Find the rotational inertia of the system below if m1 = 5.0...

Find the rotational inertia of the system below if $m_1$ = 5.0 kg, $m_2$ = 3.0 kg, $M$ = 2.0 kg, and $L$ = 1.0 m. $\frac{1}{2}L$ $\frac{1}{4}L$ $\frac{1}{4}L$ $m_1$ Rod of mass $M$ $m_2$ Rotation axis $I$ = _______ kg m$^2$

Transcript

00:01 Have a system having two identical thin rods of mass capital m and length capital l and two particles each of mass is small m and as shown in the figure so the diagram is like this so this is our first rod okay and this is our second rod and this is our axis ab okay this is our line ab which is passing through this rod okay this is suppose ab line okay now, this is a point, this is b point, and this is intersection point which is point o, okay, and this distance is l by 2 and this distance is l by 2 and here is a particle of mass small m and this distance is l, length of the raw.

00:45 And this distance total is all obviously from here to here is length l, but here is a mass particle of, here is a mass particle having mass, mass, small m okay so this is the diagram for the question okay now for the part a we have to determine the moment of inertia about an axis which passes through the point o this point and which is perpendicular to the plane containing the raw okay so we have to calculate the rotational inertia so rotational inertia about oh it will be given by moment of inertia of this mass which will be mass multiplied by distance square so this distance is l by 2 so l by 2 square okay plus moment of inertia of this particle so m multiplied by distance is square so l so l square plus moment of inertia of this rod okay so this rod is passing through the midpoint so this will be equals to m l square by 12 plus moment of inertia of this rod and this rod is passing through its end so it will be equals to ml square by 3 okay so from here after rearranging we will get rotational inertia i about o point it will be equals to 5 by 4 l square and small m plus m by 3.

02:08 So this is the answer for the part a of the problem.

02:12 Now we can calculate the moment of inertia for the part b and about axis ab...

Question

Find the rotational inertia of the system below if $m_1$ = 5.0 kg, $m_2$ = 3.0 kg, $M$ = 2.0 kg, and $L$ = 1.0 m. $\frac{1}{2}L$ $\frac{1}{4}L$ $\frac{1}{4}L$ $m_1$ Rod of mass $M$ $m_2$ Rotation axis $I$ = _______ kg m$^2$

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