11. Particle Dynamics-5

8/2/2019 11. Particle Dynamics-5

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Physics


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Session

Particle Dynamics - 5


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Session Objective

1. Circular motion

2. Angular variables

3. Unit vector along radius and tangent

4. Radial and tangential acceleration

5. Dynamics of circular motion

6. Centripetal force in circular motion

7. Circular hoops

8. Centrifugal force in circular motion


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Every day we see the sun riseand set. We see the sun movinground us. But .

Session Opener

Science says the earth movesaround its axis and in a day thesun hardly moves.

Have you asked yourselfwhy our eyes observe a

wrong phenomenon ?


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F

v

Object A moves in a circular path

radius r fixed : constrained motion

F Directed towards center

v Perpendicular toF

Constant in magnitudeF

And change directioncontinuously.

F v

For uniform motion

v is constant.

s r has a direction

Lim. 0 ds rd

Circular Motion


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Angular Variables (Constant Speed)

averages

v r

t t

avg. : average angular velocityt

vaverage= avg.rd

Lim t 0t dt

Instantaneousangular velocity

v = r

rO

r

r

r

vector relation : v r

rad:axial vector. SIunit

sec

tcons tant v cons tant

d dt

o = t

Angular kinematicalequation for constant .


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Angular Variables (Variable v)

v changes (a constant)1(t) v(t) v : along t angentr

acceleration : along t angentv = v0 + at = 0 + t : angular acceleration

20

1s v t at

2 2

01

t t2

r

r t

= 0 + t

2 20 2

is tangential.


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Class Exercise


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Class Exercise - 5

A particle moves with a constant

linear speed of 10 m/s in acircular path of radius 5 cm.What is its angular velocity?

2

v 10 m/s

r 5 10 m

= 200 radians/s.

Solution :


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Unit Vector Along Radius andTangent

r i r cos j r sin

[ox and oy : fixed reference frame].r and also define position

Transfer origin from O to Pox || px oy || pyP has acceleration Reference frame non inertial

O

p

x

y

r

(t)i

j

(t)

y

x


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r r

r can be defined asr re (e : unit vector along r)

(origin P)define a non inertialreference frame

re , e

re i cos jsin rr re

e i sin jcos

Define a unit vectorperpendicular to (along )re

e

v

i

jree

y

x


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i

jree

y

x

re icos jsin rde d d

isin jcos edt dt dt

2

r2d e d dicos jsindt dtdt

2re

Angular velocity vector isthe rate of change of radialunit vector.


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Radial and Tangential Acceleration

v constant constant

v : tangential

v rv r

drv re

dt

22

r r2

d ra re

dt

r

r

22

r

a : radial

: opposite e

va r v

r

can not change themagnitude of

ra

v

Particle moves in a circle

rr re r cons tan t

i

jree

y

x

v

is perpendicular tora v ar


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Radial and Tangential Acceleration

When changes

= particle moves in the circle(r constant)

= v changes

r

2 r

2rT

r re

drv r e

dt

dv dr e edt dt

r e r e a

ar

= tangential acceleration appearsta

at

v

a

12 22 2

t

t t2

r

a r r

aTan

a r


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Centripetal Force in CircularMotionObject in circular motion has at and ar

at only changes magnitude of v.: non-uniform circular motionar : necessity for circular motion.

: no change in magnitude of v.

22cp

vF m r m m vr

22

rv

a v rr

rO

ar

As ar exists,an externalforce Fcp must exists.

