Question 286: Centre of Mass and Rotational Motion

Question

The moment of inertia of a solid sphere, about an axis parallel to its diameter and at a distance of x from it, is 'I(x)'. Which one of the graphs represents the variation of I(x) with x correctly ?

Accepted Answer

The correct answer is
Option C
. Here's why :
The moment of inertia of a solid sphere about an axis passing through its center is given by :
$$I_0 = \frac{2}{5}MR^2$$
Where :
$I_0$ is the moment of inertia about the center
$M$ is the mass of the sphere
$R$ is the radius of the sphere
Now, using the parallel axis theorem, we can find the moment of inertia about an axis parallel to the diameter and at a distance of 'x' from it :
$$I(x) = I_0 + Mx^2$$
Substituting the value of $I_0$ :
$$I(x) = \frac{2}{5}MR^2 + Mx^2$$
This equation represents a parabola, where :
The coefficient of $x^2$ is positive (M), indicating an upward-opening parabola.
The y-intercept (the value of $I(x)$ when $x = 0$) is $\frac{2}{5}MR^2$, which is a positive constant.
Therefore, the graph of $I(x)$ vs. $x$ should be an upward-opening parabola, and
Option C
accurately represents this.

Answer

Answer

Answer

Answer

Centre of Mass and Rotational Motion

The moment of inertia of a solid sphere, about an axis parallel to its diameter and at a distance of x from it, is 'I(x)'. Which one of the graphs represents the variation of I(x) with x correctly ?

Choose the correct answer: