Please help 006 (part 1 of 5) 8.0 points An extended object has its center of ma
ID: 2099848 • Letter: P
Question
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006 (part 1 of 5) 8.0 points An extended object has its center of mass is at point C. It is pivoted about point O. Denote its moment of inertia about a rotation axis perpendicular to the paper through C to be Ic. OC = l. For the situation shown, the torque tau is given by tau = mg l. tau = mg l sin theta . tau = mg sin theta . tau = mg cos theta . tau = mg l cos theta . 007 (part 2 of 5) 7.0 points For the situation shown, the direction of the torque is counterclockwise. unknown, since the direction of swing is not given. clockwise. 008 (part 3 of 5) 7.0 points The period T is given by T = 2I/mgl. T = I/mgl. T = 2I/mgl. T = 2 pi 2I/mgl. 009 (part 4 of 5) 7.0 points Suppose at t = 0, theta = 0 degree and the pendulum swings to the left with angular velocity omega 0. Find the magnitude of the maximum angle theta for the oscillation. Given omega = 64.4 rad/sec, omega 0 = 42.5 rad/sec, where omega is the angular frequency of the small oscillation. Answer in units of rad 010 (part 5 of 5) 7.0 points If we describe the oscillation by, theta = theta max cos(omega t + delta) what is omega for the initial condition given in the previous question? delta = pi/2 delta = 5 pi/4 delta = 3 pi/4 delta = pi delta = 7 pi/4 delta = 3 pi/2 delta = pi/4 delta = 0 011 8.0 points A uniform circular disk is pivoted at its edge A. Find its period of the small oscillations. Use small angle approximation. T = 2 pi 3R/2g T = 2 pi R/g T = 2 pi 5R/2g T = 2 pi R/2g T = 2 pi R/3g T = 2 pi 5 R/3g T = 2 pi 4 R/3g T = 2 pi 2 R/g T = 2 pi 11 R/6g T = 2 pi 7 R/ 6gExplanation / Answer
6)2
7)2
8)4
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10)3
11)6