lAB 2_SainBeknaza_19193

Ministry of Education and Science Republic of Kazakhstan

International Informational Technologies University

Laboratory Task

STUDY OF MOMENTUM OF INERTIA WITH MAXWELL’S PENDULUM

Done by: Kunesbek A.,Sain B., Talgatbek A.

Cheked: Zvyaginceva Olga

STUDY OF MOMENTUM OF INERTIA WITH MAXWELL’S PENDULUM

Main tasks:

1. Experimental determination of momentum of inertia for the Maxwell’s pendulum by its fall time;

2. Calculation of momentum of inertia of Maxwell’s pendulum using the theoretical formula;

1. Theory of experiment

Maxwell’s pendulum represents a disk, whose axis is suspended on two turning on it threads (fig. 1). It is possible to study experimentally dynamics laws of translational and rotational motions of rigid body using this pendulum, as well as the main law of physics − the law of mechanical energy conservation. Having rotated the pendulum, we raise the disk to height h and let go down without push, then the disk is starting to go down and at the same time to rotate around its horizontal axis. At the same time trajectory of all points of the disk lie in parallel plane (surface). Such motion of rigid body is called planar. It can be considered as the translational motion of the body, which is occurring with the velocity of centre of mass (centre of gravity, centre of inertia) and at the same time as the rotational motion around horizontal axes, passing through this center.

The equation of motion for the centre of gravity and rotation of pendulum relatively to mentioned axes has the following form:

ma=mg-2T (1)

Iε=2Tr (2)

where m is mass of the pendulum, I is momentum of inertia, а is acceleration of

gravity center, ε – angular acceleration of the pendulum, Т is tension of the thread, r is radius of tube. Taking into account, that the accelerations in this case are connected with each other by the relation ,a=ε*r we obtain from formulae (1) and (2):

(3)

From the last relation comes that the center of mass of the pendulum moves with

constant acceleration, which depends on the body’s momentum of inertia. This

circumstance is the basis of this theory. From the relation (3) with taking into account the formula of the path for uniformly accelerated motion h = at^2/2, we obtain the calculation formula:

where D = D0 + d.

Thus, to determine the momentum of inertia of the Maxwell’s pendulum, it is

necessary to measure time t of its fall from given height h, to define its mass m, and diameter of the tube D0 and thickness d of thread.

2. Description of experimental device.

The general view of Maxwell's pendulum is shown on fig. 1. This device consists of pendulum, electromagnet, two photoelectric sensors, electronic timer which is

connected with the sensors. On the disk of the pendulum puts over one of the

removable rings, that allows to change its mass and momentum of inertia of the

pendulum. Electromagnet holds the pendulum at the upper position if the current is running through its winding. Length of the pendulums suspension (height h) is being changed by the millimeter scale, which is marked at the vertical column.

3. Procedure

1)We open the program virtual Maxwell pendulum

2) And with the help of the program counted time falling of a disk. And every

time we reduced height by 10cm. And mass of a disk 30 g. We wrote the result

on the table.

3)We counted the momentum inertia of every experiment with help this

formula. Then wrote in the table.

4)We counted the arithmetic mean value of “I”; <I>

5)Then we counted measurements х and х_i²

6)And then we find mean square error to help with this form:

S_n=0,2468*10^-5

7)Mean square error of the arithmetic mean:

S_<I>=0,0822*10^-5

8)Absolute uncertainty:

ΔI=0,0781*10^-5

9)Relative uncertainty:

ε=2,93%

10)And the last step is “We counted I_theory”:

I_theory=[m₀D₀²+m_d(D_d²+D₀²)+m_r(D_r²+D_d²)]/8

I_theory=9,03*10^-5

№	h,m	m,kg	t,sec	I,kg*m2	<I>,kg*m2	D0,m	d,m	Itheory,kg*m2
1	1,4	0.03	5,499	8,024*10-5	8*10-5	1*10-2 1*10-2 1*10-2 1*10-2 1*10-2 1*10-2 1*10-2 1*10-2 1*10-2 1*10-2	0,0101 0,0101 0,0101 0,0101 0,0101 0,0101 0,0101 0,0101 0,0101 0,0101	9,03*10-5
2	1,3		5,379	8,271*10-5
3	1,2		5,092	8,027*10-5
4	1,1		4,826	7,864*10-5
5	1,0		4,719	8,275*10-5
6	0,9		4,311	7,668*10-5
7	0,8		4,189	8,150*10-5
8	0,7		3,811	7,705*10-5
9	0,6		3,533	7,726*10-5
10	0,5		3,340	8,291*10-5

Conclusion

I I satisfied all statements of the problem. I defined the inertia moment by time of a free fall and I use a theoretical formula. Also I think that made it successfully.