Question

Ring of mass straight m and radius straight r is attached to the end of a thin rod of mass straight m and radius 2 straight r

Answer 1

For a ring of mass m and radius straight r connected to a thin rod, the total moment of inertia about an axis is $\rm I= L^2 (\frac{M}{3} +m)+\frac{2mr^2}{5}$.

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;

$\rm I = I_{rod}+I_{ring} \\\\ \rm I =\frac{ML^2}{3} +\frac{2mr^2}{5}$

According to the parallel axis theorem,

$\rm I = \frac{2mr^2}{5} + ML^2$

The total Moment of Inertia about an axis will be;

$\rm I_{total} = \frac{ML^2}{3} +\frac{2mr^2}{5}+ML^2 \\\\ \rm I_{total} =L^2 (\frac{M}{3} +m)+\frac{2mr^2}{5}$

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 $\rm I= L^2 (\frac{M}{3} +m)+\frac{2mr^2}{5}$.

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Answer 2

The total moment of inertia about an axis is : $L^{2} ( \frac{M}{3} + m ) + \frac{2mr^2}{5}$ 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

        = $L^{2} ( \frac{M}{3} + m ) + \frac{2mr^2}{5}$

 

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