Dynamics_Part77

# Dynamics_Part77 - Focusing on the vertical and horizontal...

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Focusing on the vertical and horizontal directions, clearly T cos β mg =0 and T sin β = m | a | Therefore, the magnitude of the tension force is T = mg cos β Using the equation for the acceleration of the center of mass above to determine | a | , the horizontal component of Newton’s Second Law ( T sin β = m | a | ) simplifies to g tan β =( f sin β + c sin θ ) ˙ φ 2 Finally, since T acts at Point G while gravity acts at the center of mass (and contributes no moment at G), the moment about the sphere’s center of mass is given by M G = r B/G × T Inspection of the geometry shows that r B/G = c k and T = T sin( β θ ) i + T cos( β θ ) k Therefore, the moment is M G = c k × [ T sin( β θ ) i + T cos( β θ ) k ] Now, k × i = j and k × k = 0 , wherefore M G = cT sin( β θ ) j (c) From the angular-momentum equation, we know that
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## This note was uploaded on 05/12/2010 for the course AME 301 taught by Professor Shiflett during the Spring '06 term at USC.

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