Question

Two identical solid spheres,each of radius 10cm,are kept in contact. If the mass of inertia of this system about the tangent passing through the point of contact is 0.14 kg.\(m^2\) Then mass of each sphere is:

• A. 5 kg
• B. 17.5 kg
• C. 35 kg
• D. 2.5 kg
• E. 10 kg

Answer 1

The Correct Option is A

Moment of Inertia of a Single Sphere:

• The moment of inertia of a solid sphere about its center of mass is \(I_{cm} = \frac{2}{5}MR^2\), where M is the mass and R is the radius.
• We need the moment of inertia about a tangent to the sphere. We'll use the parallel axis theorem: \(I = I_{cm} + MD^2\)
  • \(I_{cm} = \frac{2}{5}MR^2\)
  • \(D = R\) (distance from the center of mass to the tangent)
  • Therefore, \(I = \frac{2}{5}MR^2 + MR^2 = \frac{7}{5}MR^2\)

1. Moment of Inertia of the System:
   • Since there are two identical spheres, the total moment of inertia is: \(I_{total} = 2 \times \frac{7}{5}MR^2 = \frac{14}{5}MR^2\)
2. Solve for Mass (M):
   • We're given that \(I_{total} = 0.14 \text{ kg·m}^2\) and \(R = 10 \text{ cm} = 0.1 \text{ m}\). Plug these values into the equation: \[0.14 = \frac{14}{5} \times M \times (0.1)^2\]
   • Solve for M:
     • \(0.14 = \frac{14}{5} \times M \times 0.01\)
     • \(0.14 = 0.028M\)
     • \(M = \frac{0.14}{0.028} = 5 \text{ kg}\)

Answer 2

The parallel axis theorem states that the moment of inertia (I) of a body about any axis is equal to the sum of its moment of inertia about a parallel axis through its center of mass (I_cm) and the product of its mass (m) and the square of the distance (d) between the two axes:

\[I = I_{cm} + md^2\]

2. Find the moment of inertia of a solid sphere about its diameter:

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

\[I_{cm} = \frac{2}{5}mr^2\]

where r is the radius of the sphere.

3. Find the moment of inertia of one sphere about the tangent:

Using the parallel axis theorem, the moment of inertia of one sphere about a tangent touching its surface is:

\[I_{tangent} = \frac{2}{5}mr^2 + mr^2 = \frac{7}{5}mr^2\]

4. Find the moment of inertia of the system about the tangent:

Since there are two identical spheres, and the tangent passes through their point of contact, the moment of inertia of the system is simply the sum of the moments of inertia of each sphere about the tangent:

\[I_{system} = 2I_{tangent} = 2(\frac{7}{5}mr^2) = \frac{14}{5}mr^2\]

5. Substitute the given values and solve for the mass:

We are given that \(I_{system} = 0.14 \, kg \, m^2\) and \(r = 10 \, cm = 0.1 \, m\). Substituting these values:

\[0.14 = \frac{14}{5}m(0.1)^2\]

\[0.14 = \frac{14}{5}m(0.01)\]

\[0.14 = \frac{0.14}{5}m\]

\[m = \frac{0.14 \times 5}{0.14} = 5 \, kg\]

Final Answer: The mass of each sphere is 5 kg \(\boxed{A}\)