Lab partner's name:
TA’s name:
Section no
Objective:
Purpose of this lab experiment was to gain knowledge about rotational motion from an energy viewpoint and to perceive how moment of inertia depends on the rotation axis for a given object. During this experiment a constant torque created a constant angular acceleration for a rigid body rotating about its center of mass.
Experimental Data:
Dimension of steel box = 0.05m x 0.04m. X 0.03m.
Mass = 0.442 kg;
Hanging mass = 20 g.
Medium Pulley:
1) Long -ways (longer axis):
Slope = 27± 0.35 ; Slope indicates angular acceleration i.e. α.
ωi = 0.45 rad/sec ϴi = 0.157 rad
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ωf = 1.6 rad/sec ϴf = 22.148 rad
2) Short -ways (shorter axis) :
Slope = 11.9± 0.88 ; Slope indicates angular acceleration i.e. α.
Small Pulley:
1) Long -ways (longer axis) :
Slope = 9.24± 0.049 ; Slope indicates angular acceleration i.e. α.
2) Short -ways (shorter axis) :
Slope = 5.55± 0.04 ; Slope indicates angular acceleration i.e. α.
Radius of small pulley = rs = 0.0048 m.
Radius of medium pulley = rm = 0.0143 m.
Data Analysis:
Conservation of Energy:
Following notations were used in this experiment.
ωi = initial angular velocity
ωf = final angular velocity
ϴi = initial angular position
ϴf = final angular position
1st Trial-
As per conservation of energy,
0.5 (Is + mr^2) (ωf^2 - ωi^2) + mgr (- ϴf + ϴi) = Δ E (1)
( lost )
There may two values for Is.
Longer axis: 1/12 x M (a^2 + b^2) = 1/12 x (0.442) (0.04^2 + 0.03^2) = 0.00009208 m^4
Shorter axis: 1/12 x M (a^2 + b^2) = 1/12 x (0.442) (0.04^2 + 0.05^2) = 0.00015102 m^4
So, putting value of Is for longer axis and radius of small pulley in equation no. (1), we obtained,
0.5 x (0.00009208 + 0.02 x 0.0048 ^2) x (1.6^2 – 0.45^2) + 0.02 x 9.81 x 0.0048 (-22.148 + 0.157)
= -0.02 J.
Now, putting value of Is for longer axis and radius of medium pulley in equation no. (1), we obtained,
0.5 x (0.00009208 + 0.02 x 0.0143^2) x (1.6^2 – 0.45^2) + 0.02 x 9.81 x 0.0143 (-22.148 + 0.157)
= -0.06 J.
So, putting value of Is for shorter axis and radius of small pulley in equation no. (1), we obtained,
0.5 x (0.00015102 + 0.02 x 0.0048 ^2) x (1.6^2 – 0.45^2) + 0.02 x 9.81 x 0.0048 (-22.148 + 0.157)
= -0.02 J.
Now, putting value of Is for shorter axis and radius of medium pulley in equation no. (1), we obtained,
0.5 x (0.00015102 + 0.02 x 0.0143^2) x (1.6^2 – 0.45^2) + 0.02 x 9.81 x 0.0143 (-22.148 + 0.157)
= -0.06 J.
Result:
Like any other experiments, this experiment was susceptible to sources of errors. One source of error was wrong measurement of dimension and mass.
Δ mass = 0.0001 kg
Error in dimensions, Δ a, b, c = 0.00005 m.
1st trial:
ID = mr(g - rα)/α or, 0.02 x (0.0143) x (9.81- 0.0143 x 27) / 27 = 0.00009982
ID = mr(g -rα)/α = mrg/α - mr^2
Error: ΔID = [{(δID/δm) . Δm}^2 + {(δIr/δr).Δr}^2 + {(δIα/δα).Δα}^2]^0.5
= [0+ 0 + {(-mrg / α^2). Δα}^2]^0.5 [ neglecting small terms like {(δID/δm) . Δm}^2
as well as {(δIr/δr).Δr}^2 ]
= (mrg /α^2). Δα
2nd trial: ID = 0.02 x (0.0143) x (9.81- 0.0143 x11.9) / 11.9 = 0.000232
3rd trial: ID = 0.02 x (0.0048) x (9.81- 0.0048 x 9.24) / 9.24 = 0.000102
4th trial: ID = 0.02 x (0.0048) x (9.81- 0.0048 x 5.55) / 5.55 = 0.000169
Discussion and Conclusion:
Various aspects of rotational motion were studied by carrying out this experiment. From the calculation of moment of inertia about shorter and longer axes of a same object, it was found how moment of inertia depends on the rotation axis for a given object. Moment of inertia varies in accordance with geometry and orientation of an object. The moment of inertia of a symmetric object can be calculated with the formula I = βMR2, where β is a dimensionless fraction between 0 and 1 which depends on rotation axis as well as shape of that object. For 3 dimensional objects, there are three numbers of rotational axes. But in this experiment only two rotational axes were considered. In this lab, a rectangular block of mass of 0.442 kg i.e. M and dimensions 0.05 m × 0.04 m. × 0.03 m. was mounted with a rotation axis through center of one face. Another mass of 20 gram i.e. m was tied to a string wound around a mass less pulley mounted on the same axis. A torque was exerted on the rectangular block by the string and this torque resulted in rotational motion.
Another objective of this lab experiment was to check whether or not energy was conserved throughout the experiment. After carrying out the experiment and analyzing the experimental data it was found that the energy was not fully conserved and some portion of energy was lost. There might be various ways by which the energy lost. Most possible reason for energy loss is presence of friction. Because of friction some amount of energy was lost in the form of heat. The error occurred due to friction can be termed as systematic error. Moreover, pulley was assumed to be frictionless in the formula used in this experiment. Due to presence of friction motion was slowed down and this was reason behind having lower value of acceleration. Apart from this, few termed was neglected due to small value and this approximation might led to some variation from desired result.
In a nutshell, it can be said that the experiment was carried out safely and objective of this experiment was met. By performing this experiment, knowledge about various aspects of rotational motion, moment of inertia, as well as law of energy conservation was gained. Though the experiment was affected by source of errors, but this can help how to improve the experiment by removing possible source of errors. That is how the objective of the experiment was met and so it can be concluded that the experiment was successful.
