October 17, 2013 University Physics, Chapter 9 1
Longitud de arco
s = path on the perimeter of the circle for the tip of the radius vector by going from zero to θ
Define arc length (Please note: θ has to be measured in radians.)
Arc length is a vector and has the dimension of a length
Special case of above relation: Perimeter of circle: c=2πr
s = rθ
October 17, 2013 University Physics, Chapter 9 2
Example: longitud de pista en un CD (1) CDs have a track separation of 1.6 µm. Microscope pictures (1500x magnification) of a factor
pressed and a read-write CD:
Track is spiral, starts at an inner radius of 25 mm and ends at an outer radius of 58 mm.
Question: ¿Cuál es la longitud de la pista?
© Comstock/Punchstock RF
October 17, 2013 University Physics, Chapter 9 3
longitud de pista en un CD (2) AFM Picture:
October 17, 2013 University Physics, Chapter 9 4
longitud de pista en un CD(3) Answer: At a given value of r the track is almost perfectly circular and
has a length of 2πr per turn. Track density (number of groves per length unit):
Track length is result of integration
Numbers:
λ =1∆r
=1
1.6 ⋅10−6 m= 625,000 m-1
L = λ 2πrdr = λπr2
r1
r2
∫r1
r2
= λπ r22 − r1
2( )
L = (625,000 m-1)π (0.058 m)2 − (0.025 m)2( )= 5, 378 m
October 17, 2013 University Physics, Chapter 9 5
Angular Velocity
Linear motion: rate of change of displacement is velocity
Angular motion: rate of change of angular displacement is angular velocity • Average:
• Instantaneous:
Unit: Direction: right-hand rule
ω =θ2 −θ1
t2 − t1
=∆θ∆t
ω = lim∆t→0
ω =dθdt
ω[ ]= rad/s
October 17, 2013 University Physics, Chapter 9
6
Frequency
Frequency, f, measures numbers of turns around the circle Example: rpm on tachometer Since 1 turn = 2π radians:
Unit: [f]=1/s In honor of Heinrich Rudolf Hertz (1857-1894): 1/s = 1 Hz Period, T:
Relationship with angular velocity:
f = ω2π
⇔ω = 2π f
T =1f
ω = 2π f = 2πT
October 17, 2013 University Physics, Chapter 9
7
Example: Earth (1)
Question: The Earth orbits around the Sun and revolves around its own pole-to-pole
axis. What are the corresponding angular velocities, frequencies, and linear speeds?
Answer: Any point on the surface of Earth moves in circular motion around the
rotation (pole-to-pole) axis, with a rotation period of 1 day. Expressed in seconds, this period is
The Earth moves around the Sun on an elliptical path, which is very close to circular. The orbital period for the motion of the Earth around the Sun is 1 year. If we express this period in seconds, we obtain:
Tearth = 1 day ⋅24 hour
day⋅
3600 shour
= 8.64 ⋅104 s
Tsun = 1 year ⋅365 day
year⋅24 hour
day⋅
3600 shour
= 3.15 ⋅107 s
October 17, 2013 University Physics, Chapter 9 8
Example: Earth (2)
Both circular motions have constant angular velocity. Thus, we can use T=1/f and ω=2πf to obtain our answers:
Sidereal time: 24-hour period used as length of day = period for Sun to have the same position in the sky. Earth also moves around Sun during these 24 hours => it takes the Earth to complete rotation so that the stars have same position again in the night sky is only 23 hours, 56 minutes and a little more than 4 seconds, or 86,164.09074 s = (1-1/365.2425)·86,400 s.
fearth = 1 / Tearth = 1.16 ⋅10−5 Hz; ω earth = 2π fearth = 7.27 ⋅10−5 Hz
fsun = 1 / Tsun = 3.17 ⋅10−8 Hz; ω sun = 2π fsun = 1.99 ⋅10−7 Hz
leap year included
October 17, 2013 University Physics, Chapter 9
9
Example: Earth (3)
Speed of the Earth orbiting the Sun (assume circular orbit):
Orbital radius (= 1 astronomical unit):
Put in the numbers: (unos 1.07 105 km/h)
v = rω
rsun−earth = 1.49 ⋅1011 m
v = rω = (1.49 ⋅1011 m) ⋅ (1.99 ⋅10−7 s-1) = 2.97 ⋅104 m/s
!!!
October 17, 2013 University Physics, Chapter 9 10
Orbital speed of a point on the surface of the Earth due to the rotation of the Earth around itself
Radius of Earth:
Orbital radius as function of latitude:
Orbital speed:
Santander (θ=43.5˚): v=337 m/s; Miami (θ=25.7˚): v=418 m/s
Example: Earth (4)
Rearth = 6,380 km
r = Rearth cosθ
v = ωr = ωRearth cosθ= (7.27 ⋅10−5 s-1)(6.38 ⋅106 m)cosθ= (464 m/s)cosθ
October 17, 2013 University Physics, Chapter 9 11
Direction of Acceleration Vector
October 17, 2013 University Physics, Chapter 9 12
Centripetal Acceleration on Earth
Question: What is the correction to gravity due to the centripetal acceleration on the surface of Earth?
Answer:
Correction aprox. between 0.34 percent (at the equator) and 0 percent (at the poles)
ac = ω2r = ω 2Rearth cosθ
= (7.27 ⋅10−5 s-1)2 ⋅ (6.37 ⋅106 m) ⋅ cosθ= (0.034 m/s2 ) ⋅ cosθ
October 17, 2013 University Physics, Chapter 9 13
Constant Angular Acceleration
Kinematical equations for constant angular acceleration are obtained in complete analogy to those for linear motion with constant acceleration
θ = θ0 +ω0t + 12αt 2
θ = θ0 +ωtω =ω0 +αtω = 1
2 (ω +ω0 )ω 2 =ω0
2 + 2α(θ −θ0 )
October 17, 2013 University Physics, Chapter 9 14
Example: Lanzamiento de martillo(1)
Litvinov took 7 turns before releasing hammer. These took 1.52 s, 1.08 s, 0.72 s, 0.56 s, 0.44 s, 0.40 s, and 0.36 s
Frequency increased linearly with each turn => constant angular acc.
Question: What is value of α?
Answer: Total time = 1.52 s+…+ 0.36 s = 5.08 s; Total angle: rad
Constant angular acceleration: θall = 7 ⋅2π = 14π ≈ 44.0
θ = 12α t 2 ⇒α =
2θ t 2 =
2 ⋅ 44.0(5.08 s)2 = 3.41 s-2
October 17, 2013 University Physics, Chapter 9 15
Lanzamiento de martillo(2)
Question: Assuming that the radius of the circle on which the hammer moves is 1.67 m (= length of hammer + arms of the athlete), what is the linear speed with which the hammer gets released?
Answer: Con α constante, partiendo del reposo y durante 5.08 s, la ω final es
Magnitud de la velocidad:
ω = α t = (3.41 s-2 ) ⋅ (5.08 s)=17.3 s-1
v = rω = (1.67 m) ⋅ (17.3 s-1) = 28.9 m/s
October 17, 2013 University Physics, Chapter 9
16
Lanzamiento de martillo(3)
Question: What is the centripetal acceleration before the hammer gets released?
Answer: The centripetal acceleration right before
release is given by:
With a mass of 7.26 kg for the hammer, the centripetal force required is then
Same force as weight of 370 kg object!
ac = vω = (28.9 m/s) ⋅ (17.3 s-1) = 500. m/s2
Fc = mac = (7.26 kg) ⋅ (500 m/s2 ) = 3630 N
