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Kater’s Pendulum

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Kater’s Pendulum
Thomas Markovich and Kapil Chhabria
Departments of Physics University of Houston Houston, TX 77204-5006 (Dated: December 9, 2010)

We experimentally determined the local gravitational constant using Kater’s Pendulum to provide the authors with experience in data analysis. In this manuscript, we rigorously derive the relevant equations from first principles with the appropriate expressions for the experimental uncertainty. We found that by assuming the periods were equal we were able to determine that g was 9.7993 ± 0.0010 m/s2 which was 0.0653 % from the known value. If we took into account the period difference, we found that g was 9.7982 ± 0.0083 m/s2 which is 0.0552 %. Both methods gave us results well within our …show more content…

If we first begin by recognizing that Ic = mk 2 where k is the radius of gyration, we get that T1 = 2π and T2 = 2π
2 k 2 + l2 . dg 2 k 2 + l1 dg

(II.15)

(II.16)

(II.17)

FIG. 1. Kater’s pendulum. I1 is the moment of inertia about K1 with l1 is the distance from the knife edge to the small weight and I2 is the moment of inertia about K2 where w1 > w2

If we simply square our periods and solve for our radius of gyration in both equations, we find
2 4π 2 T 2 l1 − T2 l2 = 1 2 . 2 g l1 − l2

(II.18)

If we invert the system, we find that T1 = 2π and T2 = 2π
2 I + M I2 . M dg 2 Ic + M l1 M dg

(II.8)

2 g l2 − l2 = 21 2 . 2 4π T1 l1 − T2 l2

(II.19)

This is a familiar algebraic form and we solve it by separating our equation into (II.9)
2 2 l1 − l2 A B 2l − T 2l = L + l − l . T1 1 1 2 2 2

(II.20)

If we adjust the weights until the periods are equal, we simply get that
2 2 Ic + M l1 I + M I2 = M dg M dg

2 And after some algebra, we get that A+B = T1 and B2 A=−T2 and yields the following equation for g.

(II.10)

g=

8π 2
2 2 T1 +T2 L

+

2 2 T1 −T2 2l1 −L

(II.21)

3 This equation has an associated uncertainty of (∂T1 gδT1 ) + (∂T2 gδT2 ) + (∂L gδL ) + (∂l1 gδl1 ) (II.22) or δg = (
2 2 256π 4 −T2 + T1 2 2 δl
2 2 T2 +T1 L

2

2

2

2

(2l − L)4

2 2 −T2 +T1 2l−L

4

+ +

64π 4 +

2T1

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