Question #3eb2c

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Question #3eb2c

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Answer 1 · 2016-03-23T16:01:41

c) $= 0.0083 k g m^{2}$ $=0.0083kgm^2$

Explanation:

Recall that the moment of inertia of a point mass is given by the expression
$I = m r^{2}$ $I = mr^2$
In case of rod we need to consider it is to be made up of an infinite number of point masses. Each point mass to be multiplied by the square of its distance from the axis of rotation. Thereafter integral taken over the length of the rod.

Moment of inertia for a uniform rod of mass $M$ $M$ and length $L$ $L$ having negligible thickness, about its center of mass is given by the expression

$I = \int_{- \frac{L}{2}}^{\frac{L}{2}} \frac{M}{L} r^{2} . d r = \frac{1}{12} M L^{2}$ $I=int_(-L/2)^(L/2)M/Lr^2.dr=1/12ML^2$
Inserting values given in the problem

$I = \frac{1}{12} M L^{2} = \frac{1}{12} \cdot 0.1 \cdot 1^{2} = 0.008 . 3 k g m^{2}$ $I=1/12ML^2=1/12 cdot 0.1cdot1^2=0.008 dot 3kgm^2$

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Question #3eb2c

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