Question

A uniform circular disc of radius R lies in x-y plane with its centre coinciding with the origin. Its moment of inertia about an axis passing through a point on the x-axis at a distance x=2R and perpendicular to x-y plane is equal to its moment of inertia about an axis passing through a point on y-axis at a distance y =d and parallel to x-axis in x-y plane. The value of d is:

• A. $\frac{4R}{3}$
• B. $\frac{\sqrt{17}R}{3}$
• C. $\frac{\sqrt{15}R}{2}$
• D. $\frac{\sqrt{13}R}{2}$

Answer 1

The correct option is B $\frac{\sqrt{17}R}{3}$

$\begin{array}{c}\text{Moment of Inertia of disc about an axis through}\\ \text{origin and perpendicular to}x-y\text{plane is}\\ \frac{m{R}^2}{2}{\text{. Using parallel axis theorem Ic =I + md}}^2\end{array}$

$\begin{array}{c}=\frac{1}{2}m{R}^2+m(2R{)}^2=\frac{9m{R}^2}{2}\text{. By using perpendicular}\\ \text{axis theorem at}y=d\text{is}\frac{m{R}^2}{4}+m{d}^2\end{array}$

$\text{Now}\frac{m{R}^2}{4}+m{d}^2=\frac{9m{R}^2}{2}$

$d=\sqrt{\frac{17R}{2}}$