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

Answer:

L=1.49 m

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

$I=\frac{ML^2}{12}+\frac{mL^2}{4}+\frac{mL^2}{4}$

$0.831=L^2(\frac{M}{12}+\frac{2m}{4})=L^2(\frac{3.11}{12}+\frac{0.213}{2})$

$(\frac{3.11+1.278}{12}L^2=0.831$

$L^2=\frac{0.813\times 12}{3.11+1.278}$

$L=\sqrt{\frac{0.813\times 12}{3.11+1.278}}$

$L=1.49 m$

Hence, the length of rod=L=1.49 m