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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
Session Opener
Have you asked yourself
why our eyes observe a
wrong
Every
Science
day
says
we the
see earth
the sun
moves
risephenomenon ?
and
around
set. its
Weaxis
seeand
the in
sun
a day
moving
the
round
sun hardly
us. But
moves.
…….
Circular Motion
A moves in atocircular
Object
F Perpendicular
v path
radius r fixed : constrained motion
 F Directed towards center
 F
Constant in magnitude
 F And v change direction
continuously.

s  r
 has a direction
For uniform motion 
v is constant.
Lim.   0 ds  rd
F
v
Angular Variables (Constant Speed)

average
s  angular velocity


:
avg.
vaverage


r
t
t
t
vaverage= avg.r
 d
Lim t  0

  Instantaneous
t dt
angular velocity
v = r
r
vector
relation
v  tan
 rt 
cons tan
t v :cons
rad

t
:axial
vector.
SIunit
 d  
 dt
sec

r


 – o = t
r
O
r
Angular kinematical
equation for constant .
Angular Variables (Variable v)
1
v
changes
(t)  v(t) (a
v :constant)
along t angent
r

acceleration
along
t angent
v
= v0 +
+ t
at  :=
0 +
t
=
0
 : angular acceleration
1 22
1
 s  v0tt  at

t
0
2
2
2  20  2
r
 is tangential.
r
t
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?
Solution :
v
10 m/s
 
r 5  102 m
= 200 radians/s.
Unit Vector Along Radius and
Tangent
r  i r cos   j r sin 
[ox and oy : fixed reference frame].
r and  also define position
y
y’
j
p
O
(t)
r
(t)
i
x’
x
Transfer origin from O to P
ox || px’
oy || py’
P has acceleration
 Reference frame non inertial
Unit Vector Along Radius and
Tangent
r can be defined as
er  i cos   jsin  r  rer
r  rer (er : unit vector along r)
Define a unit vector e
perpendicular toer (along
v
)
e  i sin   jcos 
e
er
j
y

i

x
er ,e (origin P)
define a non inertial
reference frame
Unit Vector Along Radius and
Tangent
er  icos   jsin 
der
d
d
 isin 
 jcos 
 e
dt
dt
dt
Angular velocity vector is
the rate of change of radial
unit vector.
2
d er
dt2
e
j
y

d
d 

  icos 
 jsin  
dt
dt 

 2 er
er
i

x
Radial and Tangential Acceleration
Particle moves in a circle
r  rer
r cons tan t
  constant
v constant
2
d
dr r  ar  2rer
2v  re v : tangential
dt
dt
v r
ar : radial
v  r
: opposite er
2
v
ar  2r 
 v
r
e
er
v j
y

ar is perpendicular to v
ar
ar can not change the
magnitude of v

i
x
Radial and Tangential Acceleration
When  changes
= particle moves in the circle
(r constant)
= v changes
= tangential acceleration at appears
v
r  rer
dr
1
 v  re
dt
2 2

2
2
a  r t    r 
dv
 d


r
e  2 er 
dt
 dtat

t
Tan 
2 2
 r T ear  
r rer  a
 
 
 
a


at
ar
Centripetal Force in Circular
Motion
Object in circular motion has at and ar
art only
: necessity
changes
formagnitude
circular motion.
of v.
: no change circular
in magnitude
of v.
: non-uniform
motion
v2
ar 
 v  2r
r
Fcp
2
v
 m2r  m
 mv
r
Fcp
ar
O
r
As ar exists,an external
force Fcp must exists.
F cp 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) NA  NB  NC
(b) NB  NA  NC
(c) NC  NA  NB
(d) NB  NA  NC
C
A
B
Solution
mv 2
mg  NA 
rA
mv 2
 NA  mg 
rA
NB
mv2
rB
NA
mv2
rB
mv2

 NB = mg+ r
B
NB –mg=
B
mg
(B)
A
mg
m v2
rA
(A )
Hence answer is (b)
Solution
mv 2
mg  NC 
rC
mv 2
 NC  mg 
rC
NB is largest. From shape of
the track,
mv 2 mv 2
rA  rC 

