Circular Motion Tangential & Angular · PDF fileCircular Motion Tangential & Angular...
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Rick Field 2/6/2014 University of Florida PHY 2053 Page 1 Circular Motion Tangential & Angular Acceleration The arc length s is related to the angle θ (in radians = rad) as follows: • Tangential Acceleration: θ r s = θ ˆ ˆ t radial t radial tot a r a a a a + − = + = r r r α ω r dt d r dt dv a t t = = = dt d t t ω ω α = Δ Δ = → Δ 0 lim (radians/s 2 ) • Overall Acceleration: Tangential Velocity The tangential velocity v t is related to the angular velocity ω as follows: The tangential acceleration a t is related to the angular acceleration α as follows: a t a r Radial Axis r ω r v t = t v 2 2 t radial tot tot a a a a + = = r Radial Acceleration Tangential Acceleration
Transcript of Circular Motion Tangential & Angular · PDF fileCircular Motion Tangential & Angular...
Rick Field 2/6/2014 University of Florida PHY 2053 Page 1
Circular MotionTangential & Angular Acceleration
ωrvt =
The arc length s is related to the angle θ (in radians = rad) as follows:
• Tangential Acceleration:
θrs =
θ̂ˆ tradialtradialtot araaaa +−=+=rrr
αω rdtdr
dtdva t
t ===dtd
tt
ωωα =ΔΔ
=→Δ 0
lim (radians/s2)
• Overall Acceleration:
Tangential Velocity
The tangential velocity vt is related to the angular velocity ω as follows:
The tangential acceleration at is related to the angular acceleration α as follows:
at
ar Radial Axis
r
ωrvt =
tv
22tradialtottot aaaa +==
rRadial
Acceleration
Tangential Acceleration
Rick Field 2/6/2014 University of Florida PHY 2053 Page 2
at
ar Radial Axis
r
Angular Equations of Motion
• Angular Equations of Motion (constant α):
221
00)( ttt αωθθ ++=( )0
20
2 )(2)( θθαωω −=− tt
If the angular acceleration α is constant then
tt αωω += 0)(αα =)(t
Radial Acceleration
Tangential Acceleration
221
00)( tatvsts tt ++=
tavtv ttt += 0)(αrtat =)(
( )020
2 )(2)( stsavtv ttt −=−
)()( 2 trtaradial ω= rtvta tradial /)()( 2=
(radians/s2) (m/s2)
(radians/s)
(radians)
(m/s)
(m)
(m/s2) (m/s2)
Rick Field 2/6/2014 University of Florida PHY 2053 Page 3
Angular Equations of Motion
• Angular Equations of Motion (constant α):
221
00)( ttt αωθθ ++=( )0
20
2 )(2)( θθαωω −=− tt
Let N = Number of revolutions (rev)
tt αωω += 0)(αα =)(t
222
100 )()( ttfNtN π
α++=
tftf )()( 20 πα+=
πα2
( )022
02 )()(2)( NtNftf −=− π
α
(rad/s2)(rev/s2)
(rad/s)
(rad)
(rev/s)
(rev)
Let f = Number of revolutions per secondπ
ω2
)()( ttf =
πθ2
)()( ttN =
(frequency)
Rick Field 2/6/2014 University of Florida PHY 2053 Page 4
Angular Equations: Examples• A disk rotates about its central axis starting from rest at t = 0 and
accelerates with constant angular acceleration. At one time it is rotating at 4 rev/s; 60 revolutions later, its angular speed is 16 rev/s. Starting at t = 0, what is the time required to complete 64 revolutions?
Answer: t = 8 seconds
• An astronaut is being tested in a centrifuge. The centrifuge has a radius R and, in starting from rest at t = 0, rotates with a constant angular acceleration α = 0.25 rad/s2 . At what time t > 0 is the magnitude of the tangential acceleration equal to the magnitude of the radial acceleration (i.e. centripetal acceleration)?
Answer: t = 2 seconds
222
10 )()( tNtN π
α=−
( ) ( )2
22
0
20
2
/2602
)/4()/16()(2)(
2srev
revsrevsrev
NtNftf
=−
=−−
=πα
ssrev
revNtNt 8)/2()64(2
)())((2
22
0 ==−
=πα
ααω RatRtRta tradial ==== 222 )()(s
sradt 2
/25.011
2 ===α
Rick Field 2/6/2014 University of Florida PHY 2053 Page 5
Exam 2 Spring 2011: Problem 2• A race car accelerates uniformly from a speed of 40 m/s to
a speed of 58 m/s in 6 seconds while traveling around a circular track of radius 625 m. When the car reaches a speed of 50 m/s what is the magnitude of its total acceleration (in m/s2)?
Answer: 5% Right: 49%
smaaa
smmsm
Rva
sms
smsmttvva
rttot
r
t
/5
/4625
)/50(
/36
)/40()/58(
22
2212
12
=+=
===
=−
=−−
=
Rick Field 2/6/2014 University of Florida PHY 2053 Page 6
Gravitation: Circular Orbits (M >> m)
Fg
v M m r
3rGM
=ω
22
2 ωmrrvmma
rGmMF radialg ====
• Kepler’s Third Law:
For circular orbits the gravitational force is perpendicular to the velocity and hence the speed of the mass m is constant. The force Fg is equal to the mass times the radial (i.e. centripetal) acceleration as follows:
(speed, constant)Assume M >> m so that the
position of M is fixed!
(period of rotation)
2vGMr =
rGMv =
GMr
GMrr
vrT
3
222 πππ===
GMrT
322 4π=
(angular velocity, constant) For circular orbits r, v, and ω are also constant.
The period squared is proportional to the radius cubed.
(radius of the orbit, constant)
Fg vmM m ×
rm
CM
rMVM
In general both masses rotate about the center-of-mass and the formulas are more complicated!
Rick Field 2/6/2014 University of Florida PHY 2053 Page 7
Circular Orbits: Example
Answer: 8T
• Two satellites are in circular orbit around the Earth. The first satellite has mass M1 and is travelling in a circular orbit of radius R1. The second satellite has mass M2 = M1 is travelling in a circular orbit of radius R2 = 4R1. If the first satellite completes one revolution of the Earth in time T, how long does it take the second satellite to make one revolution of the Earth?
1
31
22
14GM
RT π= 2
11
31
2
1
31
2
2
32
22
2 64464)4(44 TGM
RGM
RGM
RT ====πππ
TTT 88 12 ==
Rick Field 2/6/2014 University of Florida PHY 2053 Page 8
Circular Orbits: Example• Two diametrically opposed masses m revolve around a
circle of radius R. A third mass M = 2m is located at the center of the circle. What is the period T of rotation for this system of three masses?
GmRT
3
34π
=
m
R
M = 2m m
Answer:
m
R
M = 2m m F( )
RvmmamM
RGm
RGm
RGmMF radialgrav
2
41
22
2
2 )2(==+=+=
GmR
GmRR
mmGRR
mMGRR
vRT
3
41
41 3
4942
)2(2
)(22 πππππ
==+
=+
==
RmMGv )( 4
1+=
