Question

A uniform thin rod of mass ????=3.41 kg pivots about an axis through its center and perpendicular to its length. Two small bodies, each of mass m=0.249 kg , are attached to the ends of the rod. What must the length L of the rod be so that the moment of inertia of the three-body system with respect to the described axis is ????=0.929 kg·m2 ?

Answer 1

The length of the rod for the condition on the question to be met is
`L = 1.5077 \ m`

Step-by-step explanation:

The Diagram for this question is gotten from the first uploaded image

From the question we are told that

The mass of the rod is
`M = 3.41 \ kg`

The mass of each small bodies is
`m = 0.249 \ kg`

The moment of inertia of the three-body system with respect to the described axis is
`I = 0.929 \ kg \cdot m^2`

The length of the rod is L

Generally the moment of inertia of this three-body system with respect to the described axis can be mathematically represented as

`I = I_r + 2 I_m`

Where
`I_r` is the moment of inertia of the rod about the describe axis which is mathematically represented as

`I_r = (ML^2 )/(12)`

And
`I_m` the moment of inertia of the two small bodies which (from the diagram can be assumed as two small spheres) can be mathematically represented as

`I_m = m * [\frac{L} {2} ]^2 = m* (L^2)/(4)`

Thus
`2 * I_m = 2 * m (L^2)/(4) = m * (L^2)/(2)`

Hence

`I = M * (L^2)/(12) + m * (L^2)/(2)`

=>
`I = [(M)/(12) + (m)/(2)] L^2`

substituting vales we have

`0.929 = [(3.41)/(12) + (0.249)/(2)] L^2`

`L = \sqrt{(0.929)/(0.40867) }`

`L = 1.5077 \ m`