Homework Answers
We know that the moment of inertia of cylinder is 1/2 MR2 and moment of inertia of sphere is 2/5 MR2 and the moment of inertia of spherical surface is 2/3 MR2 where M is the mass and R is the radius.
Using this formula, we can solve this question.
For the cylinder whose radius is R and length is L and mass is M, the moment of inertia will be = 1/2 MR2. Similarly , head has mass mh and trunk has mass mt , the moment of inertia will be 1/2 mhR2 + 1/2 mtR2 .
If you add them , we will get the below results.
Ihorizontal = 2/5mhR2 + 1/2 mtR2 + maR2 + maR2 = (2/5 mh + 1/2 mt + 2ma) R2 + 2/3 mal2 + 1/2 mar2 + 2maRL
sililary you can get the value of Ivertical. . Here the area of arms will be 3maR2 + 4maRR
(2/5 mh + 1/2 mt + 2ma) R2 + 3maR2 + 4maRR.
Hence , the result.
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