Answer:
The question is incomplete as the diagram wasn't attached. Please use the diagram in the attachement as a guide.
Moment of Inertia, [tex]I=1.61kg.m^{2}[/tex]
Step-by-step explanation:
The axis of rotation as shown in the sample diagram is the support.
Moment of Inertia, I = ∑[tex]mr^{2}[/tex]
There are two blocks: The block on the right with a mass of 2.0kg and the block on the left with a mass of 7.0kg as we were given.
Therefore I = [tex]m_{1}r_{1} ^{2} + m_{2}r_{2} ^{2}[/tex]
[tex]I=7.0kg*(0.30m)^{2} + 2.0kg * (0.70m)^{2} \\\\ I = (7.0kg * 0.09m^{2} )+(2.0kg*0.49m^{2} )\\\\I = 0.63kgm^{2} + 0.98kgm^{2}\\ \\I = 1.61kg.m^{2}[/tex]
The moment of inertia of the body is 1.61kg/m^2
Data;
The moment of inertia acting on this body is the sum total of the product of the mass with the square of the distance from it's axis of rotation.
[tex]I = \sum mr^2\\I = m_1r_1^2 + m_2r_2^2[/tex]
Let's substitute the values and solve for the moment of inertia
[tex]I = (7 * 0.3^2) + (2*0.7^2)\\I = 1.61 kg/m^2[/tex]
The moment of inertia of the body is 1.61kg/m^2
Learn more on moment of inertia here;
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