Physics Help Forum distance...?

 Kinematics and Dynamics Kinematics and Dynamics Physics Help Forum

 Nov 29th 2009, 11:58 AM #1 Senior Member   Join Date: Sep 2009 Location: St. John's, NL Posts: 124 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?
 Nov 29th 2009, 12:20 PM #2 Senior Member   Join Date: Oct 2009 Location: india Posts: 409 use a=r alpha ie a=.115*30.3 and then the formula s=1/2 a(t2^2-t1^2)=0.46 metres
 Nov 30th 2009, 01:20 AM #3 Member   Join Date: Oct 2009 Posts: 58 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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