The moment of inertia of a solid sphere is 40 kg-m² about its diametric axis. Determine the moment of inertia about any tangent.

Steps to Solve

1. Using the Parallel Axis Theorem

To find the moment of inertia about a tangent axis, we can use the parallel axis theorem, which states:

$$ I = I_C + md^2 $$

where:

• ( I ) is the moment of inertia about the tangent,
• ( I_C ) is the moment of inertia about the center of mass (diametric axis),
• ( m ) is the mass of the sphere, and
• ( d ) is the distance from the center of mass to the tangent axis.

2. Identifying the Given Values

We know:

• The moment of inertia about the diametric axis is ( I_C = 40 , \text{kg-m}^2 ).
• The distance ( d ) from the center ( C ) of the sphere to the tangent line is equal to the radius ( R ).

3. Finding the Mass of the Sphere

The moment of inertia for a solid sphere about its center is given by:

$$ I_C = \frac{2}{5} m R^2 $$

Rearranging to find ( m ):

$$ m = \frac{5}{2} \frac{I_C}{R^2} $$

4. Calculating the Moment of Inertia About the Tangent Axis

Now, substituting ( m ) and ( d ) back into the parallel axis theorem:

$$ I = I_C + md^2 $$

Since ( d = R ):

$$ I = I_C + mR^2 $$

Substituting ( m ):

$$ I = I_C + \frac{5}{2} \frac{I_C}{R^2} R^2 $$

This simplifies to:

$$ I = I_C + \frac{5}{2} I_C = \frac{7}{2} I_C $$

5. Final Calculation

Now substituting ( I_C = 40 , \text{kg-m}^2 ):

$$ I = \frac{7}{2} \times 40 = 140 , \text{kg-m}^2 $$