PHYSICS 149: Lecture 13

PHYSICS 149: Lecture 13 • Chapter 5: Circular Motion – 5.1 Description of Uniform Circular Motion – 5.2 Radial Acceleration – 5.3 Banked and Unbanked ...
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PHYSICS 149: Lecture 13 • Chapter 5: Circular Motion – 5.1 Description of Uniform Circular Motion – 5.2 Radial Acceleration – 5.3 Banked and Unbanked Curve – 5.4 Circular Orbits of Satellites and Planets

Lecture 13

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ILQ 1 When an elevator accelerates upwards, your apparent weight A)) B) C)

Lecture 13

increases. stays the same. d decreases.

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ILQ 2 Amy y and Bob release their snowballs from the same height and at the same time. Amy's is dropped y Which one hits while Bob‘s is thrown horizontally. the ground first? A) B) C)) D) Lecture 13

the "dropped" snowball the "thrown" snowball theyy hit at the same time it depends on the initial height Purdue University, Physics 149

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ILQ 3 A ball is thrown into the air and follows a parabolic trajectory Ignore air resistance. trajectory. resistance At the highest point in the trajectory, A) B) C) D)

the acceleration is zero, but the velocity is not zero. the velocity is zero, but the acceleration is not zero. both the velocity and the acceleration are zero. neither the acceleration nor the velocity are zero.

x-component: constant velocity motion y-component: constant acceleration motion Lecture 13

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Examples p of Circular Motion

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Angular g Variables • The motion of objects moving in circular (or nearly circular) paths paths, is often described by angles measured in radians rather than degrees. • The angle θ in radians, is defined as:

s θ= r • If s = r the angle is 1 rad • If s = 2πr (the circumference of the circle) the angle is 2π rad. rad (In other words, words 360 360° = 2π rad.) rad ) Lecture 13

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Example p • If a radius is 2 m and an arc length is 100 cm, what is the angle subtended by the arc? θ = s/r = 100 cm / 2 m =1m/2m = 0.5 rad Since 2π rad = 360° 360 , 1 rad = (360/2π)° Thus, θ = 0.5 rad = 0.5 × (360/2π)° = 28.65° Lecture 13

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Circular Motion • Angular displacement ∆θ = θ2-θ1 – How far it has rotated – Units: rad or °

• Angular velocity ω = ∆θ/∆t – How fast it is rotating – Units: radians/second (2π = 1 revolution)

• Period = 1/frequency – T = 1/f = 2π / ω – Time to complete 1 revolution Lecture 13

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ILQ A CD makes one complete revolution every tenth of a second. The angular velocity of point 4 is: A) B)) C) D)) E)

the same as for pt 1. twice that of pt 2. half that of pt 2. 1/4 that of pt p 1. four times that of pt 1.

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Circular to Linear • Displacement p ∆s = r ∆θ ((θ in radians)) • Speed |v| = ∆s/∆t = r ∆θ/∆t = rω |v| = 2πrf |v| = 2πr/T

• Direction of v is tangent to circle Another A th way tto express v is i by using the relation between period i d and d speed d |v| = 2πr/T Lecture 13

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ILQ A CD makes one complete revolution every tenth of a second. d Which Whi h has h the h llargest magnitude i d off linear li (tangential) velocity? A) B) C) D)

Point 1 P i t2 Point Point 3 P Point i t4

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Uniform Circular Motion • Uniform circular motion: The motion of an object j traveling g at constant (uniform) speed on a circular path – The object object’s s speed (the magnitude of the velocity) is constant. – The direction of the velocity vector changes h as th the object bj t moves along l the circle.

v v

Î Uniform circular motion is not a constant velocity motion, and there is an acceleration on the object (net force is not zero.) – There are various sources of acceleration. l ti Lecture 13

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Uniform Circular Motion • Period: – Time to travel around the circle (s)

• Frequency: F

1 T= f

– Revolution per unit time (Hz=s-1)

2π r = 2π rf v= T

• Velocity: Distance/time (m/s) • Angular velocity: ω = Δθ/Δt –H How ffastt it is i rotating t ti – Units: radians/second (2 = 1 revolution)) (2π Lecture 13

v 2π ω = = = 2π f r T

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Circular Motion A ball is going around in a circle attached to a string. If the string breaks at the instant shown shown, which path will the ball follow? 2 1

3 4

v 5 A Answer: 2

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Circular Motion Starting at the top, a ball is set in motion to travel around the inside of a ring with a hole in the side side. What will happen when the ball reaches the hole?

