ch15-p056

ch15-p056 - 56 The table of moments of inertia in Chapter...

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Unformatted text preview: 56. The table of moments of inertia in Chapter 11, plus the parallel axis theorem found in that chapter, leads to IP = 1 2 2 MR + Mh2 = 1 2 (2.5 kg)(0.21 m)2 + (2.5 kg)(0.97 m)2 = 2.41 kg·m² where P is the hinge pin shown in the figure (the point of support for the physical pendulum), which is a distance h = 0.21 m + 0.76 m away from the center of the disk. (a) Without the torsion spring connected, the period is T = 2π IP Mgh = 2.00 s . (b) Now we have two “restoring torques” acting in tandem to pull the pendulum back to the vertical position when it is displaced. The magnitude of the torque-sum is (Mgh + κ)θ = IP α, where the small angle approximation (sinθ ≈ θ in radians) and Newton’s second law (for rotational dynamics) have been used. Making the appropriate adjustment to the period formula, we have T′ = 2π IP . Mgh + κ The problem statement requires T = T′ + 0.50 s. Thus, T′ = (2.00 – 0.50)s = 1.50 s. Consequently, 4π2 κ= I – Mgh = 18.5 N·m/rad . T′ 2 P ...
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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