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PS 11 GeneralPhysics I for the Life Sciences. Rotational Motion Dr. Benjamin chan Associate Professor Physics Department february 2014. Questions and Problems for Contemplation. Chapter 8 Questions: 1, 4, 8, 10, 15, 18, 22, 24 - PowerPoint PPT Presentation
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ROTATIONAL MOTIOND R . B E N J A M I N C H A N
A S S O C I AT E P R O F E S S O RP H Y S I C S D E PA R T M E N T
F E B R UA RY 2 0 1 4
PS 11 GeneralPhysics I for the Life Sciences
Questions and Problems for Contemplation
Chapter 8 Questions: 1, 4, 8, 10, 15, 18, 22, 24 Problems: 1, 2, 4, 8, 10, 15, 22, 27, 30, 38, 45, 52, 55,
61, 64 General Problems: 72, 80, 81
Describing Rotational Motion
Angular displacement q Let O be the axis of rotation How far the object has rotated
Only 2 directions possible: clockwise(-) and counter-clockwise(+)
Measured in radians 1 radian (rad) is the angle subtended
by an arc whose length is equal to the radius of motion
rl
θ
Distance traveled
Arc length traversed
For one complete revolution
q can be expressed in revolutions
rl 2
qrl
radrev 21
Example: Bike Wheel
A bike wheel rotates 4.5 revolutions. How many radians has it rotated?
If the wheel has a diameter of 45 cm, what is the distance traveled by a point on the rim of the wheel?
radrev
radrevs 28
25.4θ
cmradcmdl 630282
452
q
Example: Bird of Prey
A bird’s eye can distinguish objects that subtend an angle no smaller than 3 10-4 rad. How many degrees is this?
How small an object can the bird just distinguish when flying at a height of 100m? For small angles (<15), arc length
and chord length are nearly the same
017.02360103θ 4
radrad
cmradmrl 3)103)(100( 4 q
Angular Velocity w
Average w
Instantaneous w Dt must be very small
Velocity v of a point on a rotating wheel
Changes direction as vector turns Increases in proportion to distance from the axis
of rotation
tDD
qw
wrv
Angular Acceleration a
Average a
Instantaneous a Make Dt as small as possible
Tangential acceleration
Radial acceleration
ara tan
tDD
wa
222 )( ww r
rr
rvaR
Review of Linear and Angular Quantities
Frequency = number of complete revolutions per second = f w = 2f
Period = time required to complete one revolution = T = 1/f
Equations of Motion
Zero angular acceleration a = 0, w = constant Uniform circular motion q = wt + qo
Linear velocity is not constant Magnitude is constant: v = wr Direction is changing
Acceleration is not constant atan= 0 but aR= rw2 = constant (centripetal) Direction is changing
Example: Earth’s Rotation
How fast is the earth’s equator turning? w = 2/T = (2 rad)/84,600s = 7.27 x 10-5 rad/s v = rw = (6,380 km)(7.27 x 10-5 rad/s) = 464 m/s
How will your speed change as you go to the North or South pole? v = (r cos f)w = (464 cos f) m/s f = 14.5°, v = 449 m/s f = 30°, v = 402 m/s f = 60°, v = 232 m/s f = 90°, v = 0 m/s
The Coriolis Effect
As you go from the equator towards the N pole, you are moving faster than the ground you are moving into: veer to your right (earth rotates west to east)
As you go from the N pole towards the equator, you are moving slower than the ground you are moving into: veer to your right Clockwise flow!
To or from the S pole: veer to the left! Counterclockwise flow!
Example: Hard Drive
The platter of the hard drive of a computer rotates at 7200 rpm. What is the angular velocity of the platter?
If the reading head of the drive is 3.00 cm from the axis of rotation, how fast is the disk moving right under the head?
srad
srevrev
f 7541202sec60min1
min7200
22 w
smsradmrv /6.22)/754)(103( 2 w
Example (continued)
If a single bit requires 0.50 mm of length along the direction of motion, how many bits per second can the writing head write when it is 3.00 cm from the axis? The number of bits passing the head per second is
or 45 megabits/s (Mbps)
sbitsbitm
sm /1045/1050.0
/6.22 66
Constant Angular Acceleration
a = constantw = wo + atq = qo + wot + ½ at2
Eliminate t between w and q w2 = wo
2 + 2aq
Total Acceleration
atotal = atan + aR
atan Constant magnitude,
changing directionaR
Variable magnitude, variable direction
Example: Centrifuge
A centrifuge motor is accelerated from rest to 20,000 rpm in 30s. Determine its angular acceleration and how many revolutions it makes while it is accelerating.
