Upload
others
View
24
Download
0
Embed Size (px)
Citation preview
1
Physics 201, Lecture 18
Today’s Topics
q Rotational Dynamics
§ Torque q Exercises and Applications
§ Rolling Motion
q Hope you have previewed Chapter 10. (really!)
Review Angular Velocity And Angular Acceleration
q Angular Velocity (ω) describes how fast an object rotstes, it has two components: § Angular speed: and
§ direction of ω: + counter clockwise - clockwise
Ø All particles of the rigid object have the same angular velocity q Angular Acceleration (α):
and
Note: the similarity between (θ,ω,α) and (x, v, a)
ωave ≡ΔθΔt
€
ω ≡Δt→ 0limΔθ
Δt=dθdt
αave ≡ΔωΔt
€
α ≡Δt→ 0limΔω
Δt=dωdt
è Angular velocity ω is a vector! (define direction next page)
è Angular acceleration α is also a vector!
axis Review: Moment of Inertia
q Moment of Inertia of an object about an axis
(unit of I : kgm2)
Ø “I” depends on rotation axis, total mass, and mass distribution.
∑≡ 2iirmI :Inertia ofMoment
€
another form: I ≡ r2dmwhole object∫
Moments Of Inertial Of Various Objects
€
I = miri2∑ (= r2dm)∫
2
Which Has Larger Moment of Inertia?
q Which of the following configurations has larger I Ø Central or side axis ?
(Trivial) Quick Quiz q A force is acting on a rigid rod around a fixed axis. Ø Which of the follow case(s) will not turn.
Ø Effective Turning: Force, direction, acting point(action length).
F F F
F
Turning counter clockwise
Turning clockwise No Turn No Turn
axis
Torque: Effect of Force on Rotation q Torque:
§ Magnitude: τ = Fsinφ r , depends on F , r, and sinφ § Direction:
• conventional: – clockwise = “-”, – counter-clockwise = “+”
• More strictly: Right Hand Rule for cross-product
q The angular acceleration of an object is proportional to the torque acting on it Στ = I α
I: Moment if inertia (Tuesday)
τ =r ×F
Lever Arm
q τ ≡ Fsinφ r = F d
axis (pivot)
acting point
Lever Arm d = r sinφ
The lever arm, d, is the perpendicular distance from the axis of rotation to a line drawn from the direction of the force
3
Alternative View of Torque q τ ≡ Fsinφ r = F⊥ r
q Only the perpendicular component of the force contributes to the torque
axis (pivot)
acting point
Summary: Two Views of Torque q τ ≡ Fsinφ r = F⊥ r = F r⊥
F⊥ r F r⊥
axis (pivot)
acting point
Torque Has a Direction q Torque is a vector. It has a magnitude and a direction. q For fixed axis rotation, the direction of torque can be described by
a sign (+/-) in one dimension; (and by right hand rule in general)
Ø Convention: Counter-clockwise : +, clockwise - .
τ > 0
τ < 0
τ > 0
τ > 0
The Acting point of Gravity: Center of Gravity
q The force of gravity acting on an object must be considered in determining equilibrium q In finding the torque produced by the force of gravity, all of the weight of the object can be considered to be concentrated at one point called center of gravity (cg) q Effectively, assuming gravitation field is uniform, the CG of an object is the same as its CM (that is usually true at the Phy103 level)
xcg =ΣmixiΣmi
and ycg =ΣmiyiΣmi
See demo: finding CG
4
pivot
L/2
θ
Quick Exercise: Calculating Torque q As shown, a pencil is falling down under gravity. What is the
torque (about the pivot) by the gravity?
q Answer: + Fgrav cosθ L/2 (note: why cosθ?)
Rotational Dynamics
Angular Linear
tαωω += 0atvv += 0
221
00 tt αωθθ ++=2
21
00 attvxx ++=
θαωω Δ+= 220
2 xavv Δ+= 220
2
Iτ
α =mFa =
Στ = I α Rotational Dynamics compared to 1-D Dynamics
KE = 12Iω 2 KE = 1
2mv2
Exercise: Pulley with Mass q A crate of mass mcrate is hanging on a pulley of mass mpulley and
radius Rpulley as shown. What is the acceleration of the crate? Solution: 1: per FBD for the crate T-mcrateg = mcratea 2: for the pulley: τ = -TR = Ipulley α 3: connection: a=αR Solve: a = - mcrateg/(mcrate+Ipulley/Rpulley
2) Get T and α yourself after class
Please make sure you understand this
Combined Translational and Rotational Motion
q Generally, the motion of an extended object is a combination of the translational motion of the CM and the rotation about the CM
KEtot = KEtrans + KErot = ½ MvCM2 + ½ Iω2
5
Combined Translation and Rotation q Combined translational and rotational motion.
