Circular and Curvilinear Motions -...
Transcript of Circular and Curvilinear Motions -...
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Chapter 4 Circular and Curvilinear Motions Here we consider particles moving not along a straight line – the curvilinear motion. We first study the circular motion, a special case of curvilinear motion. Another example we have already studied earlier is the projectile. 2.1 Circular Motion Uniform Circular Motion Revisited It is a motion with acceleration – direction of the velocity changes; The acceleration is always perpendicular to the path of the motion. The acceleration always points toward the center of the circle of
motion (it’s not even a motion of constant acceleration!). This acceleration is called the centripetal acceleration.
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The magnitude of the centripetal acceleration is given by Newton’s second law dictates that
The Period, T, is the time required
for one complete revolution.
rvac /2=
vrT /2π=
rmvamF c
2
==∑
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Example 1. Conical Pendulum
Find an expression for v and the period T. Answer: Equilibrium in vertical direction: Uniform circular motion in horizontal plane: Combine the above two eqns. and the period
0cos =−mgT θ
θθ
sinsin
22
Lmv
rvmT ==
θθ tansinLgv =
gL
LgL
vr θπ
θθθππτ cos2tansin
sin22===
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Example 2. Banked Roadway
A car of mass m travels at constant speed v round a bend of radius r on a road banked at an angle θ. The coefficient of friction between the car’s tyres and the road surface is tanλ, where λ < θ. Show that: (a) If the car travels with no tendency to slip
(b) If the car is about to slip outwards,
(c) If the car is about to slip inwards
θtan2 rgv =
)(tan2 λθ += rgv
)(tan2 λθ −= rgv
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Answer: (a) In this case there is no frictional force acting. The equation
of motion are So
and ,sin2
rmvR =θ 0cos =−mgR θ
θtan2 rgv =
RF µ=mg
v 2/r
R
α
F
(b) The frictional force acts down the slope and takes its limiting value, i.e.
The equations of motion are Hence and
and ,cossin2
rmvRR =+ θµθ 0sincos =−− mgRR θµθ
);cos(sin2 θµθ +=mrRv
θµθ sincos −=
mgR
).(tantantan1
)tan(tantan1
)(tansincos
)cos(sin2
λθλθλθ
θµµθ
θµθθµθ
+=−
+=
−+
=
−+
=
rgrgrg
rgv
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(c) The frictional force acts up the slope and takes its limiting value µR. The equations of motion are
Then
mg
v 2/r
R
α
F
and ,cossin2
rmvRR =− θµθ
0sincos =−+ mgRR θµθ
λθλθ
θµθθµθ
tantan1)tan(tan
sincos)cos(sin2
+−
=+−
=rgrgv
).(tan λθ −= rg
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Example 3. The Rotor
ω
R
fs
N
mg Along diameter
The rotor’s wall Rotor is quite often found in amusement park. It 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. Determine the minimum speed at which the bottom floor can be opened and yet the man is safe, given the static frictional coefficient µ.
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Answer: The man rotates with the rotor and the centripetal force which acts on him is provided by the wall as normal force, . As the man does not fall down, the frictional force in the upward direction balance with his weight, e.g. mg = µN. Hence, and so Note: It does not depend on the mass of the man.
RmvN
2=
)(2
Rmvmg sµ=
s
gRvµ
=
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Nonuniform Circular Motion (moving at varying speed in a circular path) In addition to radial component of acceleration, there is a tangential
component of acceleration of magnitude at =|dv/dt|; The total acceleration is then
There must be a net force exerted on particle that is inclined to :
tr aaa +=
v
tr FFF
+=
tr FF
,and are responsible for centripetal and tangential accelerations, respectively.
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Example 4. Keep you eye on the ball
A small ball of mass m is attached to the end of a cord of length R and set into motion in a vertical circle about a fixed point O. Determine the tangential acceleration of the ball and the tension in the cord at any instant when the speed of the ball is v and the cord makes an angle θ with the vertical.
Answer: The tangential force on the ball is The radial force on the ball is
tt mamgF ==∑ θsin
RvmmgTFr
2
cos =−=∑ θ
θsingat =
+= θcos
2
RgvmgT
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At top and bottom, therefore The minimum speed when the ball is at the top, such that
the tension on cord is non-vanishing
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−=
RgvmgTtop
+= 1
2
RgvmgTbottom
012
=
−=
RgvmgTtop gRvtop =
Angular displacement, velocity and acceleration
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In describing circular motion, one may use the set of quantities to specify the state of motion, they are Angular displacement θ, angular velocity ω, and angular acceleration, α. Note: and
s=rθ r θ
O
,rs
=θ ,lim0 dt
dtt
θθω =∆∆
=→∆ 2
2
dtd
dtd θωα ==
,θrs = ,ωrvt = .αrat =
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Kinematic Equation of Circular Motion at constant angular acceleration:
Circular Motions Linear kinematics
tαωω += 0 v u at= +2
012
t tθ ω α= + 212
s ut a t= +
2 20 2ω ω αθ= + 2 2 2v u as= +
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2.2 Plane Polar Coordinates
For particle moving on a plane, instead of rectangular (x, y) coordinates, it is often convenient to describe the motion by using the plane polar coordinates (r, θ). Define two Unit Vectors in the plane polar coordinates , which are perpendicular to each other (similar to vectors in (x, y) coordinates): So position can also be specified as in plane polar coordinates, or in vector form, where .
