Physics

Session

Particle Dynamics - 5

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

Every day we see the sun rise and set. We see the sun moving round us. But …….

Session Opener

Science says the earth moves around its axis and in a day the sun hardly moves.

Have you asked yourself why our eyes observe a wrong phenomenon ?

F

v

F v

Fv

F

v

F

v

F

F

v

Fv

Fv

Object A moves in a circular path radius r fixed : constrained motion F

Directed towards center

v((((((((((((((

Perpendicular to F

Constant in magnitudeF

And change direction continuously.

F

v((((((((((((((

For uniform motion v is constant.

s r has a direction

Lim. 0 ds rd

Circular Motion

Fv

F v

Angular Variables (Constant Speed)

averages

v rt t

avg. : average angular velocity

t

vaverage= avg.rd

Lim t 0t dt

Instantaneous angular velocity

v = r

rO

r

r

r

vector relation : v r

rad:axial vector. SI unit

sec

((((((((((((((((((((((((((((((((((((((((((

(((((((((((((( t

cons tant v cons tant

d dt

– o = t

Angular kinematical equation for constant .

Angular Variables (Variable v)

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

r

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

20

1s v t at

2 2

01

t t2

r

r t

r

t

r

t

= 0 + t

2 20 2

is tangential.

Class Exercise

Class Exercise - 5

A particle moves with a constant linear speed of 10 m/s in a circular path of radius 5 cm. What is its angular velocity?

2v 10 m/s

r 5 10 m

= 200 radians/s.

Solution :

Unit Vector Along Radius and Tangent

r i r cos j r sin

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

Transfer origin from O to P ox || px’ oy || py’ P has acceleration Reference frame non inertial

O

p

x

y

r(t)

i

j

(t)

y’

x’

Unit Vector Along Radius and Tangent

r r

r can be defined as

r re (e : unit vector along r)

(origin P) define a non inertial reference frame

re ,e

re i cos j sin rr re

e i sin j cos

Define a unit vector perpendicular to (along )

re

ev((((((((((((((

i

jree

y

x

Unit Vector Along Radius and Tangent

i

jree

y

x

re icos j sin

rde d d

isin j cos edt dt dt

2r

2d e d d

icos j sindt dtdt

2re

Angular velocity vector is the rate of change of radial unit vector.

Radial and Tangential Acceleration

v constant constant

v : tangential

v r

v r

((((((((((((((

((((((((((((((((((((((((((((((((((((((((((dr

v redt

((((((((((((((

22

r r2d r

a redt

r

r

22

r

a : radial

: opposite e

va r v

r

can not change the magnitude of

ra

v((((((((((((((

Particle moves in a circlerr re r cons tant

i

jree

y

x

v

is perpendicular tora

v(((((((((((((( ar

Radial and Tangential AccelerationWhen changes

= particle moves in the circle (r constant)

= v changes

r

2r

2rT

r re

drv r e

dt

dv dr e e

dt dt

r e r e a

((((((((((((((

((((((((((((((

ar

= tangential acceleration appearsta

at

v

a

1

2 22 2t

t t2

r

a r r

aTan

a r

Centripetal Force in Circular MotionObject in circular motion has at and ar

at only changes magnitude of v. : non-uniform circular motion ar : necessity for circular motion. : no change in magnitude of v.

22

cpv

F m r m m vr

22

rv

a v rr

rO

ar

As ar exists,an external force Fcp must exists.

Fcp

cpF

is a radial force, called centripetal force

Class Exercise

Class Exercise - 3

A vehicle moves with constant speed along the track ABC. The normal reaction by the road on the vehicle at A, 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

Solution

NA

Amg

mv2

rA

(A)

2

AA

mvmg N

r

2

AA

mvN mg

r

NB

B

mg

mv2

rB

(B)

N – mg = B

mv2

rB

N = mg + B

mv2

rB

Hence answer is (b)

Solution

NC

mg mv2

rC

C

(C )

2

CC

mvmg N

r

2

CC

mvN mg

r

2 2

A C A CA C

mv mvr r .So N N

r r

NB is largest. From shape of the track,

Class Exercise - 7

A pendulum, constructed by attaching a tiny mass m at the end of a light string of length L, is oscillating in a vertical plane. When the pendulum makes an angle with vertical, its speed is equal to v. Find the tension in the string and the tangential acceleration at that instant.

