Question

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).

Answer 1

`I_(edge) = 6.67 kg.m^2`

Step-by-step 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

`I_(CM)= (1)/(12)M(L)^2`

using parallel axis theorem

`I_(edge) = I_(CM) + MR^2`

`I_(edge) =(1)/(12)ML^2+ M((L)/(2))^2`

`I_(edge) =(1)/(12)ML^2+ (ML^2)/(4)`

`I_(edge) =(1)/(3)ML^2`

now, inserting all the given values

`I_(edge) =(1)/(3)* 5 * 2^2`

`I_(edge) = 6.67 kg.m^2`