( )θeer ˆ,ˆ)ˆ,ˆ( ji
+−=+=
jiejier
ˆ cosˆ sinˆ
ˆ sinˆ cos ˆθθθθ
θ
reθeP
),( yxP),( θrP rerr ˆ=
)(ˆˆ θrr ee =
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obviously, Furthermore The velocity vector is
and sincos
==
θθ
ryrx
( )
=+=
− xyyxr/tan 1
22
θ
−=−−=
=+−=
r
r
ejie
ejie
ˆ ˆ sinˆ cos d
ˆd
ˆ ˆ cosˆ sin d
ˆd
θθθ
θθθθ
θ
( )
( )
θθ
θ
θθθ
ωθθ θ
eveverev
erertere
trer
ttrv
rr
rr
te
te
r
rrr
err
ˆˆˆ ˆ ˆˆ
dˆdˆ
ddˆ
dd
dd
ˆdd
dˆd
dˆd
+=+=
=⋅=+=
+===
vr is the radial component of velocity along , and vθ is the angular component along .
reθe
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The acceleration is
( )
θθ
θ
θθθ
θθθ
θ
θθθ
θθθθθ
θθ
θθθθθ
θ
eaeaerrerr
ererererert
eret
rett
eretr
erertt
va
rr
r
rr
rr
r
ˆˆˆ)2(ˆ)(
)ˆ(ˆˆ)ˆ(ˆdd
dˆdˆ
ddˆ
ddr
dd
dˆdˆ
dd
ˆˆdd
dd
2
+=++−=
−++++=
++++=
+==
and again, ar and aθ are the radial and angular components of the acceleration . a
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),(ˆ teRr r=
θ
θ
θ
ωθ
eveR
eReRv
t
r
ˆˆ
,ˆˆ
==
+=
tangential lcentripeta
2
2
ˆˆ
ˆ)2(ˆ)(
θ
θ
αω
θθθ
eReReRReRR
r
r
+−=
++−=a
Circular motion in plane polar coordinates
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Example 5. Bug walking on a rotating wheel A ladybird set out to walk at constant speed u along a spoke of a rotating wheel with a constant angular velocity ω. The plane of the wheel is horizontal and its rotation axis is fixed vertical. Assume the ladybird starts off at r = 0, θ = 0 at t = 0. Find the magnitude of the velocity and acceleration of the bug at any other time t. Express its locus in terms of x and y. Answer: From , one has From , one gets
θω erevv rr ˆ ˆ += 22222 )()( || tuuruv ωω +=+=
θθθθ errerra r ˆ)2(ˆ)( 2
++−=
222242242 44 || ωωωω utuura +=+= ), ;0 :( ωθθ ==== urrNote
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Finally, ( )( )jtitut
jiuterr r
sin cos
sin cosˆ
ωω
θθ
+=
+==
==
tutytutx
ωω
sincos
so, 2222 tuyx =+ uyxt /22 +=
( )( )2222
2222
sin
cos Then,
yxyxy
yxyxx
u
u
++=
++=
ω
ω
General motion is not linear, nor circular, but along a curved path. The acceleration changes from point to point. At any instant, such a vector quantity can be resolved into two components based on an origin at the center of a circle (see figure) corresponding to that instant: a radial (normal) component ar along the radius, and a tangential component at perpendicular to that circle.
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2.3 Curvilinear Motion
and tr aaa +=
22 |||||| tr aaa +=
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The path is either specified by an equation
( )xfy =
and Radius of Curvature ( ) is:
2
2
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2
dd
dd1
1
xy
xy
kρ
+
==
In Cartesian coordinates
sk
ddθ
=
22
Or the path can be parameterized as in which case, the Radius of Curvature is:
tyytxx
==
)()(
( )xyyx
yxk
ρ
1 2
322
−+
==
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For the latter case, the velocity and acceleration is straight forwardly derived: and
)(
)(
=
=
tyv
txv
y
x
)(
)(
=
=
tya
txa
y
x
2222 , yxyx aa || vv || Magnitude +=+= av
=
=
x
y
x
y
aa
vv
Direction tan,tan av θθ
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For the former case, one is often given vx and/or ax . The total velocity and acceleration is then derivable: where
and dd
dd
dd
dd
xyv
tx
xy
tyv xy ===
2222 , yxyx aa || vv || Magnitude +=+= av
=
=
x
y
x
y
aa
vv
Direction tan,tan av θθ
dd
dd
dd
dd
xv
vtx
xv
tv
a yx
yyy ===
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Example 6. A car on a test track goes into a turn described by x = 20 + 0.2t3, y = 20t − 2t2, where x and y are measured in meters and t in seconds. Find the acceleration of the car at t = 3.0 seconds. Answer: Horizontal acceleration: as x = 20 + 0.2t3, so vx = dx/dt = 0.6t 2 and ax = d2x/dt2 = 1.2t At t = 3.0, ax = 3.6 Vertical acceleration: y = 20t − 2t2, so vy =dy/dt =20−4t and ay = d2y/dt2 = −4 At t = 3.0, ay = −4 Now a = (3.62 + (−4)2)1/2 = 5.38 m/s2
and θa = tan-1(ay / ax ) = tan-1(-4/3.6) = 312o from the positive x-axis.
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Example 7. A particle moves along the path described by y = x2 + 4x + 2 (cm). The horizontal velocity vx is constant at 3.0 cm/s, find the magnitude and direction of the velocity when the particle is at the point (-1, -1). Answer: It is given that vx =dx/dt = 3 cm/s. Now we need to find vy (knowing that y = x2+4x+2, and x = -1) vy = dy/dt = 2x(dx/dt) + 4(dx/dt ) + 0 = 2(−1)(3)+4(3) = 6. So we have vy = 6 cm/s. The magnitude of the velocity is then: v = [(vx)2 +(vy)2]1/2 = 6.7 cm/s. The direction of the velocity is given by: θv = tan-1(vy / vx ) = tan-1 (6/3 ) = 63.4o