Question

A rod of length 4m is rotated about an axis perpendicular to the rod and passing through a point 1 m from one of its end as shown in figure. The rod has a uniform mass density and has a total mass of 10 kg. What will be its moment of Inertia about this axis?

• A. $\frac{70}{6}Kg-m^2$
• B. $\frac{70}{3}Kg-m^2$
• C. $\frac{35}{3}Kg-m^2$
• D. $\frac{35}{6}Kg-m^2$

Answer 1

The correct option is B $\frac{70}{3}Kg-m^2$

Let's follow the similar steps as mentioned in the previous videoclip,

$dm=\frac{M}{L}dx=\frac{10}{4}dx$
$dm=\frac{5}{2}dx$
$I={\int }_{-1}^3\frac{5}{2}{x}^{2}dx$
$I=\left(\frac{5}{2}\right){\left[\begin{array}{c}\frac{x^3}{3}\end{array}\right]}^3$
$=\frac{5}{6}\left\{\begin{array}{c}27-(-1)\end{array}\right\}$
$=\frac{5}{6}\times 28=\frac{70}{3}kg-m^2.$