CHAPTER 7CHAPTER 7Rotational Motion and the Law of Rotational Motion and the Law of

GravityGravityAngular Speed and Angular Angular Speed and Angular AccelerationAcceleration

s = arc lengthΘ = arc angle (radians)Θ = s ; where r = radius r

TermTerm SymbolSymbol UnitsUnitsAngular Displacement ∆Θ radians Angular Speed ωradians/sec

(Average) Angular Acceleration radians/sec2

Average Angular Speed ω radians/sec

ω = Θ2 – Θ1

t2 – t1

ω = ∆Θ∆t

Average Angular SpeedAverage rate at which the

arc angle is changing

ω = lim

∆Θ∆t∆t0

(Instantaneous) Angular SpeedThe rate at which the arc angle is changing at a particular instant in

time

= ω2 – ω1

t2 – t1

= ∆ ω ∆t

Average Angular AccelerationThe average rate at which the

angular speed is changing

NOTE: You are not required to work with changing rates of acceleration.

Therefore, =

Linear/ Angular ComparisonLinear/ Angular Comparison

TermTerm LinearLinearAngularAngularDisplacement ∆x m ∆Θ rad

Velocity v m/s ω rad/sec Acceleration a m/s2 rad/sec2

Formulas:Formulas: ∆x = ½(vo+v)t

∆x = vot + ½at2

v = vo + at

v2 = vo2 + 2a∆x

N/A

∆Θ = ½(ωo+ω)t

∆Θ = ωot + ½ t2

ω = ωo + t

ω2 = ωo2 + 2∆Θ

s = rΘ

NOTE: All points on a rotating disc have the same values for ω and .

Example ProblemExample Problem (Rotating Disc)A disc of radius 2.50m accelerates to an angular velocity of 1.35 radians/sec in 12.5 seconds. Calculate (a) the disc’s angular acceleration; (b) the number of revolutions completed during the acceleration; (c) the distance traveled by a point on the edge of the disc.Identify the variablesIdentify the variables ωo = = ω = t = ∆Θ = Choose Appropriate Choose Appropriate FormulasFormulas ω = ωo + t

= .108 rad/sec2

[ ]

∆Θ = tωo + ω 2

∆Θ = ωot + ½ t2

∆Θ = 8.44 rad

∆Θ = 8.44 rad

1 revolution = 2 rad = 6.2 rad8.44 rad 1 rev

6.28 rad# rev = 1.34rev

∆s = r∆Θ ∆s = 21.1m

0.00 rad/sec ?1.35 rad/sec

12.5 sec ?

vt = r ω

Tangential SpeedTangential Speed (vt)

• vt = ∆s ∆t ∆s = r ∆Θ vt = r ∆Θ ∆t

Tangential Acceleration (t)

a t = r

NOTE: Not every point on a rotating disc has the same tangential speed or tangential

acceleration.

• arc length of the movement time to complete the movement

Centripetal AccelerationCentripetal Acceleration (center-seeking)

Question:Question: The tangential velocity vt illustrated above is clearly changing. How can you tell it is changing?Answer: Its direction is changing.Its direction is changing.a = ∆v = vf – vi = vf + -vi

∆t ∆t ∆tThe illustration on the right clearly shows that ∆v points towards the center of rotation.

Centripetal AccelerationCentripetal Acceleration (ac) ∆v = ∆s from similar triangles

v r∆v = v r

∆s and ∆v = ac ∆t

ac ∆t = ∆svr

ac = v ∆s and ∆s = vr ∆t ∆t

ac = = Centripetal Accelerationv2

r

Centripetal Centripetal Force:

• The net force that causes an object to accelerate towards its center of rotation (i.e. go in a circle)

• Fc = mac

• Fc is not a new force

• Fc will be either one or a combination of more than one of the following forces:

Fg

FN

Ff

Fa (T)

Example ProblemExample Problem (Vertical loop)A rollercoaster contains a loop 20m in diameter. a) What speed will cause a passenger to feel “weightless” at the top of the loop? b) If this same person has a mass of 90kg what weight will they experience at the bottom of the loop? Sketch the Force Sketch the Force

DiagramsDiagrams

Analysis:Analysis: The person is “weightless” if no forces other than gravity are acting upon them (i.e. FN=0).Hence, Fg is Fc

Analysis:Analysis: The only forces acting on the person are the normal force and force of gravity. Fc = Fnet(center) = FN – Fg

FN = Fc + Fg= mv2 + mg rFN = 1764 Newtons

mg = mv2

rv = √rg v = 9.9m/s

Example ProblemExample Problem (Banked Curve)a) For a car traveling with a speed v around a curve of radius r, determine the formula for the angle at which the road should be bank so that no friction is required. b) what is this angle for an expressway off-ramp of radius 50m at a design speed of 50km/hr?Sketch the Force DiagramSketch the Force Diagram

Part a:AnalysisAnalysis

• Since there is no friction, there is no component of friction directed to the center (right).• Only the component FNsinΘ of the normal force is directed to the center.• FN sinΘ is the net force towards the center = Fc

FN sinΘ = Fc

FN sinΘ = mv2

rmg = FN cosΘ mg · sinΘ = mv2

cosΘ rmg tanΘ = mv2

rtanΘ = v2

gr Θ = tan-1 ( )v2

grPart

b: r = 50mv = 50km 1000m 1hr hr km 3600s

= 14m/s

g = 9.8

Θ = 22°

Newton’s Universal Law of Newton’s Universal Law of GravitationGravitation

Every particle in the universe attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

Fg = G m1 m2

r2

G = universal gravitational constantG = 6.67 x 10-11 N·m2/kg2

Force of Earth’s GravityForce of Earth’s Gravity (on Earth’s Surface)

Fg = G mearth mobject

rearth2

Fg = · massobject[ ]

G mearth

rearth2

Fg = · massobject(6.67 x10-11 N·m2/kg2)(5.98x1024

kg)(6.37x106 m)2

Fg = 9.8 x massobject

Fg = mobect g

Velocity of an Earth SatelliteVelocity of an Earth SatelliteAnalysis:Analysis:

What is the only Force acting on the satellite?Answer:Answer:

FgIf satellite is in circular orbit, then…

Fg = Fc

mg = mvs2 vs = velocity of

satellite rvs = √gr g = GMearth r2vs = GMearth · r

r2√vs =

GMearth

r√

Gravitational Potential EnergyGravitational Potential EnergyPE = -GMEm

r

r = distance of object from Earth’s center

• from Calculus• PE = zero at infinite distance from Earth• PE = some negative value closer to the EarthEscape SpeedEscape Speed

• Speed at surface of Earth where the object’s KE1 + PE1 is sufficient to reach infinite distance from Earth (PE2=0J) with a final speed of 0m/s (KE2=0J)• From Conservation of Energy:

KE1 + PE1 = KE2 + PE1

mv2 + -GMearthm 2 Rearth

= 0 + 0

mv2 + -GMearthm 2 Rearth

vescape = [ ]1/22GMearth

Rearth