hw1solns - then the velocity is + B sin( t )). t B ω 3-14....

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PHYSICS 273 PROBLEM SET #1 SOLUTIONS September 1 3 , 20 11 1.7 The moment of inertia of a circular hoop of mass m and radius a rotating about a point on its circumference is I = 2ma 2 . (Use the parallel axis theorem, or integrate the mass around the hoop.) So the oscillation frequency is a g I mga 2 = = ω
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3-4. (a) Balancing the weight of the cylinder with its buoyancy, we find ρ W l π R 2 = ρ L R 2 , where W and are the mass densities of the fluid and the cylinder, respectively, and R is the radius of the cylinder. If the cylinder is submerged an amount x , then F=ma gives ( L R 2 ) x = ( L R 2 ) g W R 2 ( l+x ) g or x = ( g/l ) x and we immediately recognize the frequency of oscillation as l g = ω . (b) The position of the cylinder after being displaced by distance B at time t =0 is x ( t ) = B cos( t ). The velocity is therefore v(t) = B ω sin( t ). (We’ve defined positive x as downwards: if it is defined as upwards,
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Unformatted text preview: then the velocity is + B sin( t )). t B ω 3-14. (a) m x & & = − kx b x & , which can be written in the usual form 2 = + + x x x ω γ & & & , where m k = and m b = . (b) The frequency of oscillation with damping is given by 4 2 2 − = , so if 2 3 = then = and therefore mk m m b = = = = (0.2 kg 80 N/m) 1/2 = 4 kg/s (c) 1 = = = b mk Q . The amplitude of oscillations fall as exp( −γ t /2), so after 10 complete oscillations ( ) 2 10 π × = t , the amplitude falls by a factor ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − 3 2 10 exp 3 2 10 exp 10 exp = really small...
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This note was uploaded on 12/28/2011 for the course PHYSICS 273 taught by Professor Monroec during the Fall '11 term at Maryland.

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hw1solns - then the velocity is + B sin( t )). t B ω 3-14....

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