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Suppose a uniform slender rod has length l and mass m. the moment of inertia of the rod about about an axis that is perpendicular to the rod and that passes through its center of mass is given by icm=112ml2. find iend, the moment of inertia of the rod with respect to a parallel axis through one end of the rod. express iend in terms of m and l. use fractions rather than decimal numbers in your answer.

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TheSandman1337
The parallel axis theorem for the calculation of inertia is the following:
[tex]I=I_{CM}+Md^2 [/tex] where I is the inertia from an axis that is at distance d from the center of mass and Icm the inertia when the axis passes through the center of mass. In this case, the axis passes through the end of the rod and thus d=l/2 in this case.
We have that:
[tex]I=ml^2/12+m(l/2)^2=ml^2/12+ml^2/4=ml^2/12+3ml^2/12=ml^2/3[/tex].
This yields us the final result: it is (ml^2)/3.
onyebuchinnaji

The moment of inertia of the rod with respect to a parallel axis through one end of the rod is [tex]\frac{Ml^2 }{3}[/tex].

The given parameters:

  • Length of the rod, = L
  • Mass of the rod, = M

The parallel axis theorem for the calculation of inertia is given as follows;

[tex]I = I_{CM} + Md^2[/tex]

where;

  • I is the inertia from an axis that is at distance d from the center of mass
  • [tex]I_{CM}[/tex] the inertia when the axis passes through the center of mass.

The distance, d through the center of the rod, d = L/2

[tex]I = I_{CM} + Md^2\\\\I = \frac{Ml^2}{12} + M(\frac{l}{2} )\\\\I = \frac{Ml^2}{12} + \frac{Ml^2}{4} \\\\I = \frac{Ml^2 + 3Ml^2}{12} \\\\I = \frac{4Ml^2}{12} \\\\I = \frac{Ml^2}{3}[/tex]

Thus, the moment of inertia of the rod with respect to a parallel axis through one end of the rod is [tex]\frac{Ml^2 }{3}[/tex].

Learn more about parallel axis theorem here: https://brainly.com/question/15301243

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