Question

The solid cylinder of length 80 cm and mass M has a radius of 20 cm. Calculate the density of the material used if the moment of inertia of the cylinder about an axis CD parallel to AB as shown in figure is 2.7 kg m\(^2\). 

 

• A. 14.9 kg/m\(^3\)
• B. 1.49 \( \times \) 10\(^2\) kg/m\(^3\)
• C. 7.5 \( \times \) 10\(^1\) kg/m\(^3\)
• D. 7.5 \( \times \) 10\(^2\) kg/m\(^3\)

Hint:

Always ensure your units are consistent (preferably SI units) before starting calculations. The parallel axis theorem is fundamental for finding the moment of inertia about an axis that does not pass through the center of mass. Remember the formula: \(I = I_{CM} + Md^2\).

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We are given the dimensions of a solid cylinder and its moment of inertia about an axis CD. The axis AB is the central axis of the cylinder, and CD is parallel to it at a distance `d` shown as L/2 in the diagram. We need to find the density of the cylinder's material.
Step 2: Key Formula or Approach:
1. The moment of inertia of a solid cylinder about its central axis (AB) is \( I_{AB} = \frac{1}{2}MR^2 \).
2. The Parallel Axis Theorem states that the moment of inertia about any axis parallel to an axis through the center of mass is \( I = I_{CM} + Md^2 \), where `d` is the perpendicular distance between the axes.
3. Density is defined as \( \rho = \frac{\text{Mass}}{\text{Volume}} \). The volume of a cylinder is \( V = \pi R^2 L \).
Step 3: Detailed Explanation:
First, convert the given dimensions to SI units:
Length, \( L = 80 \) cm = 0.8 m.
Radius, \( R = 20 \) cm = 0.2 m.
Moment of inertia about axis CD, \( I_{CD} = 2.7 \) kg m\(^2\).
The distance between the parallel axes AB and CD is \( d = \frac{L}{2} = \frac{0.8}{2} = 0.4 \) m.
The moment of inertia about the central axis AB is \( I_{AB} = \frac{1}{2}MR^2 \).
Using the parallel axis theorem, the moment of inertia about axis CD is:
\[ I_{CD} = I_{AB} + Md^2 \] \[ I_{CD} = \frac{1}{2}MR^2 + M\left(\frac{L}{2}\right)^2 = M\left(\frac{R^2}{2} + \frac{L^2}{4}\right) \] Substitute the given values into this equation to find the mass M:
\[ 2.7 = M\left(\frac{(0.2)^2}{2} + \frac{(0.8)^2}{4}\right) \] \[ 2.7 = M\left(\frac{0.04}{2} + \frac{0.64}{4}\right) \] \[ 2.7 = M(0.02 + 0.16) \] \[ 2.7 = M(0.18) \] \[ M = \frac{2.7}{0.18} = \frac{270}{18} = 15 \text{ kg} \] Now, calculate the volume of the cylinder:
\[ V = \pi R^2 L = \pi (0.2)^2 (0.8) = \pi (0.04)(0.8) = 0.032\pi \text{ m}^3 \] Finally, calculate the density \( \rho \):
\[ \rho = \frac{M}{V} = \frac{15}{0.032\pi} \] Using \( \pi \approx 3.14159 \):
\[ \rho \approx \frac{15}{0.032 \times 3.14159} \approx \frac{15}{0.1005} \approx 149.2 \text{ kg/m}^3 \] This value can be written in scientific notation as \( 1.492 \times 10^2 \) kg/m\(^3\).
Step 4: Final Answer:
The calculated density is approximately 149 kg/m\(^3\), which matches option (B).