Physics LAB Report
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PHYS 163 LAB #10 The Torsion Pendulum DATA Part I: Disk Alone Amplitude = 10o Trial 10T (s) 13,19 1 13,31 2 13,22 3 Avg 10T = Avg T = 13,24 1,324 Disk MD (kg) DD (cm) ± 0,09 0,09 0,09 4,76 25,3 ± 0,01 0,05 Amplitude = 50o Trial 10T (s) 1 13,47 2 13,28 3 13,35 Deviation 0,004 0,005 0,002 0,09 0,090 Avg 10T = Avg T = ± 0,09 0,09 0,09 13,37 1,337 0,09 0,090 ± 0,09 0,09 0,09 Deviation 0,001 0,007 0,008 Part II: Disk + Ring Ring MR (kg) DRi (cm) DRo (cm) 4,3 22,3 25,3 ± 0,1 0,05 0,05 Amplitude = 10o Trial 10T (s) 1 21,5 2 21,62 3 21,29 Avg 10T = Avg T = 21,47 2,147 0,090 0,090 Part III: Disk + 2 Cylinders Each Cyl MC (kg) 2 MC (kg) RC* (cm) 2,68 5,36 10,5 ± 0,01 0,01 0,05 *RC = Distance from rod to CM of cylinder Amplitude = 10o Trial 10T (s) 1 20,78 2 20,65 3 20,94 Avg 10T = Avg T = 20,79 2,079 ± 0,09 0,09 0,09 0,09 0,09 Deviation 0,0005 0,0067 0,0072 Deviation 0,008 0,006 0,001 PHYS 163 LAB #10 The Torsion Pendulum ANALYSIS NOTE: ALL MASSES ARE IN KILOGRAMS AND ALL DISTANCES IN METERS EXP MEANS: DETERMINE USING THE MEASURED OSCILLATION PERIOD CALC MEANS: USE THE APPROPRIATE EQUATION FROM TABLE 10.2 Part I: Disk Alone Part II: Disk + Ring ± MD 4,76 RD 0,1265 T (10o) 1,324 o T (50 ) % Diff T 1,337 Agree? Yes Part III: Disk + Cylinders ± 0,01 MR 4,3 0,0005 RRi 0,1115 0,09 RRo ± 0,1 M2Cyls 5,36 0,01 0,0005 R2Cyls 0,105 0,0005 0,1265 0,0005 T (10o) 2,079 0,09 T (10 ) 2,147 0,09 IR+D 0,1001 0,0211 I2Cyls+D 0,094 0,0201 Exp IR 0,062 0,0215 Exp I2Cyls 0,056 0,0205 ± Calc IR 0,061 0,0002 Calc I2Cyls 0,059 0,0007 0,09 o 0,98 ± ± Calc ID 0,0381 0,00038 % Diff IR 1,5 % Diff I2Cyls 5,5 k 0,86 0,11 Agree? Yes Agree? Yes ...
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Final Answer

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Running Head: THE TORSION OF A PENDULUM

The Torsion of a Pendulum

Students Name
Institutional Affiliation

1

THE TORSION OF A PENDULUM

2

Objectives
To compare the experimentally determined moment of inertia of several bodies with
corresponding theoretical value
Theory
In general, the motion of a rigid body is combination of translational and rotational motion. The
description of the rotational portion of the motion involves both rotational kinematics and
rotational dynamic.
Consider a rigid body rotating with angular speed  about an axis that is fixed in a particular
reference frame. A particle of mass m at a distance r from the axis of rotation moves in a circle
of radius r with angular speed  about this axis and has linear speed v = r. Its kinetic energy is
1
1
K = 𝑚𝑣 2 − 𝑚𝑟 2 2
2
2
If the body is a rigid,  is the same for all the particles in the body but r is different from each
particle. Hence, the total kinetic energy of the rotating of body is
1
1
𝐾 = ( 𝑚1 𝑟12 + 𝑚2 𝑟22 . . ) 2 = (𝑚1 𝑟12 )2
2
2
The term 𝑚1 𝑟12 is denoted by the symbol I and I called moment of inertia or the rotational
inertia of the body with respect to the particular axis of rotation.
For a body that is not composed of that discrete point masses but is a continuous distribution of
matter, the equation
𝐼 = 𝑚1 𝑟12

THE TORSION OF A PENDULUM

3

Is replaced by 𝐼 = ʃ𝑟 2 𝑑𝑚
Consider a long uniform wire, rigidly clamped at its upper end and supporting at its lower end a
circular disk whose moment of inertia about an axis through the wire is I . If the disk is rotated
through some angles, a restoring torque, Ʈ, is a setup in the wire tending to bring the disk back to
its equilibrium position. When the disk is released the torque gives the disk an angular
acceleration
Ʈ = 𝐼𝛼
The restoring torque set up in the wire when its lower end is twisted th...

markjunior209 (5341)
UT Austin

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