Solution

mgma

L

T mv2

Ly

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

2mvT mg cos (along x) (radial)

L

2vT m(g cos )

L

Class Exercise - 10

1 m

Fixedcentre

1 m/s

M

A mass m of 50 kg is set moving in a horizontal circular path around a fixed centre O to which it is connected by a spring of unstretched length of 1 m and a spring constant of 905 N/m. Find out the amount by which the spring will stretch if the speed of the mass is 1 m/s and the spring is light.

Solution

FOM

(1 + x)

F = restoring force

As the mass moves, it tends to slip outwards, providing a stretching force. Spring provides the reaction (restoring force), which is the cause of centripetal force

(|F| = kx)

2mvkx (as spring has stretched by x)

(r x)

2 22mv mv

(r x)x x rx 0k k

Solution

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

m

221 4mv

x r r2 k

1 4 50x 1 1

2 905

= 0.05 m. (approx.)

Centripetal Force

cpF

Source of

Friction

Object moves in circular track radius r

2

cp maxv

F f mg mr

Centripetal force is friction.

fs

2maxv

g (cons tant)r

Banking of CurvesObject moves along circular track.

Track banked towards center O.

O

y axis : N cos = mg

x axis : N sin = Fc=mv2/r

mg

N

2mvNsin

r

Ncos=mg

2 2v vTan gtan

rg r

If velocity > v : object moves outward to increase r

If velocity < v : object moves inward to decrease r.

< 900

Conical PendulumP moves in horizontal circle at end of string 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 sin

v rgtan gLrg cos

p2 r L cos

t 2v g

Time tp to complete one revolution :

12 2 22 22

m vNote T m g

r

L

mg

T

Class Exercise

Class Exercise - 1Two similar cars, having masses of m1

and m2 move in circles of equal radii r. Car m1 completes the circle in time T1 and car m2 completes the circle in time T2. If the circular tracks are flat, and identical, then the ratio of T1 to T2 is

1 2

2 1

2

1

m m(a) (b)

m m

m(c) (d) 1

m

Solution

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

21

21

4 m r

T

22

2 2 22

4 m rcentripetal force F (car m )

T

As circles are identical and flat, friction supplies centripetal force in both cases.

2 21

1 2 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)

Class Exercise - 6

The driver of a car, moving at a speed of v, suddenly finds a wall across the road at a distance d. Should he apply the brakes or turn in a circle of radius d to avoid a collision with the wall? (Coefficient of kinetic friction between the road and the tyre of the car is .)

Solution

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

In applying the brakes car must stop within distance d.

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

2vd

2 g

f

In taking a circular path, the maximum radius is d.

2 2 2mv mv vf mg d

d d g

So applying the brakes is the better option.

Class Exercise - 8

A B

ra

rb A 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 a semicircular section, whose radii for track A and track B are 1 km and 2 km respectively. Friction coefficients of A and B are 0.1 and 0.2 respectively. The rules of the race requires that each of the motor cyclist must travel at constant speed without skidding. Which car wins the race? (g = 10 m/s2) (Straight sections are of equal length)

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

2

1 1T Time taken by m.c.A 10 S T10

Solution

Length of track B = 2km

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

10

Semicircular portion is negotiable in equal time.

m.c.B is faster, so will complete

straight parts faster.

Class Exercise - 9

m

Figure shows a centrifuge, consisting of a cylinder of radius 0.1 m, which spins around its central axis at the rate of 10 revolutions per second. A mass of 500 g lies against the wall of the centrifuge as it spins. What is the minimum value of the coefficient of static friction between the mass and the wall so that the mass does not slide? (g = 10 m/s2)

Solution

f

mg

1 m m

The mass will not slide if mg f

The mass will press the wall at a force equal and opposite to the centripetal force supplied by the wall which is the reaction force.

2minf m r mg

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

2g

r

210 1

0.02540100 0.1

Centrifugal Force in Circular MotionP moves in circle (radius r) with angular velocity Reference frame centered at origin O :

Reference frame is inertial2

cpF 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

Centrifugal Force in Circular Motion

P is at rest with respect to itself.

A pseudo force Fpseudo to be added 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

Class Exercise

Class Exercise - 2

A particle of mass m moves in a circular path of radius r with a uniform angular speed of in the xy plane. When viewed from a reference frame rotating around the z-axis with radius a and angular speed , the centrifugal force on the particle is equal to

2 2 20

20 0

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

(b) m a (d) m a

Solution

Centrifugal force is a pseudo force equal to –m × (Acceleration of the frame of iron inertial frame) Non inertial frame in this case has radial acceleration .2

0 a

20centrifugal force m a

Hence answer is (d)

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.

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)

Thank you