Question

A solid aluminum sphere of radius R has moment of inertia I about an axis through its center. Will the moment of inertia about a central axis of a solid aluminum sphere of radius 2R be (a) 2I, (b) 4I, (c) 8I, (d) 16I, or (e) 32I ?

Short answer

The moment of inertia of sphere when the radius is 2R is,$32I$

Step-by-step solution

Define Moment of inertia:

The measure of body resistance and angular acceleration in relation to a given axis is equal to the sum of the products of each element of body weight and the square of the distance of the element from the axis is known as moment of inertia.

Calculate inertia:

$I=\frac{2}{5}M{R}^2$

Similarly, $I\text{'}=\frac{2}{5}M\text{'}R{\text{'}}^2$ → (1)

Calculate M^',

$M=\rho V$

$M=\rho \left(\frac{4}{3}\pi R^3\right)$ → (2)

Consider R=2R,

Substitute in equation (2),

$M\text{'}=\rho \left(\frac{4}{3}\pi {\left(2R\right)}^3\right)$

$M\text{'}=8\rho \left(\frac{4}{3}\pi R^3\right)$

$M\text{'}=8M$

Substitute in (1),

$$\begin{gathered} I\text{'}=\frac{2}{5}M\text{'}R{\text{'}}^2 \\ I\text{'}=\frac{2}{5}8M(2R{)}^2 \\ I\text{'}=32\left(\frac{2}{5}\right)M{R}^2 \\ I\text{'}=32l \end{gathered}$$

The solution is (e) $I\text{'}=32l$