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• Centripetal accelerationCentripetal acceleration• • Centripetal forceCentripetal force

5.2 Centripetal 5.2 Centripetal acceleration and forceacceleration and force

• • Centripetal force experimentCentripetal force experiment


Centripetal acceleration

5.2 Centripetal acceleration and force (SB p. 170)

Change in velocity (v) = vB – vA (represented by QR)

Particle experiences change of velocity (acceleration) along direction AO (pointing to centre of circle) centripetal

acceleration

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Common Error

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Common Error




An object in uniform circular motion experiences a centripetal acceleration which is the acceleration directed towards the centre of the circle.




rvrva

22 )(on accelerati lCentripeta

pointing to centre of circle

tv

tv

tva

vv

vvvvv BA

)( takenTime

yin velocit Change

and

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Common Error


Centripetal force


Steady speed v but changing direction of motion

- force acting on it (Newton’s 1st Law)

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Common Error

No force in direction of motion (constant speed)

Force towards centre of circular path

centripetal force (Fc)


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 the body and the kinetic energy of the body remains unchanged.

mvmrrmvmaF 2

2

c )( force lCentripeta


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 the centre. It is a functional name rather than a real force. The origins of the centripetal forces may be tension, friction or reaction forces.


Centripetal force experiment


1. Procedure

- rubber bung whirled around in horizontal circle

- measure time taken for 50 revolutions of bung

- calculate angular velocity ()




2. Analysis

T cosθ = mg .......................... (1)

T sinθ = mr2 .......................... (2)

T sinθ = m ( sinθ) 2

T = m2

T is provided by Mg

Fc = mr




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


End


When a particle moves in uniform circular motion, it has constant speed but not constant velocity because its direction changes from time to time.

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TextText



Uniform circular motion is not a kind of uniformly accelerated motion since the acceleration is fixed only in magnitude, but not in direction.

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As centripetal acceleration (a) may be expressed as

It is wrong to think that

Since velocity is not a constant and v r.

Therefore,

Centripetal acceleration is actually increased with radius r.

rva

2

ra 1

rarr

rva

22Return to

TextText



The circular motion does not produce a centripetal force. The fact is that the centripetal force that causes the circular motion is actually a resultant of other forces.

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Q:Q: (a) In the centripetal force experiment mentioned above, what will happen when the string breaks while the bung is whirling?(b) What is the relationship between the vertical angle θ 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 string cannot be whirled in circle in air with the string horizontal.

Solution



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 θ increases as the bung is whirled at a higher speed.


grvmgTrmvT

2

2

tan :(2)(1)

.(2)....................cos :motion Vertical

1).........(..........sin :motion Horizontal (b)

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More to Know 1More to Know 1


Solution (cont’d):Solution (cont’d):

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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 θ with the vertical (Fig. (b)) so that the vertical component of T counteracts the weight mg.

T cosθ = mg


Fig. (a) Fig. (b)


1. The equation tanθ = is very useful in answering questions about circular motion.2. The vertical angle θ is independent of the mass m.3. A specific vertical angle θ is ideal for one speed only.

gr

grv 22

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Q:Q: (a) A pendulum bob moves in a horizontal circle 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 T in the string, and(iii) the angle θ.



Q:Q: (b) A student suggests that the value of g can be determined by measuring the period t of the revolution of the bob for various values of θ.

(i) Find an expression for t in terms of , g and θ.

(ii) Suggest a graph which could be used to obtain the value of g.

(iii) Discuss critically whether this is a good method for the determination of g.

Solution



Solution :Solution :

(a) (i) Centripetal force (Fc) = mr2 = msinθ 2

(ii) Horizontal component of T,

T sinθ = Fc = msinθ 2

∴ T = m2

(iii) Vertical component of T,


2

1

2

cos

cos

cos

g

gTmg

mgT


Solution (cont’d) :Solution (cont’d) :

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TextText


mg

gtm

gt

tg

t

g

2

22

22

2

4

4cos

)(graph ofGradient

cos4 (i), From

plotted. is cosagainst ofGraph (ii)

cos22)( Period

cos (a)(iii), From (i) (b)

(iii) It is not a good method of determining g because it is difficult to maintain the angle θ at a fixed value, and it is difficult to measure the angle θ.

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