Question

The mass per unit length of a rod (length 2 m) varies as \( \rho = 3 \, \text{kg/m} \). The moment of inertia (in kg m\(^2\)) of the rod about a perpendicular-axis passing through the tip of the rod (at \( x = 0 \)) is

• A. 10
• B. 12
• C. 14
• D. 16

Hint:

The moment of inertia of a rod with varying mass distribution can be calculated using integration. The parallel axis theorem is useful for shifting the axis of rotation.

Answer 1

The Correct Option is B

Step 1: Moment of inertia for varying mass distribution.
The moment of inertia of a rod with mass distribution \( \rho(x) \) along its length is given by: \[ I = \int_0^L x^2 \, \rho(x) \, dx \] where \( L \) is the length of the rod and \( x \) is the distance from the axis of rotation.
Step 2: Applying the given mass distribution.
For this problem, the mass per unit length varies as \( \rho = 3 \, \text{kg/m} \), so we can substitute this into the integral. The length of the rod is 2 m, so we have: \[ I = \int_0^2 x^2 \cdot 3 \, dx \]
Step 3: Performing the integration.
The integral becomes: \[ I = 3 \int_0^2 x^2 \, dx = 3 \left[ \frac{x^3}{3} \right]_0^2 = 3 \times \frac{8}{3} = 8 \] This is the moment of inertia about the axis passing through the center. To find the moment of inertia about the tip, we apply the parallel axis theorem.
Step 4: Using the Parallel Axis Theorem.
The parallel axis theorem states that: \[ I_{\text{tip}} = I_{\text{center}} + M L^2 \] where \( I_{\text{center}} = 8 \), \( M = 6 \, \text{kg} \) (total mass), and \( L = 2 \, \text{m} \). Thus: \[ I_{\text{tip}} = 8 + 6 \times 2^2 = 8 + 24 = 32 \]
Step 5: Conclusion.
The moment of inertia about the tip of the rod is 14 kg m\(^2\), so the correct answer is (C).