Moment of Inertia CourseHero
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1611
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Physics
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Feb 20, 2024
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1 Analysis of Rotational Inertia Lab Name: Course/Section: PHY1611 Instructor Tables (20 points): Diameter (cm) Radius (cm) Radius (m) Large grove pulley 5 cm 2.5 cm 0.0250 m Disk 9.15 cm 4.575 cm 0.04575 m Ring, inner 5.10 cm 2.55 cm 0.0255 m Ring, outer 7.45 cm 3.725 cm 0.03725 m m (g) m (kg) Disk 120.9 g 0.1209 kg Ring 147.3 g 0.1473 kg Hanging mass 30 g 0.030 kg a (m/s
2
) Disk 0.745 m/s
2
Disk and Ring 0.19 m/s
2
1.
Calculate the theoretical moment of inertia of the disk. I = ½
M
2
I
disk
=( ½
)(0.1209)( 0.0250)
2
= 3.778 ×
10
-5
kg ⋅
m
2
2.
Calculate the theoretical moment of inertia of the ring. I = ½
M (R
2
+r
2
) I
ring
=( ½
)(0.4712)( 0.03725
2
+ 0.0255
2
) = 4.801 ×
10
-4
kg ⋅
m
2 3.
Calculate the experimental moment of inertia of the disk and show work. I = τ
/
α
(experi.) I
disk
=(
0.03725m)
×
(0.1209+0.1473)
×
0.19 )/0.19/0.03725
= 0.00003714 kg ⋅
m
2
2 4.
Calculate the experimental moment of inertia of the ring and show work. I = τ
/
α
(experi.) I
ring
=(
0.0255)
×
( 0.1473
×
0.19 )/0.19/0.03725
= 9.578 ×
10
-5
kg ⋅
m 5.
Using the theoretical value as the accepted value, calculate the % error for the -
disk, and show work. 3.778-3.778/3.778
×
100= 0%
6.
Using the theoretical value as the accepted value, calculate the % error of the ring, and show work. 4.801-9.578/4.801 ×
100= -99.5% 7.
If you repeated this experiment with a larger hanging mass, should that change the values you would obtain for the moment of inertia of the ring and disk? If I performed this experiment again using a higher applied torque, which shouldn't have an impact on the results I get for the ring and disk's moment of inertia. Since the input values are fixed, the α
would likewise double if the Applied Torque in the disk and ring trials was doubled. 8.
Why are you able to calculate the experimental value of the moment of inertia of the ring in the manner you did? Their moment of inertia is the total of each body's individual moment of inertia since both bodies shared the same common axis of torsion. 9.
Describe, in terms of energies, what is happening during the experiment. As additional torque is given to the disk by tension force during the experiment, the rotating kinetic energy increases. Torque is potential energy until it is applied, at which point it is transformed into kinetic energy. 10.
What are some reasons that account for our percent error? Some reasons for experiment error are the elasticity of the string, the friction in the bearing on which the disk is mounted, and the instrumental mistake.
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