Fcp

cpF is a radial force, called centripetalforce


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Class Exercise


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Class Exercise - 3

A vehicle moves with constant speedalong the track ABC. The normalreaction by the road on the vehicle atA, B and C are respectively. Then

A B C B A C

C A B B A C

(a) N N N (b) N N N

(c) N N N (d) N N N

A

B

C


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Solution

N A

Am g

m v 2rA

(A )

2

A

A

mvmg N

r

2

AA

mvN mg

r

NB

B

mg

mv2rB

(B)

N mg =Bmv2

rB

N = mg +Bmv2

rB

Hence answer is (b)


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Solution

N C

m g m v 2

r C

C

( C )

2

C C

mvmg N

r

2

CC

mvN mg

r

2 2A C A C

A C

mv mvr r .So N Nr r

NB is largest. From shape ofthe track,


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Class Exercise - 7

A pendulum, constructed byattaching a tiny mass m at the endof a light string of length L, isoscillating in a vertical plane. Whenthe pendulum makes an angle withvertical, its speed is equal to v. Find

the tension in the string and thetangential acceleration at thatinstant.


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Solution

mgma

L

T mv2L

y

x

Pendulum moves along the arc of a

vertical circle.Resolving the motion along T (X-axis)

and perpendicular to T (Y-axis)

mg sinma (along y) (Tangential)

a g sin

2

mvT mg cos (along x) (radial)L

2vT m(g cos )

L


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Class Exercise - 10

1m

Fixed

centre

1m/s

A mass m of 50 kg is set moving in ahorizontal circular path around a fixed

centre O to which it is connected by aspring of unstretched length of 1 mand a spring constant of 905 N/m.Find out the amount by which thespring will stretch if the speed of themass is 1 m/s and the spring is light.


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Solution

FO M

(1+x)

F=restorinforce

As the mass moves, it tends to slipoutwards, providing a stretching

force. Spring provides the reaction(restoring force), which is thecause of centripetal force

(|F| = kx)

2mvkx (as spring has stretched by x)

(r x)

2 2

2mv mv

(r x)x x rx 0k k


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Solution

kr 1 m, v 1 m/s, m 50 kg, 905

m

221 4mvx r r

2 k

1 4 50x 1 1

2 905

= 0.05 m. (approx.)


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Centripetal Force

cpFSource of Friction

Object moves in circulartrack radius r

2

cp maxv

F f mg mr

Centripetal force is friction.

fs

2maxv g (constant)r


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Banking of Curves

Object moves along circular track.

Track banked towards center O.

O

y axis : N cos = mg

x axis : N sin = Fc=mv2/r

mg

N

2mv

Nsinr

Ncos=mg

2 2v vTan g tan

rg r

If velocity > v: objectmoves outward toincrease r

If velocity < v: objectmoves inward todecrease r.

< 900


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Conical Pendulum

P moves in horizontal circle at end ofstring OP fixed to rigid support O.

Tension T supplies Fcp

Y axis : T cos = mg

x axis : T sin = mv2 /r (r=L sin )

2 22v sinv rg tan gL

rg cos

p2 r L cos

t 2v g

Time tp to complete one revolution :

12 2 22 2

2

m vNote T m g

r

L

mg

T


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Class Exercise


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Class Exercise - 1

Two similar cars, having masses ofm1 and m2 move in circles of equal

radii r. Car m1 completes the circle intime T1 and car m2 completes thecircle in time T2. If the circular tracksare flat, and identical, then the ratioof T1 to T2 is

1 2

2 1

2

1

m m(a) (b)

m m

m(c) (d) 1

m


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Solution

Centripetal force 21 1 1 1F (car m ) m r

2 121

4 m r

T

22

2 2 2

2

4 m rcentripetal force F (car m )

TAs circles are identical and flat, frictionsupplies centripetal force in both cases.

2 21

12 21 1

m r 4 4 rm g g ............(1)

T T

2 22

2 2 22 2

m r 4 4 rm g g .................(2)

T T

2122

TDividing (2) & (1) we get, 1

T

Hence answer is (d)


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Class Exercise - 6

The driver of a car, moving at aspeed of v, suddenly finds a wallacross the road at a distance d.Should he apply the brakes or turn ina circle of radius d to avoid a collisionwith the wall? (Coefficient of kinetic

friction between the road and thetyre of the car is .)


