ALPhy Ch05b
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Transcript of ALPhy Ch05b
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7/27/2019 ALPhy Ch05b
1/23
Manhattan Press (H.K.) Ltd. 1
Centripetal accelerationCentripetal acceleration Centripetal forceCentripetal force
5.2 Centripetal5.2 Centripetalacceleration and forceacceleration and force
Centripetal force experimentCentripetal force experiment
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Centripetal acceleration
5.2 Centripetal acceleration and force (SB p. 170)
Change in velocity (v)
= vB vA (representedby QR)
Particle experiences change of velocity
(acceleration) along directionAO (pointing to
centre of circle)
centripetal
acceleration
Go to
Common Error
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Common Error
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Centripetal acceleration
5.2 Centripetal acceleration and force (SB p. 170)
An object in uniform circular motionexperiences a centripetal acceleration which is
the acceleration directed towards the centre of
the circle.
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Centripetal acceleration
5.2 Centripetal acceleration and force (SB p. 171)
r
v
rva
22
)(onacceleratilCentripeta===
pointing to
centre of circle
====
=
===
tv
tv
tva
vv
vvvvv BA
)(takenTime
yin velocitChange
and
Go to
Common Error
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Centripetal force
5.2 Centripetal acceleration and force (SB p. 171)
Steady speed vbut changing direction of motion
- force acting on it (Newtons 1st Law)
Go to
Common Error
No force in direction ofmotion (constant speed)
Force towards centre of
circular path
centripetal force (Fc)
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Centripetal force
5.2 Centripetal acceleration and force (SB p. 172)
Note:For a uniform circular motion:
1. Centripetal force is always perpendicular to
the motion of the body. It does no work on thebody and the kinetic energy of the body remains
unchanged.
mvmrr
mvmaF ====
22
c )(forcelCentripeta
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Centripetal force
5.2 Centripetal acceleration and force (SB p. 172)
Note:
2. The centripetal force (Fc) is the force required
to keep the body moving in a circle. It is provided
by the external resultant force towards thecentre. It is a functional name rather than a real
force. The origins of the centripetal forces may
be tension, friction or reaction forces.
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Centripetal force experiment
5.2 Centripetal acceleration and force (SB p. 172)
1. Procedure
- rubber bung whirled around in horizontal circle
- measure time taken for 50 revolutions of bung- calculate angular velocity ()
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Centripetal force experiment
5.2 Centripetal acceleration and force (SB p. 173)
2. Analysis
Tcos = mg .......................... (1)
Tsin = mr2
.......................... (2)Tsin = m ( sin) 2
T= m2
Tis provided by Mg
Fc = mr2
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Centripetal force experiment
5.2 Centripetal acceleration and force (SB p. 173)
3. Error
(i) there is a friction acting at the
opening of the glass tube,
(ii) the string is not inextensible,
(iii) the rubber bung is not
whirled with constant speed, and(iv) the rubber bung is not
whirled in a horizontal circle.
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Example 3Example 3
Go to
Example 4Example 4
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End
http://e/Dir/Media/PowerPoint/Chapter01/E-Ch01_02.ppthttp://e/Dir/Media/PowerPoint/Chapter01/E-Ch01_02.ppt -
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When a particle moves in uniformcircular motion, it has constant speed
but not constant velocity because its
direction changes from time to time.
Return to
TextText
5.2 Centripetal acceleration and force (SB p. 170)
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Uniform circular motion is not a kindof uniformly accelerated motion since
the acceleration is fixed only in
magnitude, but not in direction.
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TextText
5.2 Centripetal acceleration and force (SB p. 170)
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As centripetal acceleration (a) may be expressedas
It is wrong to think that
Since velocity is not a constant and vr.
Therefore,
Centripetal acceleration is actually increasedwith radius r.
r
va
2
=
ra 1
rar
r
r
v
a
=
22Return to
TextText
5.2 Centripetal acceleration and force (SB p. 171)
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The circular motion does not produce acentripetal force. The fact is that the
centripetal force that causes the circular motionis actually a resultant of other forces.
Return to
TextText
5.2 Centripetal acceleration and force (SB p. 171)
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Q:Q: (a) In the centripetal force experimentmentioned above, what will happen when the
string breaks while the bung is whirling?
(b) What is the relationship between the
vertical angle of the string and the speed
of the bung?
(c) Explain with the aid of a diagram, why a
mass at the end of a light inelastic stringcannot be whirled in circle in air with the
string horizontal. Solution
5.2 Centripetal acceleration and force (SB p. 173)
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Solution:Solution:
(a) If the string breaks, the centripetal force disappears.
The bung can no longer keep in the circular motion. It
will fly away tangentially.
The angle increases as the bung is whirled at a higherspeed.
5.2 Centripetal acceleration and force (SB p. 174)
grv
mgTr
mv
T
2
2
tan:(2)(1)
.(2)....................cos:motionVertical1).........(..........sin:motionHorizontal(b)
=
==
Go to
More to Know 1More to Know 1
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Solution (contd):Solution (contd):
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TextText
(c) If the string is horizontal (Fig. (a)), there is no vertical force
to balance the weight of the mass mg. Therefore, the string
must make an angle with the vertical (Fig. (b)) so that the
vertical component ofTcounteracts the weight mg.
Tcos = mg
5.2 Centripetal acceleration and force (SB p. 174)
Fig. (a)Fig. (b)
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1. The equation tan =is very useful in answering questions aboutcircular motion.2. The vertical angle is independent of themass m.3. A specific vertical angle is ideal for onespeed only.
gr
grv 22
=
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TextText
5.2 Centripetal acceleration and force (SB p. 174)
i l l i d f ( )
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Q:Q: (a) A pendulum bob moves in a horizontalcircle with constant angular velocity as shown in the figure. Find, in terms of m, , and g,
(i) the centripetal force acting on the bob,(ii) the tension Tin the string, and(iii) the angle .
5.2 Centripetal acceleration and force (SB p. 174)
5 2 C i l l i d f (SB 175)
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Q:Q: (b) A student suggests that the value of gcanbe determined by measuring the period tofthe revolution of the bob for various valuesof .
(i) Find an expression for t in terms of
, gand .(ii) Suggest a graph which could be used toobtain the value of g.
(iii) Discuss critically whether this is a goodmethod for the determination of g.
Solution
5.2 Centripetal acceleration and force (SB p. 175)
5 2 C t i t l l ti d f (SB 175)
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Solution :Solution :
(a) (i) Centripetal force (Fc) = mr2 = msin 2
(ii) Horizontal component ofT,
Tsin = Fc = msin 2
T= m2
(iii) Vertical component ofT,
5.2 Centripetal acceleration and force (SB p. 175)
=
==
=
2
1
2
cos
cos
cos
g
gTmg
mgT
5 2 C t i t l l ti d f (SB 175)
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Solution (contd) :Solution (contd) :
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TextText
5.2 Centripetal acceleration and force (SB p. 175)
m
g
gt
m
gt
t
gt
g
2
22
2
2
2
4
4cos
)(graphofGradient
cos4(i),From
plotted.iscosagainstofGraph(ii)
cos2
2)(Period
cos(a)(iii),From(i)(b)
=
==
=
==
=
(iii) It is not a good method of determining gbecause it is
difficult to maintain the angle at a fixed value, and it is
difficult to measure the angle .