Multiple choice

A hollow cylinder has mass M. outside radius ${ R }{ 2 }$ and inside radius ${ R }{ 1 }$ Its moment of inertia about an or axis parallel to its symmetry axis and tangential to tho outer surface is equal to :

  1. $\dfrac { M } { 2 } \left( R _ { 2 } ^ { 2 } + R _ { 1 } ^ { 2 } \right)$
  2. $\dfrac { M } { 2 } \left( R ^ { 2 } - R _ { 1 } ^ { 2 } \right)$
  3. $\dfrac { M } { 4 } \left( R _ { 2 } + R _ { 1 } \right) ^ { 2 }$
  4. $\dfrac { M } { 2 } \left( 3 R _ { 2 } ^ { 2 } + R _ { 1 } ^ { 2 } \right)$
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D Correct answer
Explanation

The moment of inertia of the hollow cylinder about its symmetry axis is M/2 x (R2^2 + R1^2). The tangential axis is displaced by R2, so the parallel-axis theorem adds M R2^2, giving M/2 x (3R2^2 + R1^2).

AI explanation

The moment of inertia of a hollow cylinder about its central symmetry axis is I = (1/2) * M * (R2^2 + R1^2). To find the moment of inertia about a parallel axis tangential to the outer surface, we apply the parallel axis theorem: I_parallel = I_cm + M * d^2, where the distance d is equal to the outer radius R2. Substituting d = R2 gives I_parallel = (1/2) * M * (R2^2 + R1^2) + M * R2^2. Combining these terms results in (M/2) * (3R2^2 + R1^2).

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