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Solution

In both cases, the friction force fsupplies the braking force. F=Nmg.Deceleration = g.

In applying the brakes car must stopwithin distance d.

2 2 2ifv v 2ax 0 v 2( g) d

2vd 2 g

f

In taking a circular path, the maximumradius is d.

2 2 2

mv mv vf mg dd d g

So applying the brakes is the better option.


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Class Exercise - 8

A B

ra

rbA B0.1 0.2

A Br 1km r 2 km

Two motor cyclists start a race along a flat race track.Each track has two straight sections connected by asemicircular section, whose radii for track A and track Bare 1 km and 2 km respectively. Friction coefficients of A

and B are 0.1 and 0.2 respectively. The rules of the racerequires that each of the motor cyclist must travel atconstant speed without skidding. Which car wins therace? (g = 10 m/s2) (Straight sections are of equallength)


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Solution

Motor cyclist B wins the race.

2vSkidding occurs for g

r

Maximun speed v rg

3Av 0.1 1 10 10 10 10 m/ s

3Bv 0.2 2 10 10 20 10 m/ s

In semicircular section: Length of track A = km

21 1T Time taken by m.c.A 10 S T

10


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Solution

Length of track B = 2km

22 1T Time taken by m.c.B 10 s T

10

Semicircular portion isnegotiable in equal time.

m.c.B is faster, so will complete

straight parts faster.


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Class Exercise - 9

m

Figure shows a centrifuge, consistingof a cylinder of radius 0.1 m, whichspins around its central axis at therate of 10 revolutions per second. Amass of 500 g lies against the wall ofthe centrifuge as it spins. What is the

minimum value of the coefficient ofstatic friction between the mass andthe wall so that the mass does notslide? (g = 10 m/s2)


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Solution

f

mg

1 m

The mass will not slide ifmg f

The mass will press the wall at a forceequal and opposite to the centripetalforce supplied by the wall which is thereaction force.

2minf m r mg

22 10 rad/ s,r 0.1,g 10 m/ s

2gr

2

10 10.025

40100 0.1


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Centrifugal Force in Circular Motion

P moves in circle (radius r) withangular velocity

Reference frame centered at origin O :

Reference frame is inertial

2cpF m r. (along PO)

Reference frame with P as origin

Reference frame is non inertial.''Y''X

P''X

''YP

''X

''YP

''X

''Y

P

Or


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Centrifugal Force in Circular Motion

P is at rest with respect to itself.

A pseudo force Fpseudo to beadded as frame is non inertial.

So in frame of P; cp pseudoF F F 0.

Fpseudo = m2r away

from center.

Fpseudo is called

centrifugal force.

P

O

Fcp Fpseudo


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Class Exercise


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Class Exercise - 2

A particle of mass m moves in acircular path of radius r with a uniformangular speed of in the xy plane.When viewed from a reference framerotating around the z-axis with radiusa and angular speed , the centrifugal

force on the particle is equal to2 2 2

0

2

0 0

(a) m ( ) a (b) m a

(b) m a (d) m a


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Solution

Centrifugal force is a pseudo forceequal to m (Acceleration of theframe of iron inertial frame) Noninertial frame in this case has radialacceleration .

20 a

20centrifugal force m a

Hence answer is (d)


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Class Exercise - 4

If the earth stops rotating, the

apparent value of g on its surface will(assuming the earth to be a sphere)

(a) decrease everywhere

(b) increase everywhere

(c) increase at pole and remain same everywhere

(d) increase everywhere but remain the same at poles.


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Solution

Equator

Apparent acceleration due to gravity:

2 2 2 2g g R sin (2g R) g

2At Equater 90 g g R (Least)

At Poll 0 g g (Largest)

When 0 g g

So except poles it increases everywhere.

Hence answer is (d)


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Thank you

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11. Particle Dynamics-5