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Chapter 5: Circular Motion Rotating Objects: Wheels, moon, earth, CDs, DVDs etc. Rigid bodies. Description of circular motion. Angular Position, Angular Displacement s r Angle (in radians) = arc length/radius = s/r. Angular displacement D = 2-1 How far from initial angular position it has rotated. D = positive going CCW, negative going CW. Magnitude D = Ds/r. If Ds = r, then D = 1 radian Angular Displacement 360o 2 rads 180o rads 90o /2 rads In 1 complete turn, Ds = circumference of the circle = 2r. In 1 turn, D = Ds/r = (2r)/r = 2 rads. 360o = 2 rads, 180o = rads, 90o = /2 rads. Angular Velocity (w) w Average Angular velocity w = D/Dt How fast it is rotating It is a vector quantity. Direction: Up for CCW rotation, down for CW rotation. [ Right Hand Rule]. Units = radians/second. Examples 1.Change 1 rpm to rad/sec. 2. What is the angular speed of the earth’s rotation? 3. What is the angular speed of the minute hand of an analogue clock? Circular to Linear Linear distance = Arc length Ds = r D ( in radians) Speed |v| = Ds/Dt = r D/Dt = rw Direction of v is tangent to circle Example: 1. A CD spins with angular speed 20 radians/second. What is the linear speed 6 cm from the center of the CD? V = r w = 0.06 x 20 = 1.2 m/s Example: 2. If you were seated at the equator, what would be your linear speed due to the rotation of the earth? (radius of the earth = 6.371 x 106 m. v = rw w = D/Dt = 2/(24x3600) = 7.27 x 10-5 rad/s V = 6.37 x 106 x 7.27 x 10-5 = 463 m/s Circular Motion Act b a c v Answer: b A ball is going around in a circle attached to a string. If the string breaks at the instant shown, which path will the ball follow? Frequency (f) Frequency (f) = number of turns a rotating object makes in one second. f = # of turns/time taken to make the turns. Unit: Cycles per second or hertz (Hz) 1 cycle/second = 1 Hz Eg 1. The frequency of the second hand of an analogue clock = 1 turn/60s = 1/60 Hz. The frequency of the moon’s rotation around the earth = 1 turn/28 days = 4.13 x 10-7 Hz. Period (T) Period (T) = Time taken to make one cycle. T = Time to make one turn. Unit: seconds Since f = # of cycles per second, Period and frequency are related by the expression f = 1/T or T = 1/f Eg 1. The period of the second hand of an analogue clock = 60 seconds. The period of the moon’s rotation around the earth = 28 days = 2.42 x 106 s. 1. An angular displacement of one radian is when A. B. C. D. Arc length > radius Arc length < radius Arch length = radius Circumference = 2r 1 2 3 4 5 6 7 8 9 10 11 12 13 14 21 22 23 24 25 26 27 28 29 30 31 32 41 42 43 44 45 46 47 48 49 50 51 0%33 34 52 54 61 62 63 64 65 A. 53 15 16 17 18 19 0%35 36 37 0% 38 39 56 57 58 59 55 B. C. 20 0%40 60 D. 2. In moving through 30 minutes, the minute hand of an analogue clock has gone through A. B. C. D. E. 1 2 21 180 rads 0.5 rads 1.57 rads 3.14 rads None of the above 3 4 5 6 7 8 9 10 11 12 22 23 24 25 26 27 28 29 30 31 41 42 43 44 45 46 47 48 49 50 51 61 62 63 64 65 14 15 16 17 18 19 20 32 0% 33 34 0% 35 36 0% 37 38 0% 39 40 0% 52 54 55 56 57 58 59 60 A. 13 53 B. C. D. E. 3. A small ball swung in a circular path takes 12 seconds to make 20 rotations. What is the frequency of its motion? A. B. C. D. 0.60 Hz 1.67 Hz 240 Hz 12.0 Hz 1 2 3 4 5 6 7 8 9 10 11 12 13 14 21 22 23 24 25 26 27 28 29 30 31 32 41 42 43 44 45 46 47 48 49 50 51 0%33 34 52 54 61 62 63 64 65 