Transcript Physics 101 Fall 02 - Youngstown State University

Chapter 5: Circular Motion
Rotating Objects:
Wheels, moon, earth, CDs, DVDs etc.
Rigid bodies.
Description of circular motion.
Angular Position, Angular Displacement
s

r
Angle (in radians)  = arc length/radius = s/r.
 Angular displacement D = 2-1
How far from initial angular position it has
rotated.
 D = positive going CCW, negative going CW.
 Magnitude D = Ds/r. If Ds = r, then D = 1 radian

Angular Displacement
360o
2 rads
180o
 rads
90o
/2 rads
In 1 complete turn,
Ds = circumference of the circle = 2r.
 In 1 turn, D = Ds/r = (2r)/r = 2 rads.
 360o = 2 rads, 180o =  rads, 90o = /2 rads.

Angular Velocity (w)
w

Average Angular velocity w = D/Dt
 How fast it is rotating
 It is a vector quantity.
 Direction: Up for CCW rotation, down for CW
rotation. [ Right Hand Rule].
 Units = radians/second.
Examples
1.Change 1 rpm to rad/sec.
2. What is the angular speed of
the earth’s rotation?
3. What is the angular speed of
the minute hand of an analogue
clock?
Circular to Linear
 Linear
distance = Arc length Ds = r D
( in radians)
 Speed
|v| = Ds/Dt = r D/Dt = rw
 Direction of v is tangent to circle
Example:
1. A CD spins with angular speed
20 radians/second. What is the
linear speed 6 cm from the center
of the CD?
V = r w = 0.06 x 20 = 1.2 m/s
Example:
2. If you were seated at the equator,
what would be your linear speed due
to the rotation of the earth? (radius
of the earth = 6.371 x 106 m.
v = rw
w = D/Dt = 2/(24x3600) = 7.27 x 10-5 rad/s
V = 6.37 x 106 x 7.27 x 10-5
= 463 m/s
Circular Motion Act
b
a
c
v
Answer: b
A ball is going around in a circle attached to a
string. If the string breaks at the instant shown,
which path will the ball follow?
Frequency (f)
 Frequency (f) = number of turns a rotating
object makes in one second.
 f = # of turns/time taken to make the turns.
 Unit: Cycles per second or hertz (Hz)
 1 cycle/second = 1 Hz
 Eg 1. The frequency of the second hand of an
analogue clock = 1 turn/60s = 1/60 Hz.
 The frequency of the moon’s rotation around
the earth = 1 turn/28 days = 4.13 x 10-7 Hz.
Period (T)
 Period (T) = Time taken to make one cycle.
 T = Time to make one turn.
 Unit: seconds
 Since f = # of cycles per second, Period and
frequency are related by the expression
 f = 1/T or T = 1/f
 Eg 1. The period of the second hand of an
analogue clock = 60 seconds.
 The period of the moon’s rotation around the
earth = 28 days = 2.42 x 106 s.
1. An angular displacement of one radian is
when
A.
B.
C.
D.
Arc length > radius
Arc length < radius
Arch length = radius
Circumference = 2r
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2. In moving through 30 minutes, the minute
hand of an analogue clock has gone through
A.
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C.
D.
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180 rads
0.5 rads
1.57 rads
3.14 rads
None of the
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3. A small ball swung in a circular path takes
12 seconds to make 20 rotations. What is the
frequency of its motion?
A.
B.
C.
D.
0.60 Hz
1.67 Hz
240 Hz
12.0 Hz
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4. A CD has a diameter of 12 cm. if the CD is
rotating at a constant frequency of 4 Hz., then
the period of its rotation is
A.
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C.
D.
E.
0.250 s
0.500 s
0.750 s
1.00 s
1.25 s
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5. The moon rotates around the earth once in
28 days. What is its average angular velocity?
A.
B.
C.
D.
0.224 rad/s
9.34x10-3 rad/s
2.60x10-6 rad/s
1.49x10-4 rad/s
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Relation between w, v, f and T
r
v
In one complete rotation:
Distance moved = 2r.
Time taken = period, T
Speed v = dist/time = 2r/T
ie v = 2r/T = 2rf
Since v = rw, we have w = 2/T = 2f
Uniform Circular Motion
• Uniform Circular Motion is motion of an
