(a) Consider an extended pendulum whose CM is a distance L from the pivot, and w
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Question
(a) Consider an extended pendulum whose CM is a distance L from the pivot, and whose moment of inertia around the pivot is I. Show that the frequency of small oscillations is w = sqrt[(mgL)/I], which gives T = 2pi/w = 2pi[sqrt(I)/(mgL)], and hence g = 4pi[I/(mLT^2)]. Therefore, by measuring I, m, L, and T, you can determine g. However, if the pendulum has an odd shape, it may be difficult to determine I. Consider the, the following alternative method of measuring g.(b) For simplicity, assume that the pendulum is planar. Pick an arbitrary point as the pivot and measure the period, T, of small oscillations. Then with the pendulum at rest, draw a vertical line through this point. By trial and error, find another pivot point on this line on the same side of the CM (you may need to extend the line with a massless extension) that yields the same period T. Let L be the sum of the lengths from these two points to the CM. Show that g is given by g = 4pi^2L/(T^2), which is independent of m and I.
Explanation / Answer
Consider a planner object of odd shape. Let A be the point of pivot, where the small oscillations to be considered, which is at distance L1 from the center of mass of the object. The frequency of the oscillation about this point is = (mgL1/I1) Where m is the mass of the object g is the free fall acceleration I is the moment of inertia Therefore the period of the oscillation will be T = 2(I/mgL1) g = 42I1/mL1T2 The moment of inertia of the object at this point will be I1 = mL12 Therefore g = 42L1/T2 Therefore g = 42L1/T2 Now consider another point B, on the same side of the center of mass with the same period, at a distance L2 from the center of mass. The period of the oscillation about point B is given by T = 2(I1/mgL2) g = 42I2/mL2T2 The moment of inertia of the object at this point will be I2 = mL22 Therefore g = 42L2/T2 Therefore the accleration due to is given by g = 42L/T2 The moment of inertia of the object at this point will be I2 = mL22 Therefore g = 42L2/T2 Therefore the accleration due to is given by g = 42L/T2 The moment of inertia of the object at this point will be I2 = mL22 Therefore g = 42L2/T2 Therefore the accleration due to is given by g = 42L/T2 (generally the average length will be considered)Related Questions
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