Download Uniform Circular Motion

PY105 (C1)
1. Assignment 5 on circular motion and impulse
will be due next Monday.
Uniform Circular
Motion
2. Lecture notes can be downloaded from the
following URL:
http://physics.bu.edu/~okctsui/PY105.html
Note that it is case-sensitive. You can also
access this website through WebCT.
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DEFINITION OF UNIFORM CIRCULAR MOTION
Uniform circular motion is the motion of an object traveling at a
constant speed on a circular path.
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Four common ways to express the speed
of a uniform circular motion
1.
2.
3.
4.
r
Specify the (uniform) speed, v.
Specify the period, τ , to complete one revolution.
Specify the angular speed, ω.
Specify the number of revolutions per unit time.
r
Notice that assumed in this definition is the circular path (with a
given radius, r) that defines the trajectory of the motion.
Also notice that the velocity, v, at any time is along the tangent
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of the circular path at where the object is.
Four common ways to express the speed
of a uniform circular motion
We can relate v and τ by the following reasoning:
Distance traveled in one revolution = 2πr
v
Secondly, ω = 2π/τ = v/r or v = rω
Lastly, number of revolutions per unit time = 1/τ
Example 1: A Tire-Balancing Machine
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= 1.2 ×10 −3 min revolution
830 revolutions min
Hence, vτ = 2πr
r
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The wheel of a car has a radius of 0.29m, and is rotating
at 830 revolutions per minute on a tire-balancing
machine. Determine the speed at which the outer edge
of the wheel is moving.
(1) speed, v, (2) period, τ , (3) angular speed, ω
and (4) number of revolutions per unit time
⇒ τ = 2πr/v or v = 2πr/τ
v
τ = 1.2 × 10 −3 min = 0.072 s
v=
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2π r
τ
=
2π (0.29 m )
= 25 m s
0.072 s
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Centripetal Acceleration
o
r
c
v at t0
p
θ
Drawing from
last page:
θ/2
θ
c
θ/2
θ
90ο − θ/2
v(t0+∆t)
(b)
r(t0)
c θ
r(t0+∆t)
90ο − θ/2
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See appendix I for a more detailed explanation.
(a)
v(t0)
θ
v(t0)
θ
v at t0
p
c θ
r at t0+∆t
v(t0+∆t)
θ
n.b. |v(t0+∆t)|=|v(t0)|
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Centripetal Acceleration
θ
∆v(t0)
o
r at t0
v at t0+∆t
∆v(t0)
θ θ
−v(t0)
What is the centripetal
acceleration, a at a
given time, t0?
Note that v, being tangential to the circular path, is
perpendicular to the radial direction.
(180ο − θ)/2
= 90ο − θ/2
90ο − θ/2
v at t+∆t
c
v(t0)
∆v(t0)= v(t0+∆t)+(−v(t0))
The fact that the direction of motion of an object in a circular
motion changes continuously with time suggests that the
velocity vector of the object varies continuously with time.
Hence the motion must have an acceleration, a. We call
accelerations giving rise to circular motions centripetal
accelerations.
∆v B
A
θ v(t0+∆t)
rθ
−v(t0)
θ
Circle drawn by using v as the
radius and the tail of –v(t0) as
the center.
v(t0+∆t)
From (a), v∆t = rθ ⇒ ∆t = rθ/v
In the limit ∆t → 0, θ → 0.
n.b. |v(t0+∆t)|=|v(t0)|=v
Acceleration, a = lim∆t →0∆v/∆t. This drawing shows
that ∆v and hence a points towards the center.
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From (b), when θ → 0, ∆v = vθ
vθ
v2
a = lim∆t →0∆v/∆t =
=
r
rθ/v
Example 2: The Effect of Radius on Centripetal Acceleration
The racing track contains turns with radii of 33
m and 24 m. Find the centripetal acceleration
at each turn for a speed of 34 m/s. Express
answers as multiples of g = 9.8 m/s2.
Use ac = v r
2
(34 m s )2
33 m
(34 m s )
= 35 m s 2 = 3.6 g
r
24 m
= 48 m s 2 = 4.9 g
Note that the centripetal acceleration required to make
a turn is bigger the sharper the turn is.
r
r
∑ F = ma = ma
2
r = 24 m: ac =
Centripetal Force
In circular motions, the object is subject to the
centripetal acceleration, ac. So, a net force must act
on an object to produce a circular motion. We call
the component of the net force that produces the
circular motion the centripetal force, FC. Accordingly,
FC = mac. Notice that the magnitude of Fc is
determined entirely by mac = mv2/r.
