I = Isphere + Irod By parallel axis theorem, I = [2/5 ms R^2 + ms R^2] + [1/3 mr

Question

I = Isphere + Irod

By parallel axis theorem,

I = [2/5 ms R^2 + ms R^2] + [1/3 mr L^2 + mr (2R)^2]

Thus,

I = 173.1072 kg m^2 [ANSWER]

AND IT DID NOT WORK!!

An object is formed by attaching a uniform, thin rod with a mass of mr = 7.36 kg and length L = 4.8 m to a uniform sphere with mass ms = 36.8 kg and radius R = 1.2 m. Note ms = 5mr and L = 4R. What is the moment of inertia of the object about an axis at the left end of the rod? If the object is fixed at the left end of the rod, what is the angular acceleration if a force F perpendicular to the rod at the center of the rod? What is the moment of inertia of the object about an axis at the center of mass of the object? (Note: the center of mass can be calculated to be located at a point halfway between the center of the sphere and the left edge If the object is fixed at the center of mass, what is the angular acceleration if a force F the rod at the end of rod? What is the moment of inertia of the object about an axis at the right edge of the sphere? Compare the three moments of inertia calculated above:

Explanation / Answer

You are doing a small mistake while applyig parallel axis theorm for stick.

I = Isphere + Irod

By parallel axis theorem,

I = Icm_sphere + ms*d1^2 + Icm_rod + mr*d2^2

= [2/5 ms R^2 + ms R^2] + [1/12 mr L^2 + mr (2R + L/2)^2]

= (7/5)*ms*R^2 + mr*L^2/12 + mr*(2*R + L/2)^2

= (7/5)*36.8*1.2^2 + 7.36*4.8^2/12 + 7.36*(2*1.2 + 4.8/2)^2

= 257.9 kg.m^2 <<<<<--------Answer

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