Chapter 10: Radians Rotational Motion s = rθ ω θ...

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1 Chapter 10: Rotational Motion Brent Royuk Phys-111 Concordia University Angular Ideas from Chp. 6 • Angular distance: Δθ = θ - θ o • Radians – The radian equation: s = rθ • Angular speed and velocity: v t = r ω ω = Δθ Δt Rolling Motion Rigid Bodies, Translations and Rotations – A rigid body is a system of particles in which the distances between the particles do not vary. – Analysis is much simpler than non-rigid body Forces acting through center of gravity produce translations. Tangential speed is equal to translational speed for a rolling object. • Examples Rolling without slipping: If ω = .89 s -1 and r = .085 cm, what is v for the rigid body? What about a book on pencils with same ω and r as above? Rolling Motion The wheel in the figure is rolling to the right without slipping. Rank in order, from fastest to slowest, the speeds of the points labeled 1 through 5. Angular Acceleration • Definition: – Uniform angular acceleration ω = ω o + αt – Units – Vector direction • Tangential Acceleration a t = rα α = Δω Δt Rotational Kinematics Translational Rotational s = r x = vt = t vt = r v = vo + at = o + t at = r x = v o t + 1 2 at 2 θ = ω o t + 1 2 αt 2 v 2 = v o 2 +2 ax ω 2 = ω o 2 + 2αθ

Transcript of Chapter 10: Radians Rotational Motion s = rθ ω θ...

Page 1: Chapter 10: Radians Rotational Motion s = rθ ω θ testrada.cune.edu/facweb/brent.royuk/phys111/docs/Chapter... · 2016-11-10 · 1 Chapter 10: Rotational Motion Δ Brent Royuk Phys-111

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Chapter 10: Rotational Motion

Brent Royuk Phys-111

Concordia University

Angular Ideas from Chp. 6 • Angular distance: Δθ = θ - θo • Radians

– The radian equation: s = rθ • Angular speed and velocity: •  vt = r ω

ω =Δθ

Δt

Rolling Motion •  Rigid Bodies, Translations and Rotations

–  A rigid body is a system of particles in which the distances between the particles do not vary.

–  Analysis is much simpler than non-rigid body –  Forces acting through center of gravity produce

translations. •  Tangential speed is equal to translational

speed for a rolling object. •  Examples

–  Rolling without slipping: If ω = .89 s-1 and r = .085 cm, what is v for the rigid body?

–  What about a book on pencils with same ω and r as above?

Rolling Motion •  The wheel in the figure is rolling to the right

without slipping. •  Rank in order, from fastest to slowest, the

speeds of the points labeled 1 through 5.

Angular Acceleration • Definition:

– Uniform angular acceleration • ω = ωo + αt

– Units – Vector direction

• Tangential Acceleration – at = rα

α =Δω

Δt

Rotational Kinematics

Translational Rotational s = r x = vt = t vt = r v = vo + at = o + t

at = r

x = vot +12

at 2 θ = ω ot +

12αt 2

v2 = vo

2 +2 ax ω2 = ω o

2 +2αθ

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Rotational Kinematics •  Examples

–  A potter’s wheel rotates from rest to 210 rpm in a time of 0.75 s. a) What is the angular acceleration of the wheel during this time? b) How many revolutions does the wheel make during this time interval?

–  How many revolutions does it take an angularly accelerating object to accelerate up to 2.8 rad/s at a rate of 0.058 rad/s2?

–  What is the angular acceleration of a spinning object that accelerates smoothly from “rest” and makes 4.0 complete revolutions while attaining an angular speed of 5.5 rad/s?

Torque •  Definition: •  Vector notation- a cross product:

– Compare 0o and 90o – Compare pushing door on outside or inside – Direction: right hand rule – Demo gadget

! τ =! r ×! F τ = rF

⊥= rF sinθ

Torque •  Lever Arms

– You can either calculate torque as the product of r and or F and . F⊥ r⊥

Rotational Inertia • What is the rotational equivalent of

mass? – Start with F = ma. – Multiply by r, use a = α r – Result: τ= m r2 α

•  I = moment of inertia: – Single particles are easy: add them

up. – Continuous mass distributions are

harder: must use calculus – See Figure 10.12 (next slide)

I = miri2∑

Moment of Inertia Rotational Dynamics

•  Using Newton’s 2nd: •  Demo rod •  Example: Wrap a string around a 1.5

kg solid sphere with radius 8.0 cm. Pull on the string with a force of 8.0 N. Find the angular acceleration

! τ NET = I

! α

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Rotational Kinetic Energy •  Equation: •  Examples

–  What is the rotational kinetic energy of a 450-g solid sphere with a diameter of 23 cm rotating at rate of 17 rpm?

–  What is the total rotational kinetic energy of a 18-kg child riding on the edge of a merry-go-round of mass 160 kg and r = 2.5 m that is rotating with a period of 2.0 s?

•  Rolling Objects –  For rolling objects: –  Examples

• What percentage of the kinetic energy of a rolling ball is rotational? How about a cylinder?

•  Compare the speed of a ball rolling down an inclined plane with its speed if it would slide.

–  Demo objects: ring and disk, eyeball, soup cans

KTOT =12

mv 2 +12

Iω2

K =12

Iω2 Angular Momentum

–  For a single particle, derive L = m v r !L = I !ω

Angular Momentum •  Conservation of Angular Momentum

–  Lb = La –  Ib ωb = Ia ωa –  Pulling string while bob rotates –  Spinning skaters –  Rotating platform

•  What if your initial angular speed is 3.74 rad/s, and by pulling your arms in you change your moment of inertia from 5.33 kg m2 to 1.60 kg m2?

–  Spacecraft and cats –  Jet engines not bolted too firmly in case of “seize-

up.”

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Angular Momentum

Lb = La

Ibωb = Iaωa

Rotational Vectors •  Direction of the L-vector •  Gyroscopic stability

–  Football spirals –  Bullets

•  Directional L conservation –  Bicycle wheel on platform

•  Precession –  ∆L = τ ∆t –  Gyroscope –  Diving airplanes –  Helicopters –  Motorcycle turns

Rotational Vectors •  Direction of the L-vector •  Gyroscopic stability

–  Football spirals –  Bullets

•  Directional L conservation –  Bicycle wheel on platform

•  Precession –  ∆L = τ ∆t –  Gyroscope –  Diving airplanes –  Helicopters –  Motorcycle turns

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The Big Picture Translational Rotational

s = rθ x = vt θ = ωt vt = rω v = vo + at ω = ωo + αt

at = rα

x = vot +12

at 2 θ = ω ot +

12αt2

v2 = vo

2 +2 ax ω2 = ω o

2 +2αθ

! F

! τ =! r ×! F

τ = rF sin θ m

I = miri2∑

! F NET =

! F ∑ = m ! a

! τ NET = I ! α

W = Fd cosθ W = τθ

K TOT =

12

mv 2 +12

Iω 2 K =

12 mv 2

K

rot= 1

2Iω 2

! p = m ! v

! L = I ! ω

F Δt = m Δv τ Δt = I Δω

Pav = ΔEΔt

P = Fv

P = τω