Question

Please, I would like some help on the attached problem

Answer 1

PART B - Problem 3

Answer 2

Thanks in advance

Answer 3

Prove that \(\bar I=M \bar x ^2 + I\). Here it goes:
Start by setting the center of mass at the origin and translate the axis of rotation. Then apply the formula for Moment of Inertia,
\[\bar I = \iint\limits_R{(x- \bar x )^2\delta dA} = \bar I = \iint\limits_{R}{(x^2 -2 \bar x x + \bar x^2) \delta dA} \]
\( \ \ \iint\limits_R{\bar x ^2\delta dA} = M \bar x^2\) because \(\bar x \) is a constant and \(\delta A = M\). Note that \(\iint dA = A\)
The second term vanishes. \(\iint\limits_R xdA = 0\) because it is the center of mass, but that is in the origin. Then we are left with \(\int \int x^2 \delta dA = I\). Thus, \[\bar I = M \bar x ^2 + I\] Completing the Proof.

Obs: The key points are the translation of the axis of rotation and the realization that since the center of mass is at the origin, and that \(\iint xdA\) is the center of mass then it must be zero.

Answer 4

thank you for your answer