physics. session particle dynamics - 5 session objective 1.circular motion 2.angular variables...
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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
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
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