ALPhy Ch05b

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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

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Common Error

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Common Error

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An object in uniform circular motionexperiences a centripetal acceleration which is

the acceleration directed towards the centre of

the circle.

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r

v

rva

22

)(onacceleratilCentripeta===

pointing to

centre of circle

====

=

===

tv

tv

tva

vv

vvvvv BA

)(takenTime

yin velocitChange

and

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Common Error

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Centripetal force


Steady speed vbut changing direction of motion

- force acting on it (Newtons 1st Law)

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Common Error

No force in direction ofmotion (constant speed)

Force towards centre of

circular path

centripetal force (Fc)

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Centripetal force


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


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


1. Procedure

- rubber bung whirled around in horizontal circle

- measure time taken for 50 revolutions of bung- calculate angular velocity ()

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2. Analysis

Tcos = mg .......................... (1)

Tsin = mr2

.......................... (2)Tsin = m ( sin) 2

T= m2

Tis provided by Mg

Fc = mr2

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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

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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.

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TextText


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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


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14

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


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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.

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TextText


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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


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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.


grv

mgTr

mv

T

2

2

tan:(2)(1)

.(2)....................cos:motionVertical1).........(..........sin:motionHorizontal(b)

=

==

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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


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


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 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 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,


=

==

=

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


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 .

ALPhy Ch05b

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