pls tell me what would be moment of inertia along body diagonal of a cube of length 'a' and mass 'm' placed on each corner ?
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Well, if the cube has side length a then the corners are at positions like (±1,±1,±1)a/2. Let’s assume the rotation axis is from (1,1,1)a/2 to (−1,−1,−1)a/2. The easiest way to find the lever arm for the other corners is to find the cross product into the unit vector for the axis direction. A representative unit vector is u=(1,1,1)/3–√. The cross product with (1,1,-1)a/2 is (1,−1,0)a/3–√ and the length of that is a2/3−−−√. The other five off-axis corners contribute equally, so the net MOI is 623ma2=4ma2.
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The acceleration due to gravity 'g' decreases if we go down from the surface of the earth towards its center.
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