s gavitational pull at a height of h = 6380 Km will be a h = ˆGM E (R E + h ) = m/s2 The acceleation you expeience on the Moon: a M = GM M R 2 M ( ˆ ) = 1.64m/s 2 g/9.8 = g/6 On the Moon you will weigh 1/6 th of that on the Eath.

9 Sec Satellite 9 A satellite is usually boosted at high speed and it is this speed that keeps it in obit. We call the velocity to maintain stable cicula obit as the citical velocity v c. If the velocity is less than v c the satellite will fall back on Eath. If the velocity is exceeds v c then the satellite can escape the Eath s gavitational field. The gavitational foce between the satellite and the Eath povides the necessay centipetal foce to sustain cicula obit. Poblem D: Calculate the speed of a satellite moving in a stable cicula obit about the Eath at a height of 3600 Km. Soln D: GM E m s = m sv 2 R 2 R Thus GME v = R = Gavitational Foce = Centipetal foce ( ) 10 3 = m/s A geosynchonous satellite is a communication satellite and it stays at the same location above the Eath s equato. The peiod of evolution of the satellite has to be equal to Eath s otational peiod (T = 1day) in ode fo it be at the same location.

10 Poblem E: Show that if a satellite obits nea the suface of a planet with Peiod T a) the density of the planet is ρ = m/v = 3π/GT 2. b) Estimate the density of the Eath, given that a satellite nea the suface obits with a peiod of about 85min Soln E: The velocity v of a satellite with peiod T in an obit of adius R is given as v = 2πR T Fo a stable cicula obit we equie GM p m s = m sv 2 GMP v = R 2 R R Thus GM P R = (2πR T )2 M P = ( 2πR T )2R G The density is given as ρ = M p /V = ( 2πR T )2R G 3 4πR = 3 Density of Eath is 3π GT 2 10 ρ E = 3π = Kg/m 3

11 Sec Keple s Laws 11 Keple s fist Law The path of each planet about the Sun is an ellipse with the Sun at one focus. Keple s second Law Each planet moves so that an imaginay line dawn fom the Sun to the planet sweeps out equal aeas in equal intevals of times. Keple s thid Law The atio of the squaes of the peiods T of any two planets aound the Sun is equal to the cube of thei mean distance s fom the Sun. Poof: Special case cicula obit. GM S m R 2 = mv2 R Gavitational Foce = Centipetal foce Substituting fo v we get v = 2πR T GM S m = m4π2 R 2 R 2 RT 2 Thus we can wite T 2 = 4π2 R 3 GM S T 2 R 3 = 4π2 GM S = constant T1 2 R1 3 = T 2 2 R 3 2

12 Poblem F: Asteoid Icaus obits aound the Sun with a peiod of 410 days. What is its mean distance fom the Sun. Soln F: 12 R 3 = GM ST 2 4π 2 R 3 = π 2 R = m Poblem G: Detemine mass of the Eath fom the peiod and distance of the Moon Soln G: The Peiod of the Moon T = 27.3days The Eath moon Distance R EM = m R 3 = GM ET 2 M 4π 2 E = 4π2 R 3 GT 2 M E = 4π 2 ( ) ( ) 2 M E = Kg

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