phy 113 c general physics i 11 am - 12:15 p m mwf olin 101 plan for lecture 11:
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PHY 113 C General Physics I 11 AM - 12:15 P M MWF Olin 101 Plan for Lecture 11: Chapter 10 – rotational motion Angular variables Rotational energy Moment of inertia Torque. Comments about Exam 1 Solutions posted online - PowerPoint PPT PresentationTRANSCRIPT
PHY 113 C Fall 2013-- Lecture 11 110/03/2013
PHY 113 C General Physics I11 AM - 12:15 PM MWF Olin 101
Plan for Lecture 11:
Chapter 10 – rotational motion
1. Angular variables
2. Rotational energy
3. Moment of inertia
4. Torque
PHY 113 C Fall 2013-- Lecture 11 210/03/2013
PHY 113 C Fall 2013-- Lecture 11 310/03/2013
Comments about Exam 1• Solutions posted online• Please review the problems while the ideas are
“fresh” in your mind
iclicker question (serious)In future exams, would you like to have an additional “take” home component? Of course it would be assumed that all portions of the exam would be subject to strict honor code guidelines
A. YesB. No
PHY 113 C Fall 2013-- Lecture 11 410/03/2013
Review:iclicker questionIn which of the following cases does gravity do positive work?
yf
yi
yi
yf
b.a.
PHY 113 C Fall 2013-- Lecture 11 510/03/2013
Review:iclicker questionIn which of the following cases does the work of gravity have a great magnitude?
rf
ri ri
rf
b.a.
PHY 113 C Fall 2013-- Lecture 11 610/03/2013
2
2
:Define
:mass ofcenter for relation General
dtdM
mMM
m
CMtotal
ii
ii
iii
CM
rFF
rr
Review of center of mass:
Example:
1 kg 3 kg
1 m
m 0.75m 4
13
:mass kg 1 torelative measured mass ofCenter
CM
ii
iii
CM m
mr
rr
PHY 113 C Fall 2013-- Lecture 11 710/03/2013
Center of mass example from Webassign
Finding the center of massFor a solid object composed of constant density material, the center of mass is located at the center of the object.
cm ˆ3333.13ˆ6667.11
cm 6
ˆ25215153ˆ25115253
1
ji
ji
rrr
i
ii
ii
iii
CM m
m
m
m
PHY 113 C Fall 2013-- Lecture 11 810/03/2013
Another Webassign question from Assignment #9
m/skgˆ6109.11ˆ8755.8
m/skgˆ99.3ˆ05.391.2
ji
jivp m
PHY 113 C Fall 2013-- Lecture 11 910/03/2013
Another Webassign question from Assignment #9
?
final initial
Tf
TfTCfCTiTCiC
ii
ii
mmmm
v
vvvv
pp
Ti
T
CfCiCTf m
mv
vvv
PHY 113 C Fall 2013-- Lecture 11 1010/03/2013
Examples of two-dimensional collision; balls moving on a frictionless surface
smsmsmv
smsmsm
vvvsmsm
vv
vmvm
vmvmvm
smv
smvkgmm
oof
o
fif
o
ff
ff
ffi
f
i
/11.188.17cos/060.1
88.17sin/342.0
88.71 060.1342.0tan
/060.120cos/1/2
coscos/342.020sin/1
sinsin
sinsin0
coscos
20 ,/1
,/2 ,06.0 :Suppose
1
o
211
21
2211
221111
o2
121
PHY 113 C Fall 2013-- Lecture 11 1110/03/2013
Angular motion
angular “displacement” (t)
angular “velocity” angular “acceleration”
dtd(t) ωα
dtd(t) θω
“natural” unit == 1 radian
Relation to linear variables: s = r (f-i)
v = r w
a = r a
so180 radians 3.14159 radians
PHY 113 C Fall 2013-- Lecture 11 1210/03/2013
Rotation at constant angular velocity w
w
r
v
v = r w
PHY 113 C Fall 2013-- Lecture 11 1310/03/2013
Special case of constant angular acceleration: a = a0:
wt = wi + a0 t
t = i + wi t + ½ a0 t2
( wt2 wi2 + 2 a0 t - i
w
r1v1=r1w
v2=r2w
r2
PHY 113 C Fall 2013-- Lecture 11 1410/03/2013
A wheel is initially rotating at a rate of f=30 rev/sec.
smRvR
πfω
/247.94)0.5m)(rad/s 495.188( 0.5m? radius aat wheel theof
rim on thedot a of speed theisWhat rad/s 188.495 rad/s )30(22
elocity?angular v theisWhat
w
R
PHY 113 C Fall 2013-- Lecture 11 1510/03/2013
A wheel is initially rotating at a rate of f=30 rev/sec. Because of a constant angular deceleration, the wheel comes to rest in 3 seconds.
