@TecQuiprnent Ltd 1999
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Educational
PRODUCTS
CONTENTS
Section
1
INTRODUCTION
1
2
DESCRIPTION OF THE APPARATUS
3
3
THEORY
5
4
EXPERIMENTAL PROCEDURE
Electrical Connections
Procedure
Results
7
7
7
8
SECTION 1 INTRODUCTION
Figure 1 Centrifugal force apparatus
The appreciationof centrifugal force has always been
regarded as one of the most important engineering
topics in the field of mechanics.Its effectscan be most
useful in somedesignapplications,but may also be the
causeof drasticfailures in rotating components.
An automaticclutch is a very good exampleof using
centrifugal force for power transmission.The unit is
usually installed between the motor shaft and the
machinerywhich it drives. It comprisestwo or more
'shoes' which are pivoted on the driving shaft. When
this shafthasreacheda certainspeed,centrifugaleffects
overcomethe force exerted by the restraining springs.
The shoesthen move outwardsto pressagainstthe inner
surfaceof the rim of an annulusmountedon the driven
shaft, and thereby rotate the machinery.The speedat
which the clutch engagesdependson the stiffnessof the
restraining springs and the mechanicaldesign of the
clutch. When the shafts are not in motion, the
restraining springs retract the 'shoes' to their Static
position and the machinery is completely disengaged
from the motor. This type of automatic clutch is
particularly useful when the driven machineryrequiresa
high startingtorque.
On high speed machinery such as turbines, the
centrifugal force due to a small out of balance
componentcan causeseriousvibrationswhich may lead
to failure of vital components. The rotating parts
thereforehave to be balancedto a fine limit and made
strong enough to withstand the centrifugal effects
involved. Since rotational speed is always a critical
factor in modem machinery,a considerableamountof
researchwork hasbeendoneto find new materialssuch
as carbon fibre plastics and titanium alloys. These
materialshavemuch lower densityto give exceptionally
high specificstrength.
The TQ Centrifugal Force Apparatus allows the
magnitudeof the centrifugal force acting on a body to
relate to its rotational speedand radius. The apparatus
hasprovisionsfor varying the massof the body, radius
and rotational speed,and the results shown to accord
with theory.
SECTION 2 DESCRIPTION OF THE APPARATUS
The TQ Centrifugal Force Apparatus is designed to
demonstratethe relationshipbetweencentrifugal force,
massof a rotating body, its distancefrom the axis and
its angular velocity. It consistsof two pivoted counter
balanced bell-cranks housed in slideable blocks, as
shown in Figure 2. Various combinationsof accurately
machinedmassesfit to the endsof the bell-crank arms.
Sufficient massesare pivoted to enable the mass on
each arm to be increasedby between25 and 175g in
incrementsof 25 g. The slideable blocks are held in
position by locating pins. Each block fits in five
different radial positions correspondingto five equally
spaced holes in each end of the horizontal rotating
member.
Horizontal
rotating member
Counter-balanced
bell-aank
Masses
A transparent safety dome covers the rotating
assembly.Removalof the dome disconnectsthe motor
from the power supply.
Figure 3 shows the horizontal rotating member
carrying two counter-balancedbell-cranks. The bellcrankspivot in brackets,which can fix at severalpoints
along the rotating member. When the unit rotates,the
upper massesM. tend to move outwards under the
actionof centrifugalforce.
With the upper massesat radius r and rotating at
rad/s,the force on eachmassis:
F = M.ro2r
Gravitational force on the lower massesMb restrains any
movement of the bell-cranks until the centrifugal force
balances this gravitational force. At this point. the upper
masses will move outwards. Taking moments about the
pivot point at the condition of balance:
MatJr x 0.04 = M.,g x 0.04
F = Ma~r
Baseunit
Figure 2
The rotating member is belt driven from a variable
speed 12 V d.c. electric motor contained in the base
unit. The motor control is via the SpeedController. An
optical tachometersensoris also incorporated,and on
the front of the module unit is an output socket for
connectingto the TachometerUnit.
= M.,g
Thus, the centrifugal force at the condition of balance is
equal to the weight of the lower mass Mb' The angular
velocity ro can be determined by measuring the speed of
rotation when the upper masses move outwards. The
theoretical centrifugal force can then be calculated and
compared with the measured value M.,g.
The independent relationships between centrifugal
force F and each of the variables: mass, angular velocity
and radius can be demonstrated. These relationships are:
F«
~ for constant Ma and r
F «M.
F« r
for constant rand ro
for constant M. and ro
It must be stressed at this point that the masses M. and
Mb are not simply the masses placed on the weight
hangers a and b. The bell crank, with its attendant
weight hangers, also has a mass subject to gravitational
acceleration and centrifugal acceleration and this acts
through the centre of gravity of the bell crank assembly.
To have any value in the experiment the mass of the bell
crank assembly must be equated to an equivalent mass
acting at the weight hangers a and b. This is calculated
at 15 g and is marked on the plate adjacent to each
weight hanger.
Hence,
M. = Mass placed on weight hanger a + 15g
Mb = Mass placed on weight hanger b + 15g
The relationshipbetweencentrifugal force and angular
velocity canbe detenninedsimply by varying the values
of Mb with both M. and r constant.To demonstratethe
other two relationshipsrequires a series of tests for
varying the massesof the bodies A and B and varying
the radius.By plotting the resultsand readingoff values
at constant angular velocity, separategraphs can be
drawn showing the variation of radial force with mass
M. and with radiusr.
SECTION 3 THEORY
Considera body moving in a circular path of radius r,
with an angularvelocity (I) as shown in Figure 1. When
the body movesthrough a small angle 09, the velocity
vector,v changesdirection asshownin Figure4.
If the changein direction of velocity takesplacein time
01, and the change of velocity is ov, representedby
vectorAB in Figure 5, then
Acceleration = Ov
~t
(1)
If the angle50 is small, we canwrite:
Of=
-50.
and
ov=v09
Substitutingthesevalues of Sv and Sf in Equation (1)
above:
Acceleration= V(J)
Finally, by noting that v = (JYwe obtain:
Acceleration= roZr
Figure 4
This accelerationis termed the centripetal, or centre
seeking,accelerationbecausethe massis continuously
acceleratedtowardsthe centre.From Newton's Second
Law of Motion, a force must act on the massm in the
direction of this accelerationi.e. a centripetal force of
magnitudemroZr.The inertia force is in the opposite
sense,i.e. acting outwardsfrom the centre.This force is
termedcentrifugalforce.