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Old Nov 29th 2009, 10:58 AM   #1
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distance...?

M, a solid cylinder (M=2.43 kg, R=0.115 m) pivots on a thin, fixed, frictionless bearing. A string wrapped around the cylinder pulls downward with a force F which equals the weight of a 0.670 kg mass, i.e., F = 6.573 N. Calculate the angular acceleration of the cylinder.


If instead of the force F an actual mass m = 0.670 kg is hung from the string, find the angular acceleration of the cylinder.


*********** How far does m travel downward between 0.550 s and 0.750 s after the motion begins?


the problem i am having with is *********** How far does m travel downward between 0.550 s and 0.750 s after the motion begins? *****
this part of questions.

i found first answer: 47.3 rad/s^2
second answer: 30.3 rad/s^2

and for the third one,

i've tried

d=(0.5)(a)(t)^2

at 0.550s
d=(0.5)(a)(t)^2
=(0.5)(30.3)(0.550)^2
= 4.58 m

at 0.750s
d=(0.5)(a)(t)^2
=(0.5)(30.3)(0.750)^2
= 8.52 m

and i subtracted them and got 3.94 m but its wrong
i've tried -3.94 as well but wrong as well

what did i do wrong?
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Old Nov 29th 2009, 11:20 AM   #2
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use a=r alpha ie a=.115*30.3 and then the formula s=1/2 a(t2^2-t1^2)=0.46 metres
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Old Nov 30th 2009, 12:20 AM   #3
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To the first question :

Calculate the Moment of Inertia of the cylinder , then the the Moment of the force ...

So M=(I)(theta'')

theta'' = M/I
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