Ø At point A (top). vrot = Rω to the right. vtop = vCM + Rω Ø At point B (bottom). vrot = Rω to the left. vbottom = vCM - Rω
Ø At arbitrary point C
CM: moving linearly (1-D)
Everything on the wheel rotation about CM
R
v = vCM +vrot
vCM
A
B
C
vrot
vrot
vCM
v
Rolling/Sliding Conditions q Recall: vbottom = vCM - Rω q Depending relative size of Vcm and Rω, there can be three
classes of rolling /sliding conditions.
vcm
ω
vbottom = vcm - Rω =0 vcm = Rω
vcm
ω
vbottom = vcm – Rω<0 vcm < Rω
vcm
ω
vbottom = vcm – Rω>0 vcm > Rω
“spinning” “pure rolling” (rolling w/o slipping)
“sliding”
“Pure” Rolling Motion (Rolling without Slipping)
q Rolling motion refers to a form of combined translational and rotational motion.
Ø Condition for rolling w/o slipping:
vCM = ωR and aCM = αR
CM: moving linearly (1-D)
wheel: rotation about CM
No slipping on road i.e. vbottom =0
R vCM
Quick Quiz and Demo q Consider a wheel in pure rolling without slipping. After a full
resolution (i.e. in period T), how far the CM moves?
q R, 2R, πR, 2πR, other
See Demo
CM: moving linearly (1-D)
R vCM = Rω=R2π/T ΔxCM =VCM*T =2πR
6
Trajectory of a Point on the Rim of a Pure-Rolling Wheel Quick Quiz: Rolling Without Slipping
q Consider a wheel rolling down a (not smooth) hill without slipping.
How many (external) forces are acting on the wheel? 2, 3, more than 3, other q In the process, the work down by friction is Positive, negative, zero q Now consider a wheel rolling on a (flat but non smooth)
horizontal plane without slipping. How many external forces are acting on it?
2, 3, more than 3, other Why? See next slide
See Demo
fs
mg
N
mg
N
Exercise: Rolling w/o Slipping Down a Slope
q A uniform disc (or wheel, or sphere) of mass M , radius R, and moment of inertia I is rolling down a slope without slipping as shown. Calculate its CM acceleration.
q Solution: Ø Step 1: FBD as shown Ø Step 2: Set up axis as shown Ø Step 3: Dynamics for CM (x direction): mgsinθ – fs = maCM Ø Step 4: Dynamics for rotation: -fsR =- Iα Ø Step 5: rolling w/o slipping: Rα=aCM Ø Solve for unknowns:
fs
mg
N
θ
x
aCM =gsinθ
1+ ImR2
, fs =mgsinθmR2
I+1
Results Discussion: Rolling Down a Slope (w/o slipping)
Consider a wheel rolling down a flat (but not smooth) slope. Ø On a slope, a friction is necessary to keep it from slipping Ø For same mR2, the larger the I, the slower it moves. Ø Spheres (or wheels or discs) of the same shape and mass
distribution roll at the same speed regardless of their size and mass. Ø On a horizontal flat surface, the friction reduces to zero and the
rolling can go forever even when the surface is not smooth!
fs
mg
N
θ
x aCM =gsinθ
1+ ImR2
, fs =mgsinθmR2
I+1
7
Moments Of Inertial Of Various Objects
€
I = miri2∑ (= r2dm)∫
See demo for rolling of wheels with different I
q General Work: W = F · Δs = Fin_direction_of_Δs Δs q For an object rotating about a fixed axis, all mass elements are
moving in the tangential direction : W = Ft Δs = rFt Δs/r = τ Δθ P = W/Δt = τ Δθ/Δt = τω Ø Rotational version of Work-Energy theorem
2212
21
if
rot
IIKEW
ωωθτ −=Δ
Δ=
Work and Power by Torque (Self Reading) Δs
Δs
Δθ