22
2
/42.31)0.5m)(rad/s 83.62(
0.5m? radius aat wheel theofrim on thedot a ofon decelerati theisWhat
rad/s 83.26 3
rad/s )30(2320
on?deceleratiangular theisWhat
smRa
R
ssπf
a
a
R
PHY 113 C Fall 2013-- Lecture 11 1610/03/2013
Example: Compact disc motion
In a compact disk, each spot on the disk passes the laser-lens system at a constant linear speed of v = 1.3 m/s.
w1=v/r1=56.5 rad/s
w2=v/r2=22.4 rad/s
What is the average angular acceleration of the CD over the time interval Dt=4473 s as the spot moves from the inner to outer radii?
a = (w2-w1)/Dt =-0.0076 rad/s2
w1
w2
PHY 113 C Fall 2013-- Lecture 11 1710/03/2013
Object rotating with constant angular velocity (a = 0)
w
v=0
v=Rw
Kinetic energy associated with rotation:
“moment of inertia”
iii
iii
iii
rmI
IrmvmK
2
22
1222
122
1
:where
;ωω
R
PHY 113 C Fall 2013-- Lecture 11 1810/03/2013
Moment of inertia: i
iirmI 2
iclicker exercise: Which case has the larger I? A. a B. b
PHY 113 C Fall 2013-- Lecture 11 1910/03/2013
Moment of inertia: i
iirmI 2
22MaI 22 22 mbMaI
PHY 113 C Fall 2013-- Lecture 11 2010/03/2013
Note that the moment of inertia depends on both
a) The position of the rotational axisb) The direction of rotation
m m
d d
I=2md2
m m
d d
I=m(2d)2=4md2
PHY 113 C Fall 2013-- Lecture 11 2110/03/2013
iclicker question:
Suppose each of the following objects each has the same total mass M and outer radius R and each is rotating counter-clockwise at an constant angular velocity of w=3 rad/s. Which object has the greater kinetic energy?
(a) (Solid disk) (b) (circular ring)
PHY 113 C Fall 2013-- Lecture 11 2210/03/2013
Various moments of inertia:
solid cylinder:
I=1/2 MR2
solid sphere:
I=2/5 MR2
solid rod:
I=1/3 MR2
R
RR
PHY 113 C Fall 2013-- Lecture 11 2310/03/2013
Calculation of moment of inertia:
Example -- moment of inertia of solid rod through an axis perpendicular rod and passing through center:
2222
31
22MRdrr
RMrdr
RMrmI
R
R
R
Riii
R
PHY 113 C Fall 2013-- Lecture 11 2410/03/2013
iclicker exercise:Three round balls, each having a mass M and radius R, start from rest at the top of the incline. After they are released, they roll without slipping down the incline. Which ball will reach the bottom first?
AB C
2MRI A
22 5.021 MRMRIB
22 4.052 MRMRIC
PHY 113 C Fall 2013-- Lecture 11 2510/03/2013
22
21
21
:object rolling ofenergy kinetic Total
CM
CMrollingtotal
MvI
KKK
w
CMvRdtdR
dtds
dtd
w
w
: thatNote
22
222
21
21
21
CM
CM
CMrollingtotal
vMRI
MvRRI
KKK
w
PHY 113 C Fall 2013-- Lecture 11 2610/03/2013
iclicker exercise:Three round balls, each having a mass M and radius R, start from rest at the top of the incline. After they are released, they roll without slipping down the incline. Which ball will reach the bottom first?
AB C
2MRI A
22 5.021 MRMRIB
22 4.052 MRMRIC
PHY 113 C Fall 2013-- Lecture 11 2710/03/2013
Review of rotational energy associated with a rigid body
iii
iii
iii
iiirot
rmI
Irm
rmvmK
2
222
22
here w21
21
21
21
:energy Rotational
ww
w
PHY 113 C Fall 2013-- Lecture 11 2810/03/2013
Note that for a given center of rotation, any solid object has 3 moments of inertia; some times two or more can be equal
ji
k
iclicker exercise:Which moment of inertia is the smallest? (A) i (B) j (C) k
d dm m
IB=2md2 IC=2md2IA=0
PHY 113 C Fall 2013-- Lecture 11 2910/03/2013
222222 24222223 mkgxmIi
iiyy
From Webassign:
PHY 113 C Fall 2013-- Lecture 11 3010/03/2013
22
21
21
:object rolling ofenergy kinetic Total
CM
CMrottotal
MvI
KKK
w
CMvRdtdR
dtds
dtd
w
w
: thatNote
22
222
21
21
21
CM
CM
CMrottotal
vMRI
MvRRI
KKK
w
CM CM
PHY 113 C Fall 2013-- Lecture 11 3110/03/2013
iclicker exercise:Three round balls, each having a mass M and radius R, start from rest at the top of the incline. After they are released, they roll without slipping down the incline. Which ball will reach the bottom first?
