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 September 12th, 2012, 02:22 AM #1 Member   Joined: Jan 2012 Posts: 82 Thanks: 0 Centre of mass I need some help with this problem. A uniform disc is projected down a plane that has incline at angle a to the horizontal. The unit vector i points down the line of greast slope of the plane and the unit vector j is perpendicular. The coefficient of frictin between the disc and the plane is 2 tan a. I have the forces acting on the disc as F = -|F| i N = |N| j W = -Mg sin a i -MG cos a j If the centre of the disc has moved a distance x down the slope at time t, I need to determine the equation of motion of the centre of mass of the disc. So far I have Newtons 2nd Law: Mx(double dot)i= N+F+W Resolving in the i-direction Mx=-|F|+Mg sin a but |F| = 2 tan a Resolving in the j-direction |N|-Mg cos a = 0 therefore |N|=Mg cos a Mx = 2 tan z + Mg cos a - Mg sin a x= 2 tan + g cos a - g sin x I need to show this becomes x= -g sin a can anyone help?
 September 12th, 2012, 04:31 AM #2 Member   Joined: Jan 2012 Posts: 82 Thanks: 0 Re: Centre of mass Actually, I get |f| = -2 tan a Hence Mx = - 2 tan a + Mg cos a + Mg sin a x= - 2 tan a + g cos a + g sin a x = -2 g sin a + g sin a x = - g sin a Does this look right?
 September 12th, 2012, 09:49 AM #3 Newbie   Joined: May 2012 Posts: 3 Thanks: 0 Re: Centre of mass I'd suggest reading the question again

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