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September 12th, 2012, 01:22 AM   #1
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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?
arron1990 is offline  
 
September 12th, 2012, 03:31 AM   #2
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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?
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September 12th, 2012, 08:49 AM   #3
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Re: Centre of mass

I'd suggest reading the question again
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