AB C
2MRI A
22 5.021 MRMRIB
22 4.052 MRMRIC
PHY 113 C Fall 2013-- Lecture 11 3210/03/2013
How to make objects rotate.
Define torque:
t = r x F
t = rF sin
r
F
αarτFraF
Imm
sinr
F sin
Note: We will define and use the “vector cross product” next time. For now, we focus on the fact that the direction of the torque determines the direction of rotation.
PHY 113 C Fall 2013-- Lecture 11 3310/03/2013
Another example of torque:
PHY 113 C Fall 2013-- Lecture 11 3410/03/2013
clockwise)(counter :T from Torque
)(clockwise :T from Torque
222
2
111
1
TR
TR
t
t
clockwise)(counter 52
(1)(5)(0.5)(15)
: torqueTotal15N 0.5m,
5N 1m, :Example
1122
22
11
Nm .Nm
TRTR
TRTR
t
PHY 113 C Fall 2013-- Lecture 11 3510/03/2013
Newton’s second law applied to rotational motion
dtdM
dtdm CM
totali
ii
ii
vFvF
Newton’s second law applied to center-of-mass motion
mi Fi
ri
dtdm
dtdm i
iiiii
iivrFrvF
axis) principalabout rotating(for αωτ
rωrτ
rωvFrτ
IdtdI
dtdm
total
iiii
ii
iii
i
iidmI 2
di
PHY 113 C Fall 2013-- Lecture 11 3610/03/2013
An example:
A horizontal 800 N merry-go-round is a solid disc of radius 1.50 m and is started from rest by a constant horizontal force of 50 N applied tangentially to the cylinder. Find the kinetic energy of solid cylinder after 3 s.
K = ½ I w2 t I a w wi at = atIn this case I = ½ m R2 and t = FR
R F
JsN
Nmg
tFgRItFt
IFRIIK
Rg
mgItI
FRtIFR
625.275)3(80050m/s8.9
/21
21
21
21
22
222
2
2222
2
w
awa
PHY 113 C Fall 2013-- Lecture 11 3710/03/2013
Re-examination of “Atwood’s” machine
T1
T2
T2T1
IT1-m1g = m1a
T2-m2g = -m2a
t =T2R – T1R = I a = I a/R
R
212
12
212
12
/RIg τ
/g a
RImmmm
RImmmm
PHY 113 C Fall 2013-- Lecture 11 3810/03/2013
Another example: Two masses connect by a frictionless pulley having moment of inertia I and radius R, are initially separated by h=3m. What is the velocity v=v2= -v1 when the masses are at the same height? m1=2kg; m2=1kg; I=1kg m2 ; R=0.2m .
m1 m2v1 v2
hh/2
sm
RImmmmv
hgmhgm
vvmvmghm
UKUK
RI
ffii
/19.02.0/112
12
/
0
:energy ofon Conservati
2
221
21
21
221
1
2212
2212
121
1 2
PHY 113 C Fall 2013-- Lecture 11 3910/03/2013
Kinetic energy associated with rotation:
i
iirot rmIIK 2221 w
Distance to axis of rotation
Rolling: rotcomtot KKK
222
1 1
:slipping no is thereIf
comtot
com
vMR
IMK
Rv
w
Rolling motion reconsidered:
PHY 113 C Fall 2013-- Lecture 11 4010/03/2013
Kinetic energy associated with rotation:
i
iirot rmIIK 2221 w
Distance to axis of rotation
Rolling: rotcomtot KKK
222
1 1
:slipping no is thereIf
comtot
com
vMR
IMK
Rv
w
Rolling motion reconsidered:
PHY 113 C Fall 2013-- Lecture 11 4110/03/2013
Newton’s law for torque:
αωτ IdtdItotal
F
fs
FfMRI
IMRFf
IRfaRIaIRf
MafF
s
s
sCMCMs
CMs
312
21
2
2
cylinder, solid aFor
/11
/
a
Note that rolling motion is caused by the torque of friction:
PHY 113 C Fall 2013-- Lecture 11 4210/03/2013
Bicycle or automobile wheel:
t
fs
τ/Rf
MRI
I/MRτ/Rf
RIaIfMaf
s
s
CMs
CMs
21
For
1
/αR-τ
2
2
PHY 113 C Fall 2013-- Lecture 11 4310/03/2013
iclicker exercise:What happens when the bicycle skids?
A. Too much torque is appliedB. Too little torque is appliedC. The coefficient of kinetic friction is too smallD. The coefficient of static friction is too smallE. More than one of these