object in a circular path with constant
(uniform) speed.
• Its linear speed is constant. Its angular
velocity (w) is constant.
• Direction always changing.
• Hence linear velocity v is not constant.
• The instantaneous direction of v is
tangential to the circular path.
• Since velocity v is not constant, an object in
uniform circular motion must have an
acceleration.
Uniform Circular Motion
v
r
w (out of page)
v = rw
Acceleration in Uniform Circular Motion
The acceleration is due to change in direction of the
velocity, not its magnitude (since magnitude of the velocity
is constant).
Dv
v2
R
DR
v1
aave= Dv / Dt
-v1 Direction of a =
v2
direction of Dv
Dv = v2 – v1
As Dt 0, v1 and v2 will be almost parallel to each other
with Dv being perpendicular to them.
Dv will point towards center of circle.
Therefore, acceleration a is directed towards the center
of the circle.
Uniform Circular Motion
(circular motion with constant speed)
• Instantaneous velocity is
tangent to circle.
aC
v
• Instantaneous acceleration
is radially inward. Hence it is
called radial or centripetal
acceleration (ar).
• The magnitude of ar = v2/r
• There must be a net force
to provide the acceleration.
ar = v2/r = rw2 = (2r/T)2/r = 42r/T2 = 42rf2
Net Force in Uniform Circular Motion
FC
Conclusion:
v
There must be a net force to
provide the acceleration.
 This net force must also be
directed radially towards the
center. Hence it is called radial
or centripetal force (Fr).
 Fnet = Fr = mar = mv2/r = mrw2
Any object moving in a circular path must have a
net force exerted on it directed towards the
center of the circle.
The magnitude of this force = mv2/r = mrw2
A spider sits on a turntable that is rotating
at a constant 33 rpm.
The radial acceleration of the spider (ar) is
(A)Greater the closer the spider is to the central
axis
(B)Greater the farther the spider is from the
central axis
(C)Zero
(D)Non zero and independent of the spider’s
location on the turn table
Two kids of equal masses are seated at positions A
and B on a merry-go-round rotating uniformly.
Compare these quantities for the two kids:
1. Angular displacement (D)
2. Angular velocity (w)
3. Linear velocity (v)
A
4. Period (T)
5. Frequency (f)
6. Centripetal acceleration (aC)
(A) A > B
(B) A < B
(C) A = B
B
Examples of Circular Motion
1. A ball, mass m, swung in a vertical circle by a
string of length L
A
L
Fnet = mar
mg
T
-TA-mg = -mv2/L
 TA = mv2/L - mg
If T = 0, the ball will not
move in the circular path.
For circular motion, minimum swing speed is
obtained by setting T = 0
0 = mv2/L – mg or v = (gL)
At the bottom of the circular path:
A
Fnet = mar
T
B
mg
TB - mg = mv2/L
 TB = mv2/L + mg
In uniform circular motion, TB > TA
2. Water in a bucket swung in a vertical circle
Fnet = mar
r
mg
N
-N-mg = -mv2/r
 N = mv2/r - mg
If N = 0, the water will
pour out.
For water not to pour out, minimum speed is
obtained by setting N = 0
0 = mv2/r – mg or v = (gr)
What is the minimum speed you must
have at the top of a 20 meter roller
coaster loop, to keep the wheels on the
track.
A 20 m roller coaster loop;
r = 10 m
v = (gr) = (9.8 x 10) =  (98)
= 9.9 m/s
3. Small ball swung in a horizontal circle
Fnet = mar
 Small mass,T >> mg
r
mg
 T = mv2/r
T
4. Car driven on valley/hill
N
v
r
N
v
mg
r
mg
Fnet = mar
 Valley: NV – mg = mv2/r and NV = mg + mv2/r
 Hill: NH – mg = -mv2/r and NH = mg – mv2/r
 On flat road, NF = mg
5. Car rounding a curve on a flat road
Fnet = mar
r
Fy: N – W = 0
 Fx: fs = mv2/r
N
W
 If v is increased, fs will
increase up till fs max (= sN)
fs
 mv2/r  sN (= smg)
 v  (srg) OR s  v2/rg
6. Car rounding a curve on a banked road
To reduce chances of skidding, roads
are banked.
Centripetal force will then come from
a component of the normal force,
reducing total reliance on friction.
 At a particular speed, all of the
centripetal force Fc will be
contributed by the normal force N.
Nx
y
N
Ny

