Homework 11 Solution

The overall torque is then mg l 2 1 sin 90 k l xo

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Unformatted text preview: attached to the plank, the spring force always acts perpendicularly to the length of the plank at distance L, and the torque is easily found. The plank’s weight still acts at a distance L/2 from the center of mass, but now at an angle 90 − θ relative to the axis of the plank. The overall torque is then τ = −mg L 2 1 sin (90 − θ) − k (Lθ − xo ) L = − mgL cos θ − kL2 θ + kxo L 2 Since the angle θ is small, we may approximate cos θ ≈ 1, and we may also make use of our earlier expression for xo . Finally, out of equilibrium the torques must give the moment of inertia times the angular acceleration. 1 1 1 τ = − mgL cos θ − kL2 θ + kxo L ≈ − mgL − kL2 θ + mgL = −kL2 θ = Iα 2 2 2 d2 θ −kL2 θ = I 2 dt (18) (19) Noting that I = 1 mL2 for a thin plank, we can put the last equation in the desired form for simple 3 harmonic motion in terms of known quantities: 3k d2 θ =− θ 2 dt m 3k ω= m Note that the length of the plank does not matter at all. ii Small enough such that we don’t have to worry about the spring bending to the left, for one. (20) (21)...
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This note was uploaded on 01/26/2014 for the course PH 105 taught by Professor Stanjones during the Fall '08 term at Alabama.

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