Download Lecture 5 - HKU Physics

Unit 6
Circular Motions
6.1
Circular Motion
6.2
Rotation with constant angular acceleration
6.3
Centripetal acceleration
6.4
The banked curve
6.5
A swinging bucket
6.6
The rotor
6.1
Circular Motion
For a rigid body to rotate around a fixed axis, each particle in the rigid body undergoes a
circular motion. Take one of the particle as an example, θ is the angular displacement, s is the
distance it moves and r is the distance between the axis and the particle.
θ=
(1)
s
r
The average angular velocity is defined as
ω av =
(2)
θ2 −θ1
t 2 − t1
∆θ
=
∆t
θ
s=rθ
O
Instantaneous angular velocity
ω = lim
∆t → 0
(3)
r
∆θ dθ
=
∆t
dt
Instantaneous tangential velocity
vT = rω .
For a circular motion, the arc length is given by s = rθ , where r is constant.
Differentiate both sides, we obtain
v=
(4)
ds
and the angular velocity is
dt
ds
dθ
. Since the instantaneous speed is
=r
dt
dt
ω=
dθ
, we obtain v = rω .
dt
Angular acceleration α
∆ω dω d 2θ
=
= 2
∆t →0 ∆t
dt
dt
α = lim
(5)
Tangential acceleration
1
aT = rα
Remark:
•
Although every point in a rigid body travels different distance during the rotation, the
angular displacements are the same. At any instant, every point in a rigid body has the
same angular velocity.
•
In advanced study, ω is quite often defined as a vector with its direction pointed by
the right-hand rule. This is common in the area of science and engineering.
6.2
Rotation with constant angular acceleration
Circular Motions
Linear kinematics
∴ω = ω 0 + α t
v= u + at
1
=
θ ω0 t + α t 2
2
s= ut +
2
ω=
ω0 2 + 2αθ
1 2
at
2
2
v=
u 2 + 2as
Remarks:
(1)
α=
dω
= constant
dt
∴ ω = ω 0 + α t , where ω0 = ω (t ) |t =0 .
(2)
ω=
dθ
dt
θ
t
θ0
0
∴ ∫ dθ = ∫ ω dt
1
∴θ = θ 0 + ω0 t + α t 2
2
(*)
1
θ ω0 t + α t 2 .
When θ0 = 0, we have =
2
(3)
Squaring on both sides of ω
= ω0 + α t gives ω 2 =
ω0 2 + 2ω0α t + α 2 t 2 .
1
Rewrite the equation as ω =
ω0 2 + 2α (ω0 t + α t 2 ) and using (*). Hence we obtain
2
ω 2 =ω0 2 + 2α (θ − θ 0 ) .
2
=
ω0 2 + 2αθ .
When θ0 = 0, we have ω
2
6.3
Centripetal acceleration

v1
Consider the case of uniform circular motion.
•
P1
P2
Uniform circular motion: Speed is a constant, e.g.


v1 = v 2 = v .
∆s
r
Obviously, triangles OP1P2 and Op1p2 are similar.

 v
∆s Δv
v
∴
or ∆v = 1 ∆s = ∆s
=
r
v1
r
r
a av =
Acceleration
a = lim a av
∆t →0

∆v
∆t
=

v2
∆φ
r
O

p1 ∆v p2


v2
v1
v ∆s
r ∆t
∆s v 2
v
= lim
=
r ∆t →0 ∆t
r
O
∆φ

v2
In fact, the centripetal acceleration is given by a = − rˆ , where r̂ is the unit vector pointed
r