A. 53 15 16 17 18 19 0%35 36 37 0% 38 39 56 57 58 59 55 B. C. 20 0%40 60 D. 4. A CD has a diameter of 12 cm. if the CD is rotating at a constant frequency of 4 Hz., then the period of its rotation is A. B. C. D. E. 0.250 s 0.500 s 0.750 s 1.00 s 1.25 s 1 2 3 4 5 6 7 8 9 10 11 12 21 22 23 24 25 26 27 28 29 30 31 41 42 43 44 45 46 47 48 49 50 51 61 62 63 64 65 14 15 16 17 18 19 20 32 0% 33 34 0% 35 36 0% 37 38 0% 39 40 0% 52 54 55 56 57 58 59 60 A. 13 53 B. C. D. E. 5. The moon rotates around the earth once in 28 days. What is its average angular velocity? A. B. C. D. 0.224 rad/s 9.34x10-3 rad/s 2.60x10-6 rad/s 1.49x10-4 rad/s 1 2 3 4 5 6 7 8 9 10 11 12 13 14 21 22 23 24 25 26 27 28 29 30 31 32 0%33 34 41 42 43 44 45 46 47 48 49 50 51 52 61 62 63 64 65 A. 53 54 15 0% 35 55 B. 16 17 18 19 36 0% 37 38 39 56 57 59 C. 58 20 0%40 60 D. Relation between w, v, f and T r v In one complete rotation: Distance moved = 2r. Time taken = period, T Speed v = dist/time = 2r/T ie v = 2r/T = 2rf Since v = rw, we have w = 2/T = 2f Uniform Circular Motion • Uniform Circular Motion is motion of an object in a circular path with constant (uniform) speed. • Its linear speed is constant. Its angular velocity (w) is constant. • Direction always changing. • Hence linear velocity v is not constant. • The instantaneous direction of v is tangential to the circular path. • Since velocity v is not constant, an object in uniform circular motion must have an acceleration. Uniform Circular Motion v r w (out of page) v = rw Acceleration in Uniform Circular Motion The acceleration is due to change in direction of the velocity, not its magnitude (since magnitude of the velocity is constant). Dv v2 R DR v1 aave= Dv / Dt -v1 Direction of a = v2 direction of Dv Dv = v2 – v1 As Dt 0, v1 and v2 will be almost parallel to each other with Dv being perpendicular to them. Dv will point towards center of circle. Therefore, acceleration a is directed towards the center of the circle. Uniform Circular Motion (circular motion with constant speed) • Instantaneous velocity is tangent to circle. aC v • Instantaneous acceleration is radially inward. Hence it is called radial or centripetal acceleration (ar). • The magnitude of ar = v2/r • There must be a net force to provide the acceleration. ar = v2/r = rw2 = (2r/T)2/r = 42r/T2 = 42rf2 Net Force in Uniform Circular Motion FC Conclusion: v There must be a net force to provide the acceleration. This net force must also be directed radially towards the center. Hence it is called radial or centripetal force (Fr). Fnet = Fr = mar = mv2/r = mrw2 Any object moving in a circular path must have a net force exerted on it directed towards the center of the circle. The magnitude of this force = mv2/r = mrw2 A spider sits on a turntable that is rotating at a constant 33 rpm. The radial acceleration of the spider (ar) is (A)Greater the closer the spider is to the central axis (B)Greater the farther the spider is from the central axis (C)Zero (D)Non zero and independent of the spider’s location on the turn table Two kids of equal masses are seated at positions A and B on a merry-go-round rotating uniformly. Compare these quantities for the two kids: 1. Angular displacement (D) 2. Angular velocity (w) 3. Linear velocity (v) A 4. Period (T) 5. Frequency (f) 6. Centripetal acceleration (aC) (A) A > B (B) A < B (C) A = B B Examples of Circular Motion 1. A ball, mass m, swung in a vertical