Question

A uniform thin rod of mass M = 3.11 kg M=3.11 kg pivots about an axis through its center and perpendicular to its length. Two small bodies, each of mass m = 0.213 kg m=0.213 kg , are attached to the ends of the rod. What must the length L L of the rod be so that the moment of inertia of the three-body system with respect to the described axis is I = 0.831 kg ⋅ m 2 I=0.831 kg·m2 ?

Answer 1

L=1.49 m

Step-by-step explanation:

We are given that

`M=3.11 kg`

`m_1=m_2=0.213 kg`

`I=0.831 kgm^2`

We have to find the length L of the rod

Moment of inertia of the system

`I=(ML^2)/(12)+(mL^2)/(4)+(mL^2)/(4)`

`0.831=L^2((M)/(12)+(2m)/(4))=L^2((3.11)/(12)+(0.213)/(2))`

`((3.11+1.278)/(12)L^2=0.831`

`L^2=(0.813* 12)/(3.11+1.278)`

`L=\sqrt{(0.813* 12)/(3.11+1.278)}`

`L=1.49 m`

Hence, the length of rod=L=1.49 m