Chapter 8: Newton s Laws Applied to Circular Motion

Circular Motion

Milky Way Galaxy Orbital Speed of Solar System: 220 km/s Orbital Period: 225 Million Years

Mercury: 48 km/s Venus: 35 km/s Earth: 30 km/s Mars: 24 km/s Jupiter: 13 km/s Neptune: 5 km/s

Differential Solar Rotation Entangled Magentic Field Lines makes Sun Spots Equatorial Rotational Period: 27.5 days

464m/s

Precession causes the position of the North Pole to change over a period of 26,000 years.

Orbital Speed of Earth: ~ 30 km/s

Although the Moon is always lit from the Sun, we see different amounts of the lit portion from Earth depending on where the Moon is located in its month-long orbit. Orbital Speed of Moon: ~ 1 km/s

155mph~70m/s

RADAR: RAdio Detecting And Ranging 200mph~90m/s

Electrons in Bohr Orbit: 2 million m/s (Speed of Light: 200 million m/s)

Maximal Kerr Black Hole Rotational Speed = Speed of Light

Horizontal Circle: Constant Speed & Acceleration Vertical Circle: Changing Speed & Acceleration

Important: Inside vs Outside the Rotating Frame

Translational vs Rotational / / m x v dx dt a dv dt = = / / I d dt d dt θ ω θ α ω = = 2 / c t s r v r a v r a r θ ω α = = = = Connection

Translational and Rotational Kinematics For CONSTANT Accelerations ONLY

Angular Velocity in Uniform Circular Motion When angular velocity ω is constant, this is uniform circular motion. In this case, as the particle goes around a circle one time, its angular displacement is θ = 2π during one period t = T. ω: angular velocity (rad/s) 2013 Pearson Education, Inc. Slide 4-87

Angular Position: Rotation Angle The rotation angle is the ratio of arc length to radius of curvature. For a given angle, the greater the radius, the greater the arc length. = θ s r θ = rotation angle s = arc length r = radius

How many radians in 360 0? Ans: 360 0 = 2 π radians Consider a circle with radius r. θ = s/r θ = 2 πr /r θ = 2 π r How many degrees in 1 radian? s =? s = C = 2πr 2 π rad = 360 0 1 rad = 360 0 /2 π = 57.3 0

θ r = 4 m s A box is fastened to a string that is wrapped around a pulley. The pulley turns through an angle of 43 0. What is the distance, d, that the box moves? d 43 = 43 x1 = 43 x 2 =.75rad π 360 First: How many radians is 43 0?

θ r = 4 m s A box is fastened to a string that is wrapped around a pulley. The pulley turns through an angle of.75 rad. What is the distance, d, that the box moves? d s = r θ d = r θ d = 4m (0.75 rad) d = 3 m rad d = 3 m 43 0 =.75rad Radian is dimensionless and is dropped!

Angular Velocity ω: angular velocity (rad/s) θ ω = d dt ( / ) = ds r dt 1 = ds r dt = v r t

Angular & Tangential Velocity ω: angular velocity (rad/s) v: tangential velocity (m/s) θ d ω = = dt v r v = ωr ω is the same for every point & v varies with r.

Problem v = ωr Calculate the angular velocity of a 0.500 m radius car tire when the car travels at a constant speed of 25.0 m/s. ω = v r = = 25 m/ s.5m 50 rad / s Insert rad for angular speed. Keep it to define units.

Tangential and Angular Acceleration changes speed a t dv = dt ( ω ) = d r dt v = ωr = d r dt ω ω α = d dt at = αr

Angular Tangential Bike A bike wheel with a radius of 0.25 m undergoes a constant angular acceleration of 2.50 rad/s 2. The initial angular speed of the wheel is 5.00 rad/s. After 4.00 s a) What angle has the wheel turned through? b) What is the final angular speed? c) What is the final tangential speed of the bike? d) How far did the bike travel? 1 2 1 2 2 θ = ω0t+ αt = 5 rad / s4s + 2.5 rad / s (4 s) = 40rad 2 2 2 ω = ω + αt = 5 rad / s + 2.5 rad / s 4s = 15 rad / s f 0 v = ωr = 15 rad / s.25m = 3.75 m / s d = s = θ r = 40 rad.25m = 10m ω Givens: 0 = 5 rad / s α = 2.5 rad / s t = 4s 2

Centripetal Acceleration v Centripetal Acceleration Only changes Direction!!! = ωr s = rθ ds d( rθ ) v = = dt dt dt = Combining: dr ( θ ) v a dv vdθ v = = = dt ( rdθ / v) r 2 2013 Pearson Education, Inc.