circle by a string of length L A L Fnet = mar mg T -TA-mg = -mv2/L TA = mv2/L - mg If T = 0, the ball will not move in the circular path. For circular motion, minimum swing speed is obtained by setting T = 0 0 = mv2/L – mg or v = (gL) At the bottom of the circular path: A Fnet = mar T B mg TB - mg = mv2/L TB = mv2/L + mg In uniform circular motion, TB > TA 2. Water in a bucket swung in a vertical circle Fnet = mar r mg N -N-mg = -mv2/r N = mv2/r - mg If N = 0, the water will pour out. For water not to pour out, minimum speed is obtained by setting N = 0 0 = mv2/r – mg or v = (gr) What is the minimum speed you must have at the top of a 20 meter roller coaster loop, to keep the wheels on the track. A 20 m roller coaster loop; r = 10 m v = (gr) = (9.8 x 10) = (98) = 9.9 m/s 3. Small ball swung in a horizontal circle Fnet = mar Small mass,T >> mg r mg T = mv2/r T 4. Car driven on valley/hill N v r N v mg r mg Fnet = mar Valley: NV – mg = mv2/r and NV = mg + mv2/r Hill: NH – mg = -mv2/r and NH = mg – mv2/r On flat road, NF = mg 5. Car rounding a curve on a flat road Fnet = mar r Fy: N – W = 0 Fx: fs = mv2/r N W If v is increased, fs will increase up till fs max (= sN) fs mv2/r sN (= smg) v (srg) OR s v2/rg 6. Car rounding a curve on a banked road To reduce chances of skidding, roads are banked. Centripetal force will then come from a component of the normal force, reducing total reliance on friction. At a particular speed, all of the centripetal force Fc will be contributed by the normal force N. Nx y N Ny Nx = N sin N x mg Ny = N cos With no dependence on friction: Along y-axis: Ny = Ncos - mg = 0 ie, Ncos = mg Along x-axis: Nsin = mv2/r OR: tan = v2/rg, ie Banking angle = tan-1(v2/rg) Speed v (rg tan ) Example 1 A 1,000 kg car rounds a curve on a flat road of radius 50 m. What should be the speed limit if the coefficient of static friction is 0.3? Example 2 A car rounds a curve on a flat road of radius 25 m at a speed of 13 m/s. What should be the banking angle of the road for the car to safely negotiate the turn if the road is icy? 1. A 4.00 kg mass is moving in a circular path of radius 2.50 m with a constant angular velocity of 5.00 rad/sec. The centripetal force 59% on the mass is A. 250 N B. 185 N C. 153 N 21% D. 107 N 12% E. 94.0 N 6% 3% A. B. C. D. E. 2. In moving through 30 minutes, the minute hand of an analogue clock has gone through 79% A. B. C. D. E. 180 rads 0.5 rads 1.57 rads 3.14 rads None of the above 11% 5% A. B. 3% C. 3% D. E. 3. A jet plane flying 6.0 x 102 m/s experiences an acceleration of 4g when pulling out of a dive. What is the radius of curvature of the 59% loop in which the plane is flying? A. B. C. D. E. 6.4 x101 m 1.2 x 103 m 7.1 x 103 m 9.2 x 103 m None of the above 14% 16% 8% 3% A. B. C. D. E. 4. You are driving through a valley whose bottom has a circular shape. If your weight is W, what is the magnitude of the normal force N exerted on you by the car seat as you drive 63% past the bottom of the hill? A. W > N B. W < N C. W = N 23% D. None of the above 15% 0% A. B. C. D. 5. A 35-kg child is seated 3.0 m from the center of a merry-go-round rotating with an angular speed (w) of 2.4 rad/s. What is the magnitude of the centripetal force Fc the child89%experiences? A. B. C. D. E. 17.3 N 604.8 N 343 N 67.2 N 1.92 N 8% 3% 0% A. B. C. 0% D. E. 6. A 1,200 kg car rounds a curve on a flat road of radius 30 m. What should be the speed limit if the coefficient of static friction is 0.45? 