Experiment 203

Moment of Inertia

Moment of Inertia

EXPERIMENT 203: MOMENT OF INERTIAVIRAY, Kristoffer K.Group No. 5ECE-PHY11L-A4 August 14, 2014

ANALYSIS

The moment of Inertia is the quantity that describes angular inertia. The Mach Principle states that Inertia here arises from mass there. Moment of inertia is abbreviated with the letter I. Based on theories, an object may be considered to be composed of many particles of mass. Each particle provides some resistance to change in angular motion. The resistance is equal to the mass of the particle times the square of the radius from the particle to the axis of rotation. The sum of all the particles resistances to the rotation is the total moment of Inertia of the object.In this experiment, the scalar form of moment of Inertia was considered.Mass moment for inertia is expressed by:I = mr2Where: m = mass of the particle r = shortest or perpendicular distance relative to the axis of rotation.The equation of the mass moment of inertia of the rigid body is:I =

Recall that p = dm / dvTherefore the equation of the body in uniform is, I = pDerivation:I = I =I =I = (r24-r14)I = (r22-r12)(r22+r12)

Recall that M = V =L(r22-r12)

Therefore,I = M(r22+r12)

Part 2 of the experiment requires solving for the Moment Inertia of the Disk. This is almost similar to the hollow disk.

Derivation:I = I =I =I = (r4)

Recall, M = V and Volume of the disk = r2LTherefore,I = MR2

Newtons Second Law for Rigid BodyMathematically, = Fd

Experimental Moment of InertiaFrom Newtons Second Law of Motionnet= I, Tr = I(mg-ma)r = I, = (mg-ma)r= I(),

Therefore, the experimental value of Inertia,I =

Where:m = mass added plus mass of pana = acceleration (from smart timers reading) r = radius of shaft which the thread is wound

In this experiment, the mass of inertia of a disk and ring was determined with the use of a rotating platform, photogate, pulley and timer. The acceleration was recorded and was used to solve for the moment of Inertia. The comparison of the moment of Inertia of solid disk rotated at two different axes was illustrated.To begin with the experiment, a disk was placed using a rotating platform and the rod and the photogate was attached for the recording of the acceleration. A mass hanger was connected to a thread and was looped along the vertical shaft. Before performing the experiment, the platform was properly leveled on the table. The picture below shows the leveling of the rotating platform. In order to level the rotating platform, notice that there is a bubble inside the tube which contains a green fluid. Adjust the rotating platform until the bubble in the tube is at the center. On the first part, a ring was placed on the disk and certain mass was added to the mass hanger. With the use of a vernier caliper, the diameter of the shaft was measured. The picture below shows the measuring of the diameter using the vernier caliper. Using the smart timer, the acceleration was determined. Three trials were performed in the first part of the experiment.

The picture below shows the measuring of the acceleration of the disk using the smart timer.

Below is the table of the data acquired from the first part of the experiment.Mass of disk (Mdisk) = 1400.2 gramsMass of ring (Mring) = 1430.0 gramsRadius of disk (Rdisk) = 11.50 cmInner radius of ring (R1) = 5.37 cmOuter radius of ring (R2) = 6.38 cmFriction mass = 25 gramsRadius (r) = 1.25 cm

(m)Acceleration (a)Experimental value of moment of Inertia

Trial 130 grams0.3 cm/s2153078.13 gcm2



The actual value of moment of Inertia of disk and ring was solved using the equation below:ITOTAL =IDISK + IRINGITOTAL =*MDISK(R2) + MRING (R12 + R22)ITOTAL = *(1400.2)(1.25)2 + *1430.0(5.372+6.382)ITOTAL = 142310.25 g-cm2The experimental value of moment of Inertia was determine by the equation below:I = ((m(g-a)r2))/aWhere g is the specific gravity which is equal to 980 cm/s2 r is the radius which is equal to 1.25 cm a is the accelerationFor trial 1 :I = ((30(980-0.3)(1.25)2)/0.3) = 153078.13 gcm2For trial 2 :I= ((40(980-0.4)(1.25)2)/0.4) = 153062.5 gcm2For trial 3 :I = ((45(980-0.5)(1.25)2)/0.5 = 137742.19 gcm2The average experimental value was computed.Average = (153078.13+153062.50+137742.19)/3 = 147960.94 gcm2The percent difference was also computed which is equal to 3.97%.By using the data acquired from the experiment, it was noticed that a friction mass is needed to overcome the kinetic friction to minimize the errors that can be incurred during the comparison. From the data acquired, acceleration and the increase of mass showed a direct relationship. As the mass increases, its acceleration is greater producing a greater moment of inertia. To cut it short, there is more resistance to rotational motion.For the second part of the experiment, a certain mass was added to the mass hanger and the ring was removed from the disk. The set up for the second part of the experiment is shown in the picture below. The values acquired from the second part of the experiment are shown below.Friction mass = 5 grams(m)Acceleration (a)Experimental value of moment of Inertia




Experimental value for Trial 1:I = ((20(980-0.33)(1.25)2)/0.33I = 92771.78 gcm2Experimental value for Trial 2:I = ((25(980-0.41)(1.25)2)/0.33I = 93329.84 gcm2Experimental value for Trial 3:I = ((25(980-0.50)(1.25)2)/0.33I = 91828.13 gcm2The average of the experimental value of moment of Inertia:Average = (92771.78 + 93329.84 + 91828.13) / 3Average = 92643.25 gcm2The percent difference is 0.06%The second part of the experiment has the similar procedure as the first part of the experiment. The only difference is that the ring which was on top of the disk in the first part of the experiment was removed from the system. The results show that the greater the mass and the acceleration, the higher the moment of Inertia. Percent difference between the actual and experimental was in an acceptable degree which only means that neglecting the possible errors, these values might become similar.The last part of the experiment was done with the same procedure but this time, the disk was placed vertically on the shaft. The set-up for the third part of the experiment is shown below.

Using the measured radius of the ring and disk, and the mass of disk, the moment of Inertia was then computed, and the percent difference was also calculated.

Below is the picture of the group measuring the acceleration of the disk.

Mass of ring (MRING) 1430.00 grams

Inner radius of ring (R1)5.37 cm

Outer radius of ring (R2)6.38 cm

The actual value of moment of inertia of ring was solved by using the equation below: IRING = MRING (R12 + R22)IRING = (1430.0)(5.372+6.382)IRING = 49722.03 gcm2The experimental value of moment of inertia is 55317.69 gcm2The percent difference is 10.12%By the values shown, the actual value of the moment of Inertia of the ring is greater than the experimental value. Errors are present while doing this part of the experiment, thus giving us a percentage difference of 10.12%.The moment of Inertia of the disk was computed in the previous part of the experiment when it was rotated about the center. In this part, the disk is rotated about the diameter of the rotating platform. The following data were acquired during the experiment.Mass of disk (MDISK)1400.20 grams

Radius of disk (RDISK)11.50 cm

Actual value of moment of Inertia of disk (IDISK)46244 g-cm2

Friction mass5 grams

Radius1.22 cm

Two trials were performed during this part of the experiment. The values acquired while performing the experiment are shown below.mAcceleration (a)Experimental value of moment of Inertia




The average experimental value is:(46851.56 + 46082.93 + 46548.59) / 3Average experimental value = 46494.35 gcm2The percent difference is 0.43%.By comparing the moment of Inertia of the disk in the second part of the experiment and to the last part of the experiment, the data shows that the moment of Inertia of the disk is greater because the distributed mass is far from the center of mass. Unlike when the disk was rotated about the diameter, the mass is much nearer to the center of mass.

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Experiment 203