Centripetal and Tangential Accelerations at = αr a c = v r 2 = ( ωr) r 2 ac 2 = ω r (r constant)

Total Acceleration 2 2 r t a = a + a a= arˆ + aθθˆ r a = a = ω 2 r = v 2 / r r c a = a = αr = dv / dt θ t

Nonuniform Circular Motion a t = d dt v a r = a = C v r 2 a = a + a 2 2 r t

Circular Motion Total Acceleration The tangential acceleration causes the change in the speed of the particle a t = d v dt The radial acceleration comes from a change in the direction of the velocity vector a r = a = C v r 2 Total Acceleration: a = a + a 2 2 r t

Total Acceleration & Force a t = d dt v 2 a a = a + a r 2 2 r t = a = C v r 2 2 r F = F + F θ

Total Acceleration a c The total acceleration of a particle moving clockwise in a circle of radius 2.50 m at a certain of time is 15.0m/s 2. At this instant, find (a) the radial acceleration, (b) the speed of the particle, and (c) its tangential acceleration. v r ac 2 ( )( ) o 2 2 = acos30.0 = 15.0 m s cos30 = 13.0 m s 2 2 2 2 = v ra ( ) c 2 2 2 = t + r a a a = = 2.50 m 13.0 m s = 32.5 m s v = 32.5 m s = 5.70 m s 2 2 2 2 2 2 2 r 15.0 m s 13.0m s 7.50 m s ( ) ( ) a = a a = = t

Centrifugal Force & Acceleration Center FLEEing Suppose there is a lady bug in the can. There is a centripetal force acting on the bug, transmitted to her feet by the can. Her feet push back on the can producing a centrifugal fictitious force, that acts like artificial gravity.

The Earth rotates once per day around its axis as shown. Assuming the Earth is a sphere, is the angular speed at Santa Rosa greater or less than the speed at the equator?

The Earth rotates once per day around its axis as shown. Assuming the Earth is a sphere, is the tangential speed at Santa Rosa greater or less than the speed at the equator? 366 m/s 464 m/s

Is the centripetal acceleration greater at the Equator or at Santa Rosa? a c v r 2 0.027 m/ s = 2 2 0.034 m/ s

The Earth rotates once per day around its axis. Assuming the Earth is a sphere with radius 6.38 x 10 6 m, find the tangential speed of a person at the equator and at 38 degrees latitude (Santa Rosa!) and their centripetal accelerations. At the equator, r = 6.38 x 10 6 m: 6 2πr 2 π(6.38x10 m) v= = = t 86, 400s a c ( 464 m/ s) 2 2 v = = = 6 r 6.38x10 m 464 m/ s.034 m/ s At Santa Rosa, r = 6.38 x 10 6 m cos38: 2 a c a c 6 2πr 2 π(6.38x10 m)cos38 v= = = t 86, 400s a c ( 366 m/ s) 2 366 m/ s 2 v = = =.027 m/ s 6 r cos 38 6.38x10 m cos 38 2

What is the total acceleration acting on a person in Santa Rosa? The vector sum. a c g

Is your apparent weight as measured on a spring scale more at the Equator or at Santa Rosa? a c g

Since you are standing on the Earth (and not in the can) the centrifugal force tends to throw you off the Earth. You weigh less where the centripetal force is greatest because that is also where the centrifugal force is greatest the force that tends to throw you out of a rotating reference frame.

Centrifugal Force & Acceleration Center FLEEing F actual = Centripetal Force F fictitious = Centrifugal Force

Centripetal & Centrifugal Force Depends on Your Reference Frame Center-seeking Outside Observer (non-rotating frame) sees Centripetal Force pulling can in a circle. Center-fleeing Inside Observer (rotating reference frame) feels Centrifugal Force pushing them against the can.

Centrifugal Force is Fictitious? The centrifugal force is a real effect. Objects in a rotating frame feel a centrifugal force acting on them, trying to push them out. This is due to your inertia the fact that your mass does not want to go in a circle. The centrifugal force is called fictitious because it isn t due to any real force it is only due to the fact that you are rotating. The centripetal force is real because it is due to something acting on you like a string or a car.

In Physics, we use ONLY CentriPETAL acceleration NOT CentriFUGAL acceleration!

Coriolis Force This is an apparent force caused by changing the radial position of an object in a rotating coordinate system The result of the rotation is the curved path of the ball

Coriolis Effect

Artificial Gravity How fast would the space station segments A and B have to rotate in order to produce an artificial gravity of 1 g? v = 56 m / s ~ 115mph A v = 104 m / s ~ 210mph B Can the two segments be connected?