from the origin. The acceleration can be rewritten as a = −ω 2 r rˆ , since v = r ω . When we
a
consider only the magnitude of the centripetal force, we have=
v2
= rω 2 .
r
Remarks:
For a non-uniform circular motion, the acceleration consists of two parts.
(1) Tangential component of acceleration
a tan ≡
dv
dω
=r
= rα
dt
dt
(Non-uniform circular motion)
Its direction is always tangential to the circular path of the particle.
(2) Radial component of acceleration
a rad =
v2
= rω 2 .
r
3
Example
On a wet day the coefficient of friction between a car’s tyres and the ground is 0.2. At what
speed is it safe for a driver to round a corner of radius 25 m, if the road surface is horizontal?
R
Answer
It is the frictional force F, which makes it possible for
v 2/r
the car to move in a circle.
F
The equations of motions are
F
=
mv 2 mv 2
,
=
r
25
mg
R = mg.
Since F ≤ µ R , we have
mv 2
≤ 0.2 mg , v 2 ≤ 0.2 (25) (10) ,
25
v 2 ≤ 50 , v ≤ 7.07 m / s .
The maximum speed, if skidding is to be avoided, is 7.07 m/s.
6.4
The banked curve
A car of mass m travels at constant speed v round a bend of radius r on a road banked at an
angle α. The coefficient of friction between the car’s tyres and the road surface is tan λ,
where λ < α . We have the following cases.
(a)
R
if the car travels with no tendency to slip,
v 2 = rg tan α ,
(b)
if the car is about to slip outwards,
v = rg tan (α + λ ) ,
2
(c)
if the car is about to slip inwards,
v 2/r
α
mg
v 2 = rg tan (α − λ ) .
Explanations:
(a)
In this case there is no frictional force acting. The equation of motion are
R sin α =
mv 2
, R cos α − mg = 0 .
r
Therefore, v 2 = rg tan α .
4
The frictional force acts down the slope and takes its limiting value, i.e. F = µR .
(b)
The equations of motion are
R sin α + µR cos α =
mv
,
r
R cos α − µR sin α − mg = 0 ,
Therefore, v 2 =
R=
R
2
v 2/r
F
rR
(sin α + µ cos α ) ,
m
mg
α
mg
.
cos α − µ sin α
Hence when the car is about to slip outwards,
rg (sin α + µ cos α )
cos α − µ sin α
rg (tan α + µ ) rg (tan α + tan λ )
=
=
= rg tan (α + λ ).
1 − tan α tan λ
1 − µ tan α
v2 =
The frictional force acts up the slope and takes its limiting value µR. The
(c)
equations of motion are
R sin α − µR cos α =
mv 2
r
R
R cos α + µR sin α − mg = 0 .
rg (sin α − µ cos α )
Therefore, v =
cos α + µ sin α
v 2/r
rg (tan α − tan λ )
= rg tan (α − λ ).
1 + tan α tan λ
α
F
2
v2 =
mg
Example
A particle of mass m is connected by an inextensible light string of length l to a fixed point on
a smooth horizontal table. The string breaks when subject to a tension whose magnitude
exceeds mg. Find the maximum number of revolutions per second that the particle can make
without breaking the string.
5
Answer
For maximum revolution ω, we have the maximum tension
ω
of string equals mg, i.e. mg = mω 2 l , which implies
ω=
The frequency f=
m
g
rad / s .
l
1
1
1
=
=
T 2π / ω 2π
l
g
rev / s .
l
Example
A star is believed to rotate at about 1 rev s . If such a star has a radius of 20 km (a typical
value), (a) what is the speed of a point on the equator of the star and (b) what is the
centripetal acceleration of this point?
Answer
Angular velocity = 2π
a)
Speed = 2π × 20 =
125.7 km s
Centripetal acceleration = ( 2π ) × 20 =
789.6 km s 2
2
b)
Centripetal acceleration:
∴=
F ma
=
a=
v2
R
mv 2
R
Example
Cord (Length = L)
Find the period of the conical pendulum as shown in
z
figure.
θ
T cosθ
Answer
Period: The time required to make a complete
x
T sin θ
revolution.
•
y
T
mg
The particle moves in a plane x-y, so there is
R
ω
no motion along z-direction
∑F
z
•
=0
or
T cos θ − mg = 0 ⇒ T =
mg
cos θ
In x-y plane, it is a uniform circular motion
6
F=
mv 2
R
The period
τ
=
or
T sin θ =
mv 2
L sin θ
2π R 2π L sin θ
L cos θ
.
=
= 2π
v
v
g
Example
Find the tension T before and after the string is cut.
T
Answer
(a) Before the cut, the system is in equilibrium
