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Sagot :
The total moment of inertia about an axis is : [tex]L^{2} ( \frac{M}{3} + m ) + \frac{2mr^2}{5}[/tex] for a ring of mass m and radius straight r attached to a thin rod.
Determine the Total moment of Inertia about an axis
Given data:
mass of ring --> m
radius of ring --> r
mass of rod --> M
Length of rod ---> L ( 2 * radius )
Total Moment of Inertia about an axis = Irod + Iring
where : Irod = moment of inertia of rod, Iring = moment of inertia of ring
Irod = ML² / 3
Iring = 2mr² / 5
moment of inertia around an axis by Iring = I
where ; I = 2mr² / 5 + ML² according to parallel axis theorem
Hence the Total moment of Inertia about an axis is :
Itotal = 2mr²/5 + ML² + ML² / 3
= [tex]L^{2} ( \frac{M}{3} + m ) + \frac{2mr^2}{5}[/tex]
Learn more about Moment of inertia : https://brainly.com/question/6956628
For a ring of mass m and radius straight r connected to a thin rod, the total moment of inertia about an axis is [tex]\rm I= L^2 (\frac{M}{3} +m)+\frac{2mr^2}{5}[/tex].
What is a moment of inertia?
The sum of the products of the mass of each particle in the body with the square of its distance from the axis of rotation expresses a body's tendency to resist angular acceleration.
The given data in the problem is;
m is the ring's mass.
r is the radius of a ring.
L is the rod length = 2r
The total Moment of Inertia about an axis;
[tex]\rm I = I_{rod}+I_{ring} \\\\ \rm I =\frac{ML^2}{3} +\frac{2mr^2}{5}[/tex]
According to the parallel axis theorem,
[tex]\rm I = \frac{2mr^2}{5} + ML^2[/tex]
The total Moment of Inertia about an axis will be;
[tex]\rm I_{total} = \frac{ML^2}{3} +\frac{2mr^2}{5}+ML^2 \\\\ \rm I_{total} =L^2 (\frac{M}{3} +m)+\frac{2mr^2}{5}[/tex]
Hence for a ring of mass m and radius straight r connected to a thin rod, the total moment of inertia about an axis is [tex]\rm I= L^2 (\frac{M}{3} +m)+\frac{2mr^2}{5}[/tex].
To learn more about the moment of inertia refer to the link;
https://brainly.com/question/15246709
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