Download Experiment No : M3 Name of the Experiment: MOMENT OF INERTIA

Experiment No : M3
Name of the Experiment:
ACCELERATION
MOMENT
OF
INERTIA
AND
ANGULAR
Purpose of the Experiment : Experimental observation and understanding of moment of
inertia in rotational motion.
Theoretical Background:
In daily life, we know that moving massive objects is harder and requires more energy
compared to lighter objects. This simple observation shows that mass is the unwillingness or
more generally resistance of a body against the motion. In nature, objects can do rotational
and translational motion (motion along a straight line).
Imagine that you are trying to give a mass an acceleration of a by applying a force F. The
acceleration will depend on the mass, i.e. the greater the mass the smaller the value of a. The
reason for this is the resistance of mass against its motion. Despite the applied force, the
object wants to remain still. Hence, mass is defined as a measure of inertia in linear kinetics.
This is why Newton used the word ‘inertia’ to describe the mass.
Now, try to rotate the same mass around an axis or its center of mass. You will see that the
object puts up a resistance against rotation depending on its mass and the rotational axis. For
example, the effort for rotating a heavy bar around its center of mass will not be the same as
the effort for rotating the bar around its end points. Thus, can we say that objects also show
reluctance against rotational motion as in the linear (translational) motion?
This difficulty in rotation is called the moment of inertia or rotational inertia. While the
difficulty in linear motion depends only on mass, the resistance of object against rotation or
rotational inertia in rotational motion will depend on both mass and axis of rotation at the
same time. The law of motion for rotational motion is expressed as;
4.1
where τ is the torque (the force of rotation) and its unit is N.m (SI). The unit of angular
acceleration is radian/s2 while the unit of moment of inertia (I) is kg.m2.
Let’s look at the equation 4.1 closely. It is known by definition that α = dω/dt (ω is angular
velocity). Therefore, if the torque on a body is zero, then;
dω
dt
ω
dt
dL
dt
Hence, in this case angular momentum of body L=Iω is conserved! This conservation law is
different from that you studied in the 3rd experiment. If the there is no torque on the body,
then the angular momentum is a constant of motion.
In general, the moment of inertia of a system consisting of point mases can be formulated as
4.2
Equation 4.2 should be rearranged for continuous mass distributions. For a mass chosen close
enough to a point in a coordinate system, writing mi as mi → ρ(ri)ΔVi , equation 4.2 becomes:
4.3
Here, Ri is the perpendicular distance of the mass to the rotational axis. So, for a rigid body
with continuous mass distribution, it is possible to convert this sum into an integral as follow:
d
4.4
In general, the moment of inertia is determined not just by a number, but more than one
number depending on the geometry of the object. For example, there are two symmetry axes
of the rectangular sheet shown below and each has a different moment of inertia relative to
the axes (Ix and Iy). These axes are called the principal axes.
Figure 4.1
There are two different principal axes for a cylinder which has its axis of symmetry oriented
along the z axis. First two axes are lying on the x and y, and the third one is lying on the z
axis. The corresponding moments of inertia of these axes are Ix = Iy and Iz.
It will be easily understandable by similar reasoning that there is only one component of the
moment of inertia for a sphere.
Theoretical moment of inertia expressions of some symmetric bodies which are calculated
according to the basal axis through center of mass (CM) with the help of Equation 4.4 are as
follow:
a) A smooth disc with mass M and radius r or solid cylinder;
ICM= mr
b) A bar with mass M and length L;
ICM= mL
c) Point mass with M which has a r distance from the rotational axis;
ICM=m
d) Solid sphere;
ICM = mr2
e) Thin-walled hollow sphere;
ICM= mr2
f) Thin-walled hollow cylinder;
ICM= mr2
Experimental Procedure:
Figure 4.2. The experimental setup.
The experimental setup you will use is shown in Figure 4.2. In this experiment, the
theoretical and experimental moment of inertia will be determined for the disc, cylinder,
sphere, and rod separately. At the end of the experiment, theoretical and experimental
moment of inertia will be calculated for each cylinder, sphere and bar separately. Please
follow the steps given below during the experiment.
Conducting the Experiment and Obtaining the Data:
1. Place the sphere whose period is to be determined on the rotation axis.
2. Turn the sphere for 90 degrees such that black stripe on the sphere passes through the light
barrier, and let it go and measure the period. Then measure the circumference and mass of the
sphere.
Tsphere=............ Csphere=............. msphere=...............
3. Calculate moment of inertia of the sphere (Isphere) by using period measured in step 2 and
equation 4.5:
where D denotes angular displacement factor peculiar to the experimental setup and its value
is D=0,023 Nm/rad.
Calculate theoretical moment of inertia of the sphere with the help of the expression given in
the theory section and compare your finding with the experimental value.
4. In the following step, measure the diameter and mass of the disk and perform the same
calculations described for the sphere.
Tdisk=............ Rdisk=............. mdisk=...............
5. In this step, first of all measure the period of the iron bar. Then, place cylindrical masses
on the iron bar equally from the rotation axis: r1=r2. Measure the period of the whole system
and register this value. By using these values and equations 4.5 and 4.6 find the moment of
inertia of the system. In order to use in calculations, measure the positions and of masses of
the cylinders and the length of the bar and masses.
Isystem=Ibar+Imasses= (M1+M2)r2+Ibar
Tbar=………………
Lbar
Tsystem=………………r1=r2
………m bar
4.6
………
…… m1=m2 ………
Compare the experimental and theoretical moment of inertia values for this system.
6. Repeat the last step only by placing cylindrical masses on different distances from the
rotation axis:
. Also repeat all calculations for this case. Measure the positions and
masses of cylinders and length of bar for calculations. Note that in this case the moment of
inertia of the mass system has to be taken as Imasses=M1r12+M2r22.
Tbar =……………… L bar
……… m bar ………
Tsystem=………………r1
………r2
………
m1=m2 ………
IEbar=………………
IEsystem=………………
IEmasses=………………
ITbar=………………
ITmasses=………………
ITsystem=………………
Relative Error=…………..
Comments and Results:
(Hints: What have you learned from the experiment? Possible errors and their reasons.
Explain in detail. Summarize the experiment)
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Questions:
1) An ice skater starts to rotate vertically with open arms on a flat ground with an initial
angular velocity (i.e. no torque on him/her). After a while, she/he suddenly closes
her/his arms. Is there any change in the angular velocity of ice skater as a result of this
action? If your answer is positive, does it increase or decrease? Explain.
2) How could you measure moment of inertia in a different manner? Propose an
experimental setup.
3) Show that in a closed system, the total angular momentum of a system of particles is
conserved.
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