θ
A m
ΣF = 0
∑F
∑F
x
y
=0 ⇒ T0 =T sin θ
=
0 ⇒ mg =
T cos θ
Hence, we have T =
T
T0
mg
.
cos θ
θ
mg
(b) Once the string is cut, it is not in equilibrium anymore. It will undergo a circular motion
(non-uniform).
For circular motion, say, it is at an angle α with the vertical.
Resultant force along radial direction = m
v2
.
R
2
v
Radial acceleration: T − mg cos α =
m =
mar ,
R
α
T
Radial
direction
Or=
T mg cos α + mar .
Tangent
direction
mg
Tangential acceleration: mat = mg sin α .
Example
A disk of mass m on a frictionless table is attached to a hanging
cylinder of mass M by a cord through a hole in the table (see figure).
Find the speed with which the disk must move in a circle of radius r for
the cylinder to stay at rest.
7
Answer
For M to remain at rest the tension T in the cord must equal the gravitational force Mg exerted
on M. The tension supplies the centripetal force that keeps m in its circular orbit, so
T = mv 2 r . Thus Mg = mv 2 r and we obtain v = Mgr m .
6.5
A swinging bucket
Example
Explain the following description.
A bucket of water can be swung round in a vertical circle without falling spilling. If the
weight mg of the water is less than mv2/r, the normal force N of the bottom of the bucket on
the water provides the rest of the force required to maintain the water in its circular path.
However, if the bucket is swung more slowly, then mg will be greater than mv2/r and the
‘unused’ part of the weight causes the water to leave the bucket.
Answer
Consider the bucket in position A, the circular
A
motion is maintained by the centripetal force
N
mg
which is contributed by the normal reaction N and
the weight of water mg, i.e.
mv 2
N + mg = ,
r
if water is still in the bucket, N > 0, and
mg <
B
Water stays in
bucket if mg < mv2/r
N
r
mg
2
mv
.
r
When the bucket is in position B, the centripetal
v
force is provided solely by the normal reaction N,
i.e. N =
mv 2
.
r
C
N
mg
Lastly, when the bucket is in position C, the
difference of the normal reaction N and the weight of water mg contributes to the centripetal
mv 2
force, i.e. N − mg = .
r
8
Example
An airplane is moving along a vertical circle of radius r with a constant speed v as shown in
figure (a), compute the reaction force exerted on the pilot from the seat when the flight is at
the bottom and top of a vertical circle.
Answer
r
O
Figure (a)
Figure (b)
Figure (c)
When the airplane is at the bottom of the vertical circle (figure b):
nbot − mg =
mv 2
,
r
 v2 
thus =
nbot mg 1 +  .
 rg 
When the airplane is at the top of the vertical circle (figure c):
mv 2
ntop + mg = ,
r
 v2

ntop mg  − 1 .
thus =
 rg

9
6.6
The Rotor
Rotor is quite often found in amusement park for visitors to enjoy with. A rotor is a hollow
cylindrical room that can be set to rotate about the central vertical axis. A person enters the
rotor and stand against the wall. The rotor gradually increases its rotating speed up to a preset
one and the floor below the person is opened downward. The person does not fall down but
remains rotating with the rotor.
ω
The rotor’s wall
fs
N
Along diameter
mg
R
The man rotates with the rotor and the centripetal force which acts on him is provided by the
wall as normal force, N =
mv 2
. As the man does not fall down, the frictional force in the
R
upward direction balance with his weight, e.g. f s = mg = µ s N . Hence, we obtain
mg = µ s (
mv 2
) and thus v =
R
gR
µs
.
It should be noted that the velocity does not depend on the mass of the man. A practical data
for the coefficient of friction between the textile material of clothing and a rotor wall is about
0.40. For a rotor of radius of 2.0 m, v is about 7.0 m/s or more. That is the frequency f is
given by
f
=
7.0 m / s
v
1
1
=
= =
= 0.56 revolution/s or about 33 rpm.
Period 2π r / v 2π r 2π (2.0 m)
10