Solution:
r = 33 m: ac =
10
c
Fc
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r
+ maother
This term is non-zero
if the net force is
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bigger than Fc
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(Conceptual) Exercise 3 Object on a turntable
Centripetal Force
In Example 3, the centripetal
forces, Fc, required by the
racing car in making the turns
are provided by the static
frictional forces between the
road and the tires of the wheels.
For an object placed on top of
a rotating turntable, what is
the source of the centripetal
force on the object if it rotates
with the turntable?
fS
fS
fS
vr
r
Ans. Static friction
Turntable
fS’
What will happen if the static
friction is not large enough to
provide the centripetal force
needed for making the turn?
fS’
fS’
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If the radial position, r, of the object
on the turntable is doubled, how
much does the centripetal force
change?
Ans. Fc = mv2/r = m(2πr/τ)2/r = 4π2mr/τ2
Hence, if r is doubled, Fc will be doubled.
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Example 4: Centripetal Force Caused by a Tension
(Conceptual) Exercise 3 Object on a turntable (cont’d)
The model airplane has a mass of 0.90
kg and moves at a constant speed of 19
m/s on a circle that is parallel to the
ground. The path of the airplane and the
guideline lie in the same horizontal plane
because the weight of the plane is balanced by
the lift generated by its wings. Find the tension
in the 17 m guideline.
If static friction is not enough to provide the
centripetal force required by the object to
rotate with the turntable, which way will the
object go?
ω
Ans. It will follow the straight
path along which the velocity vr
r
vector, vr, is pointed.
Object
ω
Object
FBD at an instant:
Solution:
v2
Fc = T = m
r
Turntable
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Example 5: Centripetal Force by a Banked Curve
On a banked curve that makes an angle θ with the horizontal,
the centripetal force is mostly provided by the horizontal
component of the normal force. The vertical component of the
normal force balances the car’s weight. Ignore friction, what is
the relation between θ and v?
T = (0.90 kg )
(19 m s )2
17 m
T
= 19 N
a
L (Lift from
the wings)
mg
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Example 5: Centripetal Force by a Banked Curve (cont’d)
From the above, we deduce that the car remains
in the circular path if
Solution:
FN sinθ = ma = m
FN cos θ − mg = 0
v2
r
v2
tan θ =
rg
v2
= tan θ ⇒ v = rg tan θ .
rg
What will happen if the driver increases the speed
of the car from v = (rgtanθ)1/2?
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Vertical Circular Motions
Example 5: Centripetal Force by a Banked Curve (cont’d)
fsinθ f
fcosθ
But if friction is not negligible and the condition, tanθ = v2/rg
is not satisfied (i.e., v2 ≠ rgtanθ), static friction comes into
play to make the net force equal to the required centripetal
force = mv2/r.
v2
Note that if v2 > rgtanθ,
x: FN sinθ − f cos θ = m
f < 0 meaning that f
r
y: FN cos θ + f sinθ − mg = 0
points towards the
mg
v 2 center to help provide
⇒ f =
(tan θ − ) the centripetal force.19
rg
sinθ tan θ + cos θ
Consider the vertical circular motion
as shown.
1 FN 1 − mg = m
v22
r
v42
=m
r
2
FN 2 = m
4
FN 4
v
r
v2
3 FN 3 + mg = m 3
r
FN3<FN2=FN4<FN1.
If FN = 0, the cyclist loses contact
with the track. Amongst all positions it
happens most probably at 3 .
v2
FN 3 + mg = m
r
FN3 = 0 ⇒ v = (rg)1/2
Maximum
negative
contribution
No
contribution
Maximum
positive
contribution
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If the speed is uniform,
Contribution
of mg to ac:
2
1
Motions confined to a circular
path but under gravitational
acceleration, g. At different
points of the circular path, mg
contributes to the centripetal
force, Fc, differently, so the
normal force, FN, which
provides the difference
between Fc and mg varies
with position.
If v is less than (rg)1/2, FN = 0 and part of mg would
cause the object to fall at 3 :
mg = mv2/r + may,fall
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ΣF
may,total
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Another example of vertical circular motions
Consider an object tied to a string
v3
and being maneuvered into a
T
vertical circular motion as shown at
mg
ac
right. The criterion for the object to
remain in the circular trajectory is
r
that the tension in the string
remains non-zero throughout the
motion. As before, if speed v3 at the highest point is
such that mv32/r < mg, tension T will be zero and
the difference, mg – mv2/r, will cause the object to
fall vertically and leave the circular path.
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