85% A. B. C. D. E. 294 m/s 17.1 m/s 132 m/s 11.5 m/s None of these 10% 5% 0% A. 0% B. C. D. E. Circular Orbits Satellites orbit the earth in circular path due to gravitational force. Its speed must be just right. Too low – falls to the earth. Too high – escapes into space m v r Fr = GmM/r2 = mv2/r v2 = GM/r = GM/(rE + h) M For Geosynchronous satellites, V = 2r/T, where T = 24 hours Example 3 The Hubble telescope moves in a circular orbit 613 km above the earth’s surface. The radius of the earth is 6.37 x 103 km and the mass of the earth is 5.98 x 1024 kg. (a) What is the speed of the satellite in this orbit? (b) Is the telescope geosynchronous? G = 6.67 x 10-11 N.m2/kg2 Non-Uniform Circular Motion • Uniform circular motion – circular motion in which speed stays constant and direction changes. • Centripetal acceleration – changes only direction of velocity, not its magnitude. • ac – points along radius towards the center. • Non-uniform circular motion – when both magnitude and direction of the velocity changes. • The acceleration is not along the radius. Non-Uniform Circular Motion Non-uniform circular motion – when both magnitude and direction of the velocity changes. The acceleration is not along the radius. This acceleration (a) can be resolved into tangential (at) and centripetal (aC) components a at ac a = ar + at ar = changes direction of v. at = changes magnitude of v Angular Acceleration Angular acceleration is the change in angular velocity w divided by the change in time. = Dw/Dt = (w - wo)/Dt Units: rad/s2 Note: tangential acceleration at = r Example If the speed of a roller coaster car is 15 m/s at the top of a 20 m loop, and 25 m/s at the bottom. What is the car’s average angular acceleration if it takes 2 seconds to get from the top to the bottom. = Dw/Dt = (w - wo)/Dt = vo/r = 15/10 = 1.5 rad/s w = v/r = 25/10 = 2.5 rad/s = Dw/Dt = (2.5 – 1.5)/2 = 0.5 rad/s2 wo Kinematics for Circular Motion with constant Linear Variables Angular Variables When a = constant When v = vo + at w = wo + t v2 = vo2 + 2a (x – xo) w2 = wo2 + 2 ( - o) x = x0 + vot + ½ at2 = o + w o t + ½ t2 = constant CD Player Example The CD in your disk player spins at about 20 radians/second. If it accelerates uniformly from rest with angular acceleration of 15 rad/s2, how many revolutions does the disk make before it is at the proper speed? w2 = wo2 + 2 ( - o) 1 Revolution = 2 radians Example Example A small ball of mass 200 g is rotated in a horizontal circle of radius 0.5 m. If the ball makes 10 revolutions in 5 seconds, what is its centripetal acceleration? Example The moon orbits the earth in a nearly circular path once every 27.3 days. It the distance from the moon to the earth is 384,000 km, what is the moon’s acceleration? Example Calculate the velocity of a satellite moving in a stable circular orbit about the Earth at a height of 3600 km. Example A hypothetical planet has a radius 2.5 times that of earth but has the same mass. What is the acceleration due to gravity near its surface? Summary of Concepts Circular Motion = angular position radians w = angular velocity radians/second = angular acceleration radians/second2 Linear to Circular conversions s=r Uniform Circular Motion Speed is constant Direction is changing Acceleration toward center a = v2 / r Newton’s Second Law F = ma Uniform Circular Acceleration Kinematics Similar to linear!