Consider a rod of length L rotated about one of its ends instead of about its center of mass. If the mass of the rod is 5 kg, and the length is 2 meters, calculate the magnitude of the moment of inertia (I).

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wagonbelleville wagonbelleville · Answer 1 · 2019-12-04T09:34:03+01:00

Answer:

[tex]I_{edge} = 6.67 kg.m^2[/tex]

Explanation:

given,

mass of rod = 5 Kg

Length of rod = 2 m

R = 1 m

moment of inertial from one edge of the rod = ?

moment of inertia of rod through center of mass

[tex]I_{CM}= \dfrac{1}{12}M(L)^2[/tex]

using parallel axis theorem

[tex]I_{edge} = I_{CM} + MR^2[/tex]

[tex]I_{edge} =\dfrac{1}{12}ML^2+ M(\dfrac{L}{2})^2[/tex]

[tex]I_{edge} =\dfrac{1}{12}ML^2+ \dfrac{ML^2}{4}[/tex]

[tex]I_{edge} =\dfrac{1}{3}ML^2[/tex]

now, inserting all the given values

[tex]I_{edge} =\dfrac{1}{3}\times 5 \times 2^2[/tex]

[tex]I_{edge} = 6.67 kg.m^2[/tex]