Nx = N sin
N
x
mg
Ny = N cos

With no dependence on friction:
Along y-axis: Ny = Ncos - mg = 0 ie, Ncos = mg
Along x-axis: Nsin = mv2/r
OR: tan  = v2/rg, ie Banking angle  = tan-1(v2/rg)
Speed v  (rg tan )
Example 1
A 1,000 kg car rounds a curve on a flat
road of radius 50 m. What should be the
speed limit if the coefficient of static
friction is 0.3?
Example 2
A car rounds a curve on a flat road of
radius 25 m at a speed of 13 m/s. What
should be the banking angle of the road
for the car to safely negotiate the turn if
the road is icy?
1. A 4.00 kg mass is moving in a circular path
of radius 2.50 m with a constant angular
velocity of 5.00 rad/sec. The centripetal force
59%
on the mass is
A. 250 N
B. 185 N
C. 153 N
21%
D. 107 N
12%
E. 94.0 N
6%
3%
A.
B.
C.
D.
E.
2. In moving through 30 minutes, the minute
hand of an analogue clock has gone through
79%
A.
B.
C.
D.
E.
180 rads
0.5 rads
1.57 rads
3.14 rads
None of the above
11%
5%
A.
B.
3%
C.
3%
D.
E.
3. A jet plane flying 6.0 x 102 m/s experiences
an acceleration of 4g when pulling out of a
dive. What is the radius of curvature of the
59%
loop in which the plane is flying?
A.
B.
C.
D.
E.
6.4 x101 m
1.2 x 103 m
7.1 x 103 m
9.2 x 103 m
None of the above
14%
16%
8%
3%
A.
B.
C.
D.
E.
4. You are driving through a valley whose
bottom has a circular shape. If your weight is
W, what is the magnitude of the normal force
N exerted on you by the car seat
as
you
drive
63%
past the bottom of the hill?
A. W > N
B. W < N
C. W = N
23%
D. None of the above
15%
0%
A.
B.
C.
D.
5. A 35-kg child is seated 3.0 m from the center
of a merry-go-round rotating with an angular
speed (w) of 2.4 rad/s. What is the magnitude of
the centripetal force Fc the child89%experiences?
A.
B.
C.
D.
E.
17.3 N
604.8 N
343 N
67.2 N
1.92 N
8%
3%
0%
A.
B.
C.
0%
D.
E.
6. A 1,200 kg car rounds a curve on a flat road
of radius 30 m. What should be the speed limit
if the coefficient of static friction is 0.45?
85%
A.
B.
C.
D.
E.
294 m/s
17.1 m/s
132 m/s
11.5 m/s
None of these
10%
5%
0%
A.
0%
B.
C.
D.
E.
Circular Orbits
Satellites orbit the earth in circular path
due to gravitational force. Its speed
must be just right.
Too low – falls to the earth.
Too high – escapes into space
m
v
r
Fr = GmM/r2 = mv2/r
v2 = GM/r = GM/(rE + h)
M
For Geosynchronous satellites,
V = 2r/T, where T = 24 hours
Example 3
The Hubble telescope moves in a circular
orbit 613 km above the earth’s surface.
The radius of the earth is 6.37 x 103 km
and the mass of the earth is 5.98 x 1024
kg.
(a) What is the speed of the satellite in this
orbit?
(b) Is the telescope geosynchronous?
G = 6.67 x 10-11 N.m2/kg2
Non-Uniform Circular Motion
• Uniform circular motion – circular motion in
which speed stays constant and direction
changes.
• Centripetal acceleration – changes only
direction of velocity, not its magnitude.
• ac – points along radius towards the center.
• Non-uniform circular motion – when both
magnitude and direction of the velocity
changes.
• The acceleration is not along the radius.
Non-Uniform Circular Motion
Non-uniform circular motion – when both
magnitude and direction of the velocity changes.
The acceleration is not along the radius. This
acceleration (a) can be resolved into tangential
(at) and centripetal (aC) components
a
at
ac
a = ar + at
ar = changes direction of v.
at = changes magnitude of v
Angular Acceleration
 Angular
acceleration is the change in
angular velocity w divided by the
change in time.
= Dw/Dt = (w - wo)/Dt
 Units: rad/s2
 Note: tangential acceleration at = r

Example
If the speed of a roller coaster car is 15
m/s at the top of a 20 m loop, and 25
m/s at the bottom. What is the car’s
average angular acceleration if it takes
2 seconds to get from the top to the
bottom.
 = Dw/Dt = (w - wo)/Dt
= vo/r = 15/10 = 1.5 rad/s
 w = v/r = 25/10 = 2.5 rad/s
  = Dw/Dt = (2.5 – 1.5)/2 = 0.5 rad/s2
 wo
Kinematics for Circular Motion with constant 
Linear Variables
Angular Variables
When a = constant
When
v = vo + at
w = wo + t
v2 = vo2 + 2a (x – xo)
w2 = wo2 + 2 ( - o)
x = x0 + vot + ½ at2
 =  o + w o t + ½  t2
 = constant
CD Player Example
The CD in your disk player spins at
about 20 radians/second. If it
accelerates uniformly from rest
with angular acceleration of 15
rad/s2, how many revolutions does
the disk make before it is at the
proper speed?
w2 = wo2 + 2 ( - o)
1 Revolution = 2  radians
Example
Example
A small ball of mass 200 g is rotated in
a horizontal circle of radius 0.5 m. If
the ball makes 10 revolutions in 5
seconds, what is its centripetal
acceleration?
Example
The moon orbits the earth in a nearly
circular path once every 27.3 days. It the
distance from the moon to the earth is
384,000 km, what is the moon’s
acceleration?
Example
Calculate the velocity of a
satellite moving in a stable
circular orbit about the Earth at
a height of 3600 km.
Example
A hypothetical planet has a
radius 2.5 times that of earth but
has the same mass. What is the
acceleration due to gravity near
its surface?
Summary of Concepts



Circular Motion
  = angular position radians
 w = angular velocity radians/second
  = angular acceleration radians/second2
 Linear to Circular conversions
s=r
Uniform Circular Motion
 Speed is constant
 Direction is changing
 Acceleration toward center a = v2 / r
 Newton’s Second Law F = ma
Uniform Circular Acceleration Kinematics
 Similar to linear!