Solution Assuming constant angular acceleration
2/7030
0/2100 srads
sradt
o
wwa
Example continued
Where the final angular velocity w is
The angular displacement in 30s is then
We divide by 2 to convert to revolutions
sradms
revrevradf /2100
/60min/2000022
w
radssrad 422 1015.3)30)(/70)(2/1(0 q
revrevradrad 34
100.5/21015.3
q
Rolling Motion
Translational + rotational motion
No Slipping Static friction between object
and rolling surfacewrv
Example: Bicycle
A bicycle slows down uniformly from a velocity of 8.40 m/s to rest over a distance of 115 m. The overall diameter of the tire is 68.0 cm. Determine the initial angular velocity of the wheels.
sradmsm
rvo
o /7.24340.0
/40.8w
Example (continued)
Determine the number of revolutions each wheel undergoes before stopping. The rim of the wheel turns 115m before stopping.
Thus,
Determine the angular acceleration of the wheel
revm
mrm 8.53
)340.0(2115
2115
2222
/902.0)8.53)(/2(2
)/7.24(02
sradrevrevrad
srado
q
wwa
Example continued
Determine the time it took the bicycle to stop
Note: when the bike tire completes one revolution, the bike advances a distance equal to the outer circumference of the tire (no slipping or sliding).
ssradsradt o 4.27/902.0/7.2402
aww
Announcements
FINAL EXAM Wednesday, March 19 7.30 - 10.30 F-113
Long Test 4 Thursday, March 13 6.00 – 7.30 Room TBA c/o Paulo
Center of Mass
You can reduce an object to a point and describe its translational motion by considering the motion of this point (called its center of mass)
Determining Center of Mass
Consider masses m1, m2, m3, … with coordinates (x1, y1), (x2, y2), (x3, y3), …
ii
iii
cm m
xm
mmmmxmxmxx
321
332211
ii
iii
cm m
ym
mmmmymymyy
321
332211
CM for a Leg
Determine the center of mass of a leg when a) stretched out and b) bent at 90°. Assume the person is 1.70 m tall.
Solution a) Straight leg
Essentially 1-D Measure distance from hip joint
CM is 52.1-20.4 = 31.7 units from base of foot For a height of 172 cm, xcm = 54.5 cm above the bottom of
the foot
4.204.36.95.21
)3.50)(4.3()9.33)(6.9()6.9)(5.21(
cmx units
CM of Leg
b) Bent leg
For a height of 172 cmxcm = (172 cm)(0.149) = 25.6 cmycm = (172 cm)(0.23) = 39.6 cm Center of mass of bent leg is 39.6 cm
above the floor and 25.6 cm from thehip joint!
9.144.36.95.21
)6.23)(4.3()6.23)(6.9()6.9)(5.21(
cmx units
0.234.36.95.21
)5.28)(5.21()2.18)(6.9()8.1)(4.3(
cmy units
CM Trajectory
Center of mass of swimmer in flight follows projectile motion (parabolic) trajectory
Center of mass of wrench follows constant velocity trajectory
Torque
What causes an object to rotate?Torque = force x lever arm
q sinrFFr
More Torque
Units: Nm (Newton-meter) Reserve J for work and energy
Torque is a vector quantity Direction determined by the right hand rule
Newton’s First law
Translational EquilibriumAll forces cancel out: SF = 0
Rotational EquilibriumTorques must balance out: SG = 0
When is an object in equilibrium?
Newton’s Second Law
F = maG = Ia
I = moment of inertia a = angular acceleration Only two possible directions
Counter-clockwise rotation Clockwise rotation
Moment of Inertia of Particles
For a single moving object with mass m = rF = rma = rmra
=mr2a I = mr2
For several objects rigidly attached to each other S = (Smiri
2)a I = Smiri
2
Changing Moment of Inertia
Determine the change in the moment of inertia of a particle as the radius of its orbit doubles
Solution
It increases by
41
4)2( 2
2
2
2
mRmR
RmmR
II
f
i
%300%100114%100
i
if
III
Changing Your I
Vertical axis of rotationArms on the side
R = 25 cm, M = 9.6 kg
Raise your arms in a crucifixion pose R = 57.5 cm, M = 9.6 kg
433% increase
222 60.0)25.0)(6.9( mkgmkgMRIside
222 20.3)575.0)(6.9( mkgmkgMRIcross
Moments of Inertia for Various Objects
Rotational Kinetic Energy
22
2122
212
21 )()()( ww iiiiii rmrmvmKE
221 wIKErotational
2221
21
CMCM MvIKETotal w
Example: Ball Rolling Down an Inclined Plane
Determine the speed of a solid sphere of mass M and radius R when it reaches the bottom of an inclined plane if it starts from rest at a height H and rolls without slipping. Assume no slipping occurs. Compare the result to an object of the same mass sliding down a frictionless inclined plane.
