Experiment -1
Aim:To study the dynamics of a fly wheel
Apparatus: Fly Wheel Rope Stopwatch Weight
Introduction:Inertia is a property of a body to resist the change in linear state of motion. It is measured by the body. The moment of inertia I of a body is a measured of its ability to resist change in its rotational state of motion. If we compare for rotational motion with corresponding relations for translator motion, we find that moment of inertia plays the same role in rotatory motion as is played by mass in translator motion, i.e., if a body has large moment of inertia, it is difficult to start rotation or to stop it if rotating. Large moment of inertia also helps in keeping the motion uniform. Moment of inertia of a particle about an axis is defined as I
= mr2
Figure.1.1 Where r is the perpendicular distance of the particle from the axis of rotation and m is mass of object.
For a system of particles of masses m1, m2, m3,, mn at perpendicular distance r1, r2, r3,., rn from the axis of rotation, the moment of inertia is
I = m1r12 + m2r22 +..+ mnrn2 =miri2
m1 m2
r1 r3 m3
r2
r4 m4
Figure.1.2 Moment of inertia of a rigid body about an axis is given by
I = r2dm
where r is the
perpendicular distance of a particle of mass dm of rigid body from the axis of rotation.
Theory:When a weight is suspended through the hook which connected to the free end of string which is wrapped round tin: shaft and the other end of which is tied to the shaft and it is allowed to fall to touch the ground then potential energy possessed by the falling weight has partly been used to give motion to the fly wheel and partly used in overcoming frictional resistance present in the bearings For a system:1. Initial potential energy (P.E) = W h . Where W is the falling weight and h is the height of the weight from the level ground. And initial kinetic energy of a system is equal to zero. 2. When falling weight just heating the ground then potential energy of a system is zero and kinetic energy of a system = kinetic energy of the falling weight + kinetic energy of the fly wheel. Final (K.E) =
1 1 Mv2 + I 2 . Where M is the falling mass, v is the final velocity 2 2
of falling weight, is the final angular velocity of the wheel or of the shaft and I is the moment of inertia of a fly wheel.
3. Work done due to frictional resistance = F h . Where F is the force of friction acting tangentially to the shaft. From law of conservation of energy we can write: W h =
1 1 Mv2 + I 2 + F h .. (1) 2 2
Moment of inertia from equation (1)
I 2(W F )h Mv 2
(2)
Now average velocity can be calculated as
vu h For a falling weight initial velocity ( u ) = 0 2 t
So, v
2h t
Angular velocity can be calculated as
v Where r is the radius of fly wheel. r
(W F )t 2 r 2 2 Mhr 2 I 2h
Procedure:1. Wrap the cotton string round the shaft. 2. Suspend a small weight to the free end of the string. 3. Increase the applied weights till the shaft just begins to rotate. 4. Note down the total applied weight (M0). 5. Note the distance between the applied weight and horizontal surface ( h). 6. Place the weight more than M0 (M), let it fall through distance (h) and note down the time taken, with the help of stop watch. 7. Repeat the experiment for different weights
Diagram:-
Fly wheel
Shaft
Figure.1.3
Observation:S.no 1. 2. 3. 4. Mass (kg) 2.5 2 1.5 1 Time (seconds) 4.16 5.3 5.41 7.31
Calculation:D= 19.3mm, D= 19.3 10 -3 m D= 1.93 10 -2 m F= 0.981N
h = 173cm = 1.73m, M0= 0.132kg
r D 19.3 mm 9.65mm 2 2
F = M0 g F = 0.1329.81 = 1.295 kgm/s2
W m g, kgm/ s 2
I (W F )t 2 r 2 2mhr 2 kg.m 2 2h
Reading -1
m1 2.5kg , W1 m g 2.5 9.81 24 .52 N Kg-m/s2,
t = 4.16sec I1 =
(W F )r 2 t 2 2Mhr 2 2h (24 .52 1.295 ) (4.16 ) 2 (9.65 10 3 ) 2 2 2.5 1.73 (9.65 10 3 ) 2 2 1.73
I1 =
I1 = 0.011 kgm2 Reading -2 m2 = 2 kg W2 = m2g W2 = 2 9.81 = 19.62 kg-m/s2, t = 5.3 sec I2 =
(W F )r 2 t 2 2Mhr 2 2h
I2 =
(19 .62 1.295 ) (5.3) 2 (9.65 10 3 ) 2 2 2 1.73 (9.65 10 3 ) 2 2 1.73
I2 = 0.0136 kgm2 Reading -3 m3 = 1.5kg W3 = m3g = 1.5 9.81 = 14.715 kg-m/s2, t = 5.41sec I3 =
(W F )r 2 t 2 2mhr 2 2h (14 .715 1.925 ) (5.41) 2 (9.65 10 3 ) 2 2 1.5 1.73 (9.65 10 3 ) 2 2 1.73
I3 =
I3 = 0.01039 kgm2 Reading -4 M4 = 1kg W4 = m4g = 1 9.81 = 9.81 kg-m/s2, t = 7.31sec I4 =
(W F )r 2 t 2 2mhr 2 2h
(9.81 1.925 )( 7.31) 2 (9.65 10 3 ) 2 2 1 1.73 (9.65 10 3 ) 2 I4 = 2 1.73
I4 = 0.0121 kgm2 Iaverage =
I1 I 2 I 3 I 4 0.011 0.0136 0.0.014 0.0121 = 4 4
= 0.0126 kgm2
Nomenclatures:-
Result:Reading. Inertia (I) kg-m2
1. 2. 3. 4.
0.011 0.0136 0.014 0.0121
Iaverage= 0.0126 kgm2
Precaution:1. The turns of strings must not overlap the others. 2. Stop watch must be handled with the care. 3. Oil the bearings to reduce friction. 4. Over-lapping of the string should be avoided.
Conclusion:The development and testing of a simple laboratory device for the determination of moment of inertia of a flywheel has been carried out by this work. The method of rotating the flywheel has been adopted from this development.
