Chapter 5: Circular Motion

Transcription

1 Page 1 Chapter 5: Circular Motion Rotating Objects: Wheels, moon, earth, CDs, DVDs etc. Rigid bodies. Description of circular motion. Angular Position, Angular Displacement θ r s Angle (in radians) θ = arc length/radius = s/r. Angular displacement θ = θ 2 -θ 1 How far from initial angular position it has rotated. θ = positive going CCW, negative going CW. Magnitude θ = s/r. If s = r, then θ = 1 radian

2 Page 2 Angular Displacement 360 o 2π rads 180 o π rads 90 o π/2 rads In 1 complete turn, s = circumference of the circle = 2πr. In 1 turn, θ = s/r = (2πr)/r = 2π rads. 360 o = 2π rads, 180 o = π rads, 90 o = π/2 rads. Angular Velocity (ω) ω Average Angular velocity ω = θ/ t 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.

3 Page 3 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 s = r θ (θ in radians) Speed v = s/ t = r θ/ t = rω Direction of v is tangent to circle

4 Page 4 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 ω = 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 = x 10 6 m. v = r ω ω = θ/ t = 2π/(24x3600) = 7.27 x 10-5 rad/s V = 6.37 x 10 6 x 7.27 x 10-5 = 463 m/s

5 Page 5 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.

6 Page 6 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 10 6 s. Relation between ω, v, f and T r v In one complete rotation: Distance moved = 2πr. Time taken = period, T Speed v = dist/time = 2πr/T ie v = 2πr/T = 2πrf Since v = rω, we have ω = 2π/T = 2πf

7 Page 7 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 (ω) is constant. Direction constantly 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 ω (out of page) r v = rω

8 Page 8 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). v 2 v a ave = v / t R R v 1 v 2 -v 1 Direction of a = direction of v v = v 2 v 1 As t 0, v 1 and v 2 will be almost parallel to each other with v being perpendicular to them. v 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. a C v Instantaneous acceleration is radially inward. Hence it is called radial or centripetal acceleration (a C ). The magnitude of a C = v 2 /r There mustbe a net force to provide the acceleration. a C = v 2 /r = rω 2 = (2πr/T) 2 /r = 4π 2 r/t 2 = 4π 2 rf 2

9 Page 9 Net Force in Uniform Circular Motion F C 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 (F C ). F net = F C = ma C = mv 2 /r = mrω 2 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 = mv 2 /r = mrω 2 A spider sits on a turntable that is rotating at a constant 33 rpm. The centripetal acceleration of the spider (a C ) 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

10 Page 10 Examples of Circular Motion 1. A ball, mass m, swung in a vertical circle by a string of length L L A mg T F net = ma C -T A -mg = -mv 2 /L T A = mv 2 /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 = mv 2 /L mg or v = (gl) At the bottom of the circular path: A F net = ma C B T mg T B -mg = mv 2 /L T B = mv 2 /L + mg In uniform circular motion, T B > T A

11 Page Water in a bucket swung in a vertical circle F net = ma C r mg N -N-mg = -mv 2 /r N = mv 2 /r - mg If N = 0, the water will pour out. For water not to pour out, minimum speed is obtained by setting T = 0 0 = mv 2 /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

12 Page Small ball swung in a horizontal circle F net = ma C r Small mass,t >> mg T = mv 2 /r mg T 4. Car driven on valley/hill r N v N v mg r mg F net = ma C Valley: N V mg = mv 2 /r and N V = mg + mv 2 /r Hill: N H mg = -mv 2 /r and N H = mg mv 2 /r On flat road, N F = mg

13 Page Car rounding a curve on a flat road F net = ma C W N r f s F y : N W = 0 F x : fs = mv 2 /r If v is increased, f s will increase up till f smax (= µ s N) mv 2 /r µ s N (= µ s mg) v (µ s rg) OR µ s v 2 /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 F c will be contributed by the normal force N.

14 Page 14 y N mg θ N y N x θ N N x = N sinθ N y = N cosθ x With no dependence on friction: Along y-axis: N y = Ncosθ - mg = 0 ie, Ncosθ = mg Along x-axis: Nsinθ = mv 2 /r OR: tan θ = v 2 /rg, ie Banking angle θ = tan -1 (v 2 /rg) Speed v (rg tan θ) 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?

15 Page 15 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? 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 r v M F c = GmM/r 2 = mv 2 /r v 2 = GM/r = GM/(r E + h) For Geosynchronous satellites, V = 2πr/T, where T = 24 hours

16 Page 16 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 10 3 km and the mass of the earth is 5.98 x kg. (a) What is the speed of the satellite in this orbit? (b) Is the telescope geosynchronous? G = 6.67 x N.m 2 /kg 2 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. a c 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.

17 Page 17 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 (a t ) and centripetal (a C ) components a = a c + a t a a t a c = changes direction of v. a c a t = changes magnitude of v Angular Acceleration Angular acceleration is the change in angular velocity ω divided by the change in time. α = ω/ t = (ω - ω o )/ t Units: rad/s 2 Note: tangential acceleration a t = rα

18 Page 18 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. α = ω/ t = (ω - ω o )/ t ω o = v o /r = 15/10 = 1.5 rad/s ω = v/r = 25/10 = 2.5 rad/s α = ω/ t = ( )/2 = 0.5 rad/s 2 Kinematics for Circular Motion with constant α Linear Variables Angular Variables When a = constant When α = constant v = v o + at v 2 = v o2 + 2a (x x o ) x = x 0 + v o t + ½ at 2 ω = ω ο + αt ω 2 = ω ο2 + 2 α(θ θ ο ) θ = θ ο + ω o t + ½ α t 2

19 Page 19 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/s 2, how many revolutions does the disk make before it is at the proper speed? ω 2 = ω ο2 + 2 α(θ θ ο ) 1 Revolution = 2 π radians 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?

20 Page 20 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.

21 Page 21 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?