Space Station Rotation A space station of diameter 80 m is turning about its axis at a constant rate. If the acceleration of the outer rim of the station is 2.5 m/s 2, what is the period of revolution of the space station? a. 22 s b. 19 s c. 25 s d. 28 s e. 40 s

Horizontal Circle: Constant Speed & Acceleration Vertical Circle: Changing Speed & Acceleration

Important: Inside vs Outside the Rotating Frame

Motion in a Horizontal Circle The speed at which the object moves depends on the mass of the object and the tension in the cord. It is constant! The centripetal force is supplied by the tension. Looking down: c F = T = ma = c mv r 2 v = Tr m

Motion in a Horizontal Circle

Horizontal Circle: Conical Pendulum y F = T cosθ = mg c F = T sinθ = ma = c mv r 2 Combining : r = Lsinθ v = Lg sinθ tanθ

Horizontal (Flat) Curve The force of static friction supplies the centripetal force Fc = f = The maximum speed at which the car can negotiate the curve is mv r 2 v µ gr Note, this does not depend on the mass of the car = y F = N mg =0 f = µ mg

Horizontal (Flat) Curve 3. A highway curve has a radius of 0.14 km and is unbanked. A car weighing 12 kn goes around the curve at a speed of 24 m/s without slipping. What is the magnitude of the horizontal force of the road on the car? What is μ? Draw FBD. a. 12 kn b. 17 kn c. 13 kn d. 5.0 kn e. 49 kn

Banked Curve These are designed with friction equaling zero - there is a component of the normal force that supplies the centripetal force that keeps the car moving in a circle. Fr y 2 mv = nsinθ = r F = n cosθ mg = 0 Dividing: tanθ = 2 v rg

Banked Curve 4. A race car travels 40 m/s around a banked circular (radius = 0.20 km) track. What is the magnitude of the resultant force on the 80-kg driver of this car? a. 0.68 kn b. 0.64 kn c. 0.72 kn d. 0.76 kn e. 0.52 kn

Hints for HW Problem Determine the range of speeds a car can have without slipping up or down the road when it is banked AND has friciton. If the car is about to slip down the incline, f is directed up the incline. This would happen at a minimum speed. When the car is about to slip up the incline, f is directed down the incline. This would happen at a maximum speed.

Vertical Circle with Non-Uniform Speed Where is the speed Max? Min? Where is the Tension Max? Min?

Vertical Circle with Non-Uniform Speed The tension at the bottom is a maximum The tension at the top is a minimum Look at radial and tangential: t F = mg sinθ = ma t at = g sinθ r F = T mg cosθ = mv R 2 2 v T = m + g R cosθ

Vertical Circle: Mass on a String A 0.40-kg mass attached to the end of a string swings in a vertical circle having a radius of 1.8 m. At an instant when the string makes an angle of 40 degrees below the horizontal, the speed of the mass is 5.0 m/s. What is the magnitude of the tension in the string at this instant? Draw the FBD. a. 9.5 N b. 3.0 N c. 8.1 N d. 5.6 N e. 4.7 N

Minimum Speed for Vertical Circular Motion What is the minimum speed so that the ball can go in the circle? That is, when T = 0 at the top? At the top: θ = 180 2 v T = m + gcosθ = 0 R v = gr Minimal Speed to JUST get around the circle only depends on R! ROOT GRRRRRRRR

Loop d Loops: Inside the Vertical Loop Minimum Speed to get to the Top. What is the minimum speed so that the car barely make it around the loop the riders are upside down and feel weightless? R = 10.0m

QuickCheck 8.11 Loop d Loops: Inside the Vertical Loop A roller coaster car does a loopthe-loop. Which of the free-body diagrams shows the forces on the car at the top of the loop? Rolling friction can be neglected. Slide 8-82

QuickCheck 8.10 Humps in the Road: Outside the Vertical Loop A car that s out of gas coasts over the top of a hill at a steady 20 m/s. Assume air resistance is negligible. Which free-body diagram describes the car at this instant? Slide 8-80

Humps in the Road Outside the Vertical Loop A roller-coaster car has a mass of 500 kg when fully loaded with passengers. The car passes over a hill of radius 15 m, as shown. At the top of the hill, the car has a speed of 8.0 m/s. What is the force of the track on the car at the top of the hill? a. 7.0 kn up b. 7.0 kn down c. 2.8 kn down d. 2.8 kn up e. 5.6 kn down

Loop d Loops: Inside the Vertical Loop A roller-coaster car has a mass of 500 kg when fully loaded with passengers. At the bottom of a circular dip of radius 40 m (as shown in the figure) the car has a speed of 16 m/s. What is the magnitude of the force of the track on the car at the bottom of the dip? a. 3.2 kn b. 8.1 kn c. 4.9 kn d. 1.7 kn e. 5.3 kn

What is the maximum speed the vehicle can have at B and still remain on the track?

Maximum Speed for Vertical Circular Motion Humps in the Road What is the maximum speed the car can have as it passes this highest point without losing contact with the road? : Max speed without losing contact MEANS: n Take : n = 0 Therefore: mg = mv r 2 v = gr mg Maximum Speed to not loose contact with road only depends on R! ROOT GRRRRRRRR

What is the maximum speed the vehicle can have at B and still remain on the track?