A

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B

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C

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Radial Acceleration • Magnitude of the velocity vector is constant, but direction is constantly changing • At any instant of time, the direction of the instantaneous velocity is tangent to the path • Therefore: nonzero acceleration

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Radial Acceleration Δv Δv

v2 R

ΔR

Magnitude of radial acceleration: l ti

v1

v2

v1

aave= Δω / Δt acceleration inward

2

v a= R

centripetal acceleration

Acceleration is due to change in direction, direction not speed speed. Since turns “toward” center, must be a force toward center. Lecture 13

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Angular g Acceleration • Angular acceleration is the change in angular velocity l it ω divided di id d by b the th change h iin titime. ω f − ω0 α≡ Δt • 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 cars average angular acceleration if it takes 1.6 seconds to go from the top to the bottom? v 15 25 ω= ω0 = = 1.5 ωf = = 2.5 10 10 r 2.5 − 1.5 α≡ 1.6 Lecture 13

= 0.64 0 64 rad/s2

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CD Player y Example p The CD in your disk player spins at about 20 radians/second. If it accelerates uniformlyy from rest with angular g acceleration of 15 rad/s2, how many revolutions does the disk make before it is at the proper speed?

ω 2 − ω02 = 2αΔθ

ω 2f − ω02 = Δθ 2α 202 − 02 = Δθ 2 ×15 Lecture 13

Δθ = 13.3 radians 1 Revolutions = 2 π radians

Δθ = 13.3 radians = 2.12 revolutions Purdue University, Physics 149

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Uniform Circular Motion Circular motion with constant speed R a

v

Recall: v=ωR

2

v 2 ar = = ω R R centripetal i l acceleration

• Instantaneous velocity is tangent to circle • Instantaneous acceleration is radially inward • There Th mustt be b a fforce to t provide id th the acceleration l ti Lecture 13

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Circular Motion Angular velocity

Linear velocity

v = rω

Δθ ωav = Δt

Linear velocity is tangent to circle Centripetal acceleration

R

2

v 2 ac = = ω r r

a

v

Direction: toward the center Centripetal Acceleration is radially inward Lecture 13

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Rolling g without Slipping pp g • An object that is rolling is both rotating (around its axis) and d translating l i ((the h axis i iis moving) i ) • If an object is rolling without slipping, as it turns through one complete rotation rotation, the axle a le moves mo es a distance equal eq al to the circumference of the object. vaxle

vaxle

2r

vaxle = Lecture 13

2πr = | ω | r (ω in i radians di per unit it time ti ) T Purdue University, Physics 149

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Rotation and Translation Often objects are both rotating and translating. The wheel rotates, rotates axle translates. translates If there is not skidding or slipping: 2π r vaxle =

T= vaxle

T



ω = ωr

Kevin is riding g is bike at 13 m/s. If the radius of his tire of 32.5 cm, what is the angular speed of the rear wheel?

v 13m / s ω= = = 40rad / s r 0.325m Lecture 13

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ILQ An object is in uniform circular motion. Which of the following statements is true? A) Velocity = constant B) Speed = constant C) Acceleration = constant

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Examples p • The wheel of a car has a radius of 0.29 m and is being rotated d at 830 revolution l i per minute i ((rpm)) on a tirei balancing machine. Determine the speed (in m/s) at which the outer edge of the wheel is moving:

The speed could be obtained by |v| = 2πr/T 1 T= = 1.2 × 10−3 min/revolutions=0.072 s 830revolutions/min 2π r 2π (0.29m ) v= = = 25m / s T 0.072 s

• A CD spins with an angular frequency 20 radians/second radians/second. What is the linear speed 6 cm from the center of the CD?

v = r ω = 0.06 × 20 = 1.2 m/s / Lecture 13

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ILQ You are driving a car with constant speed around a horizontal circular track. On a p piece of p paper, p , draw a Free Body Diagram (FBD) for the car. How many forces are acting on the car? FN A) 1 B) 2 correct C) 3 f R D) 4 W E) 5 ΣF = ma = mv2/R Fn = Normal Force, W = Weight, the force of gravity, f = Centripetal p force Lecture 13

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ILQ You are driving a car with constant speed around a horizontal circular track. On a p piece of p paper, p , draw a Free Body Diagram (FBD) for the car. The net force on the car is FN A) Zero B) Pointing P i ti radially di ll iinward d C) Pointing radially outward W

f

R

ΣF = ma = mv2/R

Answer: B)

“Because the centripetal acceleration is always pointing inward towards the center off the h circle.” i l Lecture 13

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ACT Suppose you are driving through a valley whose bottom has a circular shape. If your mass is m, what is the magnitude it d off th the normall fforce FN exerted t d on you by b th the car seat as you drive past the bottom of the hill. A) FN < mg B) FN = mg C) FN > mg

a=v2/R R

correct

Since there is centripetal acceleration, the normal force is greater than simply mg

FN v

ΣF = ma mg

FN - mg = mv2/R FN = mg + mv2/R / Lecture 13

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Roller Coaster Example p What is the minimum speed you must have at the top of a 20 meter diameter roller coaster loop, to keep the wheels on the track. y-direction: F = ma -N – mgg = m a Let N = 0, just touching -mg mg = m a -mg = -m v2/R g = v2 / R v = sqrt(g*R) q (g ) = 10 m/s Lecture 13

N

mg

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Kinematics for Circular Motion with constant acceleration Linear Variables x,v,a. x v a Constant a

