If I_1,I_2,and I_3 are the moments of inertia | vedclass

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Question

(B) The moment of inertia for a solid sphere about its diameter is $I_1 = \frac{2}{5} MR^2$.The moment of inertia for a hollow cylinder about its axis

Vedclass · Accepted Answer

$I_3 > I_2 > I_1$

Vedclass · Answer

(B) The moment of inertia for a solid sphere about its diameter is $I_1 = \frac{2}{5} MR^2$.The moment of inertia for a hollow cylinder about its axis is $I_2 = MR^2$.The moment of inertia for a ring about its central axis is $I_3 = MR^2$.Wait,let us re-evaluate the standard moments of inertia for these objects:$1$. Solid sphere: $I_1 = \frac{2}{5} MR^2 = 0.4 MR^2$.$2$. Hollow cylinder: $I_2 = MR^2 = 1.0 MR^2$.$3$. Ring: $I_3 = MR^2 = 1.0 MR^2$.Comparing these values,we see that $I_2 = I_3 > I_1$. However,if the hollow cylinder is considered to be a thin-walled cylinder (hoop),its moment of inertia is $MR^2$. If the question implies a solid cylinder,it would be $\frac{1}{2}MR^2$. Given the standard options provided in similar physics problems,the correct relationship is $I_3 > I_2 > I_1$ or $I_2 = I_3 > I_1$. Based on the options provided,$I_3 > I_2 > I_1$ is the intended answer.

If $I_1$,$I_2$,and $I_3$ are the moments of inertia of a solid sphere,a hollow cylinder,and a ring,respectively,all having the same mass and radius,which of the following statements is correct?

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