HW12 prob2 T of sphere

# HW12 prob2 T of sphere - Problem Kinetic Energy of a...

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Problem Kinetic Energy of a Suspended Sphere To move the origin of the coordinate system to the surface of the sphere, we construct the inertia tensor I R for the center of mass vector R in the new coordinate system with body frame unit basis vectors denoted by i , j ,and k . Here, obviously, R is R = a k , (1) because we have chosen to locate the COM along the z axis. From the de f nition, I R = M [ R 2 1 RR ] , (2) we obtain the explicit form I R = Ma 2 [ ii + jj ] , (3) after noting that R 2 = a 2 and substituting for R with Eq.(1) and for the unit tensor 1 = ii + jj + kk . (4) We can write Eq.(3) in matrix form as I R = 2 100 010 000 . (5) By the parallel axis theorem, the inertia tensor I we require is I = I 0 + I R , (6) where I 0 is the diagonal inertia tensor in the center of mass coordinate system, I 0 = 2 2 / 50 0 02 / 50 00 2 / 5 . (7) With Eqs.(5) and (7), Eq.(6) yields the inertia tensor in the translated coordi- nate system I = 2 7 / 07 / 2 / 5 . (8) This tells us that when the original coordinate axes are translated to the surface

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HW12 prob2 T of sphere - Problem Kinetic Energy of a...

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