R.R. Mishra

Department Of Physics

BITS Pilani

Mechanics, Oscillations

and Waves (MEOW)

1. Mechanics (R.R. Mishra)

20 ~ 22 Lectures

2. Oscillations & Waves (D.D. Pant)

20 ~ 22 Lectures

Major Division

Textbooks :

1. An Introduction to Mechanics :

Daniel Kleppner & Robert

Kolenkow

2. The Physics of Vibrations &

Waves : A. P. French

Mechanics

Chapter No. 2 : Review of Newton’s

Equations

Chapter No. 3 : Linear Momentum

Chapter No. 4 : Work, Energy & Power

Chapter No. 6 : Angular Momentum

Chapter No. 8 : Non-inertial systems and

Fictitious Forces

A world simple enough to be

understood, would be too simple

to produce a mind that can

understand it. J.D. Barrow

Chapter 2

• Constrained Motion

• Newton’s Equations in Polar

Coordinates

Constrained Motion

Examples

1. Pulley-mass system

h

1x

2x

Constraint eq.

)xh()xh( 21

0aa 21

2. Block on a fixed wedge

x

y

Constraint equation :

h

1h

yx

xy a

ha

Prob. 2.16

450

A Angle of wedge : 450

Acceleration of

wedge : A, to right

No friction between block and

wedge

Q : What is the acceleration of the

block w.r.t. the ground

Block on accelerated wedge

450

A

N

mg

2tA2

1hyx Aaa yx

mAmg2

N

2

N

)gA(

2

mN

x

y

2At2

1 h

h

Constraint Equation :

)gA(2

1a;)gA(

2

1a yx

The block will climb up the wedge iff A > g

450

N

mg

x

y

h

h

X

Suppose the wedge is left to itself and

is free to move on a frictionless surface.

Xhyx xyx Aaa

2

Nmg

2

N

2

N

3

mg

2

N

3

gA;

3

g2a;

3

ga xyx

(Mass of wedge : m)

1M

2M

3M

Prob. 2.20

Consider the “pedagogic machine”. All

surfaces are frictionless. Find the

acceleration of block when the system

is released. 1M

The Pedagogic Machine

Constraint Equation :

)yy()xx( 021

0aaa 321

1M

2M

3M

1x

0y

2x

y

x

y

22aMT

333 aMgMT

131 a)MM(T

0aaa 321

Four unknowns, , and four

equations ! 321 a&a,a,T

1M

2M

3M

T

T

T

All four unknowns can be solved for.

In particular,

gMMM2MMMM

MMa

2

3323121

321

Newton’s Equations in Polar Co-ordinates

Review of Newton’s Eq. in Cartesian

Co-ordinates

The Cartesian System

i

j F

i

j

jFiFF yx

jvivv yx

Equations of Motion

j)t(yi)t(x)t(r

jyixdt

rd)t(v

jyixdt

vda

m

x

y

r F

Fam

jFiF)jyix(m yx

yx Fym;Fxm

Polar Co-ordinates

P r

,r

The Polar Grid and Unit Vectors

r

r

1ˆˆrr

0ˆr

Resolving Vectors in Polar Coordinates

r

A

ˆArAA r

Unit vectors in the polar co-ordinates

vary from point to point, unlike unit

vectors in the Cartesian co-ordinates.

Expressing in terms of .

)ˆ,r(

)j,i(

)ˆ,r( )j,i(

jsinicosr θ+θ=

jcosisin-ˆ θ+θ=θ

r

x

y

θ

θ

r

Newton’s Equations of Motion in Polar

Coordinates

r

x

y

rθ

θ

jsinicosr θ+θ=

jcosisin-ˆ θ+θ=θ

r-θ&θr

rrr =

We have,

ˆrrr

Another differentiation leads to

rr-ˆrˆrˆrrrva 2

ˆ)r2r(r)r-r( 2

ˆaraawithComparing r

rrrrrv

rv&rvr

r2ra&r-ra 2

r

r

2 F)r-r(m∴

F)r2r(m } Newton’s Eqs. in

polar coordinates

Prob. 2.29

A car moving radially

outward on revolving

platform.

Coeff. of friction = μ

Ang. Velocity of

platform = ω (constant)

Velocity of car w.r.t.

platform = v0

Car starts from centre of platform

v0

ω

Find :

a) Acceleration of car as a function of

time using polar coordinates. Show by

vector diagram

b) The time at which the car starts to

skid

c) Direction of frictional force at the

time of skidding

a) We have :

tvr 0

t

22

r rrra

0v2r2ra

v0

x

y

x

y

ra

aa

2

r ra

0v2a

b) Net force on car :

22

r FFF 22

r aam

2

0

242 v4rm

22

0

2

0

2242

0 t4vmv4tvm

The car will start to skid when

gmF 2

0

2

0

222

v

v4gt

c) Direction of frictional force at the

time of skidding

rF

F

F

2

r rmF

0vm2F

Direction of

skidding

Prob. 2.33

x

r

θ

y ω

m

Rod with a mass m

on it. Rod rotates

with constant

angular velocity

ωon a horizontal

plane and mass free

to slide.

i) Show that motion is given by tt- eBeA)t(r

Find β.

ii) Show that for a particular choice of

initial conditions, it is possible to obtain

a solution such that r continually

decreases and that for all other choice, r

will eventually increase.

x

r

θ

y ω

m

a) The Polar equations

are :

0)rr( 2

NFrm2

N

rdt

rd 2

2

2

The coordinate Should be such a

function of that twice differentiation of

w.r.t. will be proportional to itself.

rrt

t

An intelligent guess :

te)t(r

Substituting this into the LHS

The radial equation is

rr 22

tt e&eBoth are solutions for r

The given equation being a linear

equation, the most general solution is

tt eBeA)t(r

where, A & B are constants to be determined

ii) Let the initial conditions be :

00 v)0(r&r)0(r

The complete solution is then :

t

00

t

00 evr2

1evr

2

1)t(r

Now,

t00

t

00 evrevr2

1

dt

dr

Additional Part

rm2N

What is the normal force of the rod on

the bead?

t0

0

t00

2 ev

rev

rm

For to be negative, the coeff. of

must be negative.

dtdrte

Prob. 2.35

v0

A block of mass m slides

on the inside of a ring

fixed to a frictionless

table. It is given an initial

velocity of v0. Coefficient

of friction between the ring and the block

is μ. Find the velocity and position at a

later time t.

N

μN

N-Fr =

N-F μ=θ

ANS :

NmR)rr(m 22

NmR)r2r(m

The polar equations are :

Substituting N from the first equation in

the second

2

Putting

2

dt

d

t

0

2td

d

0

t11

0

t1 0

0

tvR

vRv

0

0

t1dt

d

0

0

0

t

0 0t1

tdd

t1n1

0

0

Prob. 2.37

A bowl shaped racing

track, on which a

racing vehicle can

move in horizontal

circles without

friction.

Q : What should be the equation of the

vertical cross section of the track, so

that it takes the same time T to circle the

track, whatever be its elevation

x

y

θ

x

y

θ

N

mg

mgcosN

x

mvsinN

2

xg

v

dx

dytan

2

,T

x2v,Since

xkTg

x4

dx

dy2

2

2

xky

2

(Parabola)