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NON-uniform Circular Motion
* The NET acceleration is no longer pointing towards
the centre of the circle.
* There are TWO components of acceleration:
Radial / centripetal : due to the change in direction of velocity
Tangential : due to the change in magnitude of velocity
Radial acceleration
centre
Tangential acceleration
NET acceleration
Speed is changing
Examples of non-uniform circular motions
Vertical circle with a string and bob
string
bob
w
Roller Coaster
v
Free body diagram
Change in direction
Vertical circle with a string and bob
Radial direction :
T - mg cos q = mac = mv2 / r
string
q
T
bob
Tangential direction :
mg sin q = mat
mg sin q
q
mg cos q
Change in speed
mg
Can an object (mass m) go round a vertical circle of radius l
if the initial speed at the bottom is u?
C
Can go round the circle :
(1) Have enough energy to reach point C.
(2) Have sufficient high centripetal force
to maintain the circular motion at C.
D
B
l
m
u
A
Consider Conservation of energy ;
1
1 2
2
mu  mv  mg (2l )
2
2
u 2  v 2  2 g (2l )
u 2  v 2  4 gl
v 2  u 2  4 gl  0
u 2  4 gl
u  4 gl
Can an object (mass m) go round a vertical circle of radius l
if the initial speed at the bottom is u?
C
Can go round the circle :
v
D
(1) Have enough energy to reach point C.
(2) Have sufficient high centripetal force
to maintain the circular motion at C.
T
mg
B
l
m
A
Consider force at point C ;
mv2
mg  T 
l2
u
mv
T
 mg  0
l
v 2  gl
By Conservation of energy, v 2  u 2  4 gl
u 2  4 gl  gl
u  5gl
Can an object (mass m) go round a vertical circle of radius l
if the initial speed at the bottom is u?
C
Can go round the circle :
(1) Have enough energy to reach point C.
D
B
l
m
(2) Have sufficient high centripetal force
to maintain the circular motion at C.
u
A
u  4 gl
u  5gl
The object can go round the circle if the initial speed
is greater than
5 gl
hat happens if u < 5 gl ?
W
What happens if u <
C
(1)
5 gl ?
4 gl < u <
5 gl
Can reach C (as u > 4 gl )
D
B
l
m
No more circular motion can
be processed (as T = 0 but mg is
greater than mv2/l)
u
A
Projectile motion due to gravity
What happens if u <
C
(2)
D
B
2 gl < u <
4 gl
Between B and C(as u < 4 gl )
Projectile motion due to gravity
l
m
5 gl ?
u
(3) u <
2 gl
A
Cannot reach B
For reaching B,
1/2 mu2 = 1/2mvB2 + mgl
u2  2gl
u  2 gl
Swing about A between B and D
More about Circular Motion
* A astronaut feels weightless in a spaceship which is moving with
uniform circular motion about the Planet, say the Earth.
R
man
Mg + mg = (M+m) v2 / r
v2 = g r
R
Consider the man only,
R
Mg
for weightless
Consider the whole system (spaceship and man),
mg
v
R=0
mg
mg -R = mv2 / r
r
mg -R = m(g r) / r
mg -R = mg
R=0
More about Circular Motion
R
R = mg’
* Artificial gravity made for Space stations
mg’
man
Rotating axis
w
R
r
w
No weight as it is far away from all planets
There is only normal contact reaction force due to contact N.
R  mrw  mg '
rw 2  g '  9.8ms 2
2
More about Circular Motion
* Working principle of a centrifuge
* Working principle of a centrifuge
(1) Assume it is horizontally aligned with liquid of density
P1 = P
r inside.
P2 = P+P
(P2 - P1)A
Pressure gradient as centripetal force
FC = P A = (P2 - P1 )A = mrw2
The pressure gradient increases
with the distance from the rotating axis
* Working principle of a centrifuge
(2) Consider an element of the liquid of density
r inside.
All liquid rotates with uniform speed
Net force = (P2 - P1 )A
= [m] r w2
= [r V] r w2
= r(A r) r w2
Net force due to pressure gradient
= r r A w2 r
* Working principle of a centrifuge
(2) Consider an element of other substance of density
r’ inside.
r’
r’< r for less dense object
Net force Fnet = (P2 - P1 )A = r r A w2 r
Required centripetal force Fc = [m’] r w2
= [r’ V] r w2
= r’(A r) r w2
Move towards the axis
= r’ r A w2 r
r’> r for denser object
Move away from the axis
More about Circular Motion
* Why centrifuge ?
Excess force for separation 
= weight - upthrust
= (r’ A  r g )  (r A  r g)
FC = r’ r A w2  r
Fnet = r
r A w2
= (r’  r) A g  r
r
Assume r’ > r
Excess force for separation 
= (r’  r) r A w2  r
Fg
Fc
2
Fc ( r ' r )rw 2 Ar
r
w

(r ' r )Agr  g
Fg
Typical : r = 10 cm, w = 2500 rev min-1
Fc ~ 700 / 1
Fg