1

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.

2

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

==∑

3

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===

4

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

5

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

6

(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

7

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.

8

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

=

9

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.

10

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

11

At top and bottom, therefore The minimum speed when the ball is at the top, such that

the tension on cord is non-vanishing

,12

−=

RgvmgTtop

+= 1

2

RgvmgTbottom

012

=

−=

RgvmgTtop gRvtop =

Angular displacement, velocity and acceleration

12

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 =

13

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= +

14

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 =

15

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

16

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

17

),(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

18

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

19

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.

20

2.3 Curvilinear Motion

and tr aaa +=

22 |||||| tr aaa +=

21

The path is either specified by an equation

( )xfy =

and Radius of Curvature ( ) is:

2

2

23

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

−+

==

23

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

24

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 ===

25

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.

26

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

27

Example 8. A jet plane travels along a vertical parabolic path defined by y=0.4x2. At point A, the jet has a speed of 200 m/s, and is increasing at the rate of 0.8 m/s2. Find the total acceleration of the jet at point A. Answer: