Question

A solid sphere of mass $5$ kg and radius $10$ cm is kept in contact with another solid sphere of mass $10$ kg and radius $20$ cm. The moment of inertia of this pair of spheres about the tangent passing through the point of contact is ___________ kg$\cdot$m$^2$.

• A. $0.18$
• B. $0.63$
• C. $0.72$
• D. $0.36$

Hint:

For axes touching the surface of a sphere, always use the parallel axis theorem with $d=R$.

Answer 1

The Correct Option is A

Step 1: Use moment of inertia of a solid sphere.
Moment of inertia of a solid sphere about its center: \[ I_{\text{cm}}=\frac{2}{5}MR^2 \] Step 2: Apply parallel axis theorem.
For rotation about the tangent through the point of contact: \[ I = I_{\text{cm}} + Md^2 \] where $d=R$.
Step 3: Calculate for the first sphere.
\[ M_1=5\text{ kg}, \quad R_1=0.10\text{ m} \] \[ I_1=\frac{2}{5}(5)(0.1)^2 + 5(0.1)^2 =0.04+0.05=0.09 \] Step 4: Calculate for the second sphere.
\[ M_2=10\text{ kg}, \quad R_2=0.20\text{ m} \] \[ I_2=\frac{2}{5}(10)(0.2)^2 + 10(0.2)^2 =0.16+0.40=0.56 \] Step 5: Total moment of inertia.
\[ I_{\text{total}} = I_1 + I_2 = 0.09 + 0.09 = 0.18 \] Final Answer: $\boxed{0.18\text{ kg}\cdot\text{m}^2}$