Solution
Initial mechanical energy PE = MgH KEtrans = 0 KErot = 0
Final mechanical energy PE = 0 KEtrans = ½ Mv2
KErot = ½ Iw2
Conservation of EnergyMgH = ½ (Mv2 + Iw2)
Solution (continued)
I = (2/5)MR2 for a solid sphere rotating about an axis through its center of mass
w = v/RThus
MgH = ½ Mv2 + ½(2/5)MR2(v/R)2
(1/2 + 1/5) v2 = gHv = [(10/7)gH]1/2
v does not depend on the mass and radius of the sphere!!
Frictionless Incline
Ball slides down the incline and does not rollThus,
½ Mv2 = MgHv = (2gH)1/2
The speed is greater! None of the original PE is converted into rotational
energy.
Work Done on a Rotating Body
W = FDl = F rDqW = Dq
PowerP = W/DtP = Dq/Dt = w
Angular Momentum L
L = IwNewton’s second law becomes
Thus, tII
tII o
D
DD
wwwa
tL
DD
Conservation of Angular Momentum
If the net torque acting on a rotating object is zero, then its angular momentum remains constant.
fi
if
LLLL
tL
DD
0
0
ffii II ww
The Ice Skater
How can the ice skater spin so fast?
if
if I
I ww
The Diver
How can the diver make somersaults?Does she have to rotate initially?What trajectory does she follow?
The Hanging Wheel
Why is the wheel standing up?Why does it turn around about the point of
support?
Rotating Disk Demo
What happens when you tilt the rotating disk?
dtdL
HINT:
Drunk Driver Test/Tightrope Artist
Follow the line walkIncrease your moment of inertia to minimize
rotations
Quiz 8
1. A 4 kg mass sits at the origin, and a 10 kg mass sits at x = + 21 m. Where is the center of mass on the x-axis?
(a) + 7 m (b) + 10.5 m (c) + 14 m (d) + 15 m2. An object moving in a circular path experiences
(a) free fall. (b) constant acceleration. (c) linear acceleration. (d) centripetal acceleration.3. A boy and a girl are riding on a merry-go-round which is turning
at a constant rate. The boy is near the outer edge, and the girl is closer to the center. Who has the greater angular velocity?
a) The boy b) The girl c) Both have the same non-zero angular velocity. d) Both have zero angular velocity.
Quiz 8
4. A wheel starts at rest, and has an angular acceleration of 4 rad/s2. Through what angle does it turn in 3 s?
a) 36 rad b) 18 rad c) 12 rad d) 9 rad 5. A wheel of diameter 26 cm turns at 1500 rpm. How far will a point on
the outer rim move in 2 s? a) 314 cm b) 4084 cm c) 8995.5 cm d) 17990.8 cm
6. What is the centripetal acceleration of a point on the perimeter of a bicycle wheel of diameter 70 cm when the bike is moving 8 m/s?
a) 91 m/s2 b) 183 m/s2 c) 206 m/s2 d) 266 m/s2
7. A bicycle is moving 4 m/s. What is the angular speed of a wheel if its radius is 30 cm?
a) 0.36 rad/s b) 1.2 rad/s c) 4.8 rad/s d) 13.3 rad/s
Quiz 8
8. An ice skater is in a spin with his arms outstretched. If he pulls in his arms, what happens to his kinetic energy?
a) It increases. b) It decreases. c) It remains constant but non-zero. d) It remains zero. 9. What is the quantity used to measure an object's resistance to
changes in rotation? a) mass b) moment of inertia c) linear momentum d) angular momentum
10. A wheel of moment of inertia of 5.00 kg-m2 starts from rest and accelerates under a constant torque of 3.00 N-m for 8.00 s. What is the wheel's rotational kinetic energy at the end of 8.00 s?
a) 57.6 J b) 64.0 J c) 78.8 J d) 122 J