.So NA  NC
rA
rC
N
C
C
m g
(C )
m v2
rC
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
Pendulum moves along the arc of a
vertical circle.
Resolving the motion along T (X-axis)
and perpendicular to T (Y-axis)
mg sinma (along y)
(Tangential)
 a  g sin 

mv2
T  mg cos  
(along x) (radial)
L
 T  m(g cos  
m v2
L
y
L

2
v
)
L
T
ma
mg
x
Class Exercise - 10
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.
1m
M
F
ix
e
d
c
e
n
tre
1m
/s
Solution
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)
mv 2
 kx 
(as spring has stretched by x)
(r  x)
(1+
x
)
mv2
mv2
2
 (r  x)x 
 x  rx 
0 O
k
k
M
F
F
=
re
s
to
rin
g
fo
rc
e
Solution
k


r

1
m,
v

1
m/s,
m

50
kg,

905


m



2
1
2 4mv 
x
r  r 
2
k 


x
1
4  50 

1

1



2 
905 
= 0.05 m.
(approx.)
Centripetal Force
Source of F cp
Friction
Centripetal force is friction.
fs
Object moves in circular
track radius r
Fcp  fmax
v2
 mg  m
r
v2max
 g (cons tan t)
r
Banking of Curves
Object moves along circular track.
Track banked towards center O.
N
Ncos=mg
y axis : N 2cos = 2mg
v
v
 Tan 

 g tan 
x axis : Nrgsin = rFc=mv2/r
mv2
 Nsin 
r

O
mg
If
If velocity
velocity <
> vv:: object
object
moves
moves outward
inward toto
decrease
increase rr.
 < 900
Conical Pendulum
P moves in horizontal circle at end of
string OP fixed to rigid support O.
Tension T supplies Fcp
2
Y axis
:T
cos
=
mg
v2
sin


 v2  rg tan   2gL
x axis
(r=L
sin )
rg : T sin = mv /r cos

Time tp to complete one revolution :
2r
L cos 
tp 
 2
v
g
1

 2 2 m2v2  2 

Note T   m g 

2 

r

 



L

T

mg
Class Exercise
Class Exercise - 1
Two 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
(a)
m1
m2
m2
(b)
m1
(c)
m2
m1
(d) 1
Solution
Centripetal force F1 (car m1)  m112r

centripetal force F2 (car m2 ) 
42 m1r
T12
42 m2r
T22
As circles are identical and flat, friction
supplies centripetal force in both cases.
  m1g 
  m2g 
m1r 42
T12
m2r 42
T22
 g 
42r
 g 
T12
............(1)
42r
T22
Hence answer is (d)
Dividing (2) & (1) we get,
.................(2)
T12
T22
1
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=Nmg.
Deceleration = g.
In applying the brakes car must stop
within distance d.
2
v
v 2f  vi2  2ax  0  v 2  2( g) d  d 
2 g
In taking a circular path, the maximum
radius is d.
mv 2
mv 2
v2
f
  mg 
d
d
d
g
So applying the brakes is the better option.
f
Class Exercise - 8
ra
A  0.1 B  0.2
rb
rA  1km rB  2 km
A
B
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.
v2
Skidding occurs for
 g
r
Maximun speed v  rg
 v A  0.1 1 10  103  10 10 m/ s
 vB  0.2  2  10  103  20 10 m/ s
In semicircular section: Length of track A = km
T1  Time taken by m.c.A 

 102 S  T1
10
Solution
Length of track B = 2km
T2  Time taken by m.c.B 

 102 s  T1
10
Semicircular portion is
negotiable in equal time.
 m.c.B is faster, so will complete
straight parts faster.
Class Exercise - 9
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)
m
Solution
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.
 fmin  m2r  mg
  2  10 rad / s,r  0.1,g  10 m / s2

g
1m
2r

f
10
2  100  0.1

1
 0.025
40
m
mg
Centrifugal Force in Circular Motion
P moves in circle (radius r) with
angular velocity 
Reference frame centered at origin O :
Reference frame is inertial
 Fcp  m2r. (along PO)
Y' '
X' '
X' '
P
r
Reference frame with P as origin
Y' '
Reference frame is non inertial.
X' '
P
O

X' '
P
P
Y' '
Y' '
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;  F  Fcp  Fpseudo  0.
Fcp
P
O
Fpseudo
Fpseudo = m2r away
from center.
Fpseudo is called
centrifugal force.
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
(a) m (2  02 ) a
(b) m2a
(b) m  0a
(d) m 02a
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
2
acceleration 0 a .
 centrifugal force  m02 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
Apparent acceleration due to gravity:
g  g2  2R sin2  (2g  2R)  g
At Equater   90  g  g  2R (Least)
At Poll   0  g  g (Largest)
When   0 g  g
So except poles it increases everywhere.
Hence answer is (d)

Equator
Thank you