Δx = x − x0 = vavg Δt

Δθ = θ − θ 0 = ωavg Δt

v − v0 = at

ω − ω0 = α t ω0 + ω

v0 + v vavg = 2 1 2 x − x0 = v0t + at 2

1 2 θ − θ 0 = ω0 t + α t 2

v − v = 2aΔx

ω 2 − ω02 = 2αΔθ

2

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Angular Variables θ,ω,α. θ ω α Constant α

2 0

ωavg =

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2

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Circular vs. Linear Motion with comparison to 1-D kinematics Angular

Linear

α = constant

a = constant

ω = ω0 + α t

v = v0 + at 1 2

1 x = x0 + v0t + at 2 2

θ = θ 0 + ω0t + α t 2

And for a point at a distance R from the rotation axis: x = Rθ

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v = ωR R

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a = αR R

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Merry-Go-Round y ACT Bonnie sits on the outer rim of a merry-go-round with radius 3 meters meters, and Klyde sits midway between the center and the rim. The merry-go-round makes one complete revolution every two seconds. Klyde Bonnie Klyde’s speed is: A) the same as Bonnie’s Bonnie s B) twice Bonnie’s C) half Bonnie’s Bonnie s

VKlyde Bonnie travels 2 π R in 2 seconds

1 = VBonnie 2

vB = 2 π R / 2 = 9.42 m/s

Klyde travels 2 π (R/2) in 2 seconds vK = 2 π (R/2) / 2 = 4.71 m/s Lecture 13

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Merry-Go-Round y ACT Bonnie sits on the outer rim of a merry-go-round, and Klyde sits midway between the center and the rim. rim The merrymerry go-round makes one complete revolution every two seconds. Klyde Bonnie Klyde’s angular velocity is: A) the same as Bonnie’s Bonnie s B) twice Bonnie’s C) half Bonnie’s Bonnie s

• The angular velocity ω of any point on a solid object rotating about a fixed axis is the same. – Both Bonnie and Klyde go around once (2π radians) every two seconds. Lecture 13

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Unbanked Curve What force accelerates a car around a turn on a level road at constant speed? A) it is not accelerating B) the th road d on th the titires C) the tires on the road D) the engine on the tires

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Unbanked Curve What is the maximum velocity a car can go around an unbanked curve?

y : N − mg = 0 N = mg

mv 2 x : Fc = f s = μ s N = μ s mg = r fs ≤ μs N The maximum velocity v = μ s gr to go around an unbanked curve depends only on μs (g and r are fixed) Dry road: μs=0.9 Icy road: μs=0.1 Lecture 13

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Banked Curve A car drives around a curve with radius 410 m at a speed of 32 m/s. The road is banked at 5.0°. The mass of the car is 1400 kg. A) What is the frictional force on the car? B) At what speed could you drive around this curve so that the force of friction is zero?

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Banked Curve

y

θ

x

θ =5

y-direction y

Σ F y = ma y = 0 N cos θ − mg − f sin θ = 0

(1)

0

r = 410 m v = 32 m / s N

x-direction

ΣF x = max = ma 2

v N sin θ + f cos θ = ma = m r Lecture 13

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((2))

W

f 37

Banked Curve 2 equations and 2 unknown we can solve for N in (1) and substitute in (2)

f sin θ + m g v2 N = N sin θ + f cos θ = m r co s θ 2 f sin θ + mg mv ⎛ ⎞ ⎜ ⎟ sin θ + f cos θ = r cos θ ⎝ ⎠ 2 mv 2 2 f (sin θ + cos θ ) = cos θ − mg sin θ r ⎛ v2 ⎞ f = m ⎜ cos θ − g sin θ ⎟ = 2300 N ⎝ r ⎠ Lecture 13

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Banked Curve A car drives around a curve with radius 410 m at a speed of 32 m/s. The road is banked at 5.0°. The mass of the car is 1400 kg. A) What is the frictional force on the car? B) At what speed could you drive around this curve so that the force of friction is zero? Like an airplane

f =0 2

v cos θ = g sin θ r v = gr tan θ = 19m / s Lecture 13

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Conical Pendulum Net force must point towards the center of the circle x : ∑ Fx = T sin φ = ma x = mω 2 r r = L sin φ T sin i φ = mω 2 L sin i φ T = mω 2 L y : ∑ Fx = T cos φ = mg mω 2 L cos φ = mg g ω = L cos φ 2

ω= Lecture 13

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g L cos φ 40

Kepler’s p Laws • First law: The orbit of a planet l t about b t the th Sun S is i an ellipse with the Sun att one focus. f • Second law: A line joining a planet and the Sun sweeps out equal areas in equal intervals of time.

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Kepler’s p Third Law • The square q of a p planet's orbital p period is proportional to the cube of the length of its orbit's j axis. semimajor • Or simply… simply T2 = R3 if T is measured in years and R is measured in astronomical units. • An AU is the average distance of the Earth from the Sun. 1 AU = 93,000,000 miles = 8.3 lightminutes Lecture 13

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Kepler’s p Laws

Elliptical orbits orbits…

Equal areas in equall time ti

T2 = R3

These were empirical laws Lecture 13

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Drawing g an Ellipse p

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