Question

A uniform cylinder of radius 10 cm and mass 20 kg is mounted so as to rotate freely about a horizontal axis that is parallel to and 5.0 cm from the central longitudinal axis of the cylinder. What is the rotational inertia of the cylinder about the axis of rotation?

• A. \(0.15 \text{ kg}\cdot\text{m}^2\)
• B. \(0.21 \text{ kg}\cdot\text{m}^2\)
• C. \(0.26 \text{ kg}\cdot\text{m}^2\)
• D. \(0.17 \text{ kg}\cdot\text{m}^2\)

Hint:

Always double-check that your units are converted to meters before squaring: \[ R = 0.1\text{ m} \rightarrow R^2 = 0.01, \quad d = 0.05\text{ m} \rightarrow d^2 = 0.0025 \] This prevents decimal alignment errors during the final summation.

Answer 1

The Correct Option is A

Concept: The moment of inertia (rotational inertia) of a body depends on how its mass is distributed relative to the axis of rotation. When calculating the moment of inertia about an axis parallel to one passing through the center of mass, we use the Parallel Axis Theorem: \[ I = I_{\text{cm}} + Md^2 \] where \(I_{\text{cm}}\) is the moment of inertia about the parallel central axis, \(M\) is the total mass, and \(d\) is the perpendicular distance between the two parallel axes.

Step 1: Converting all physical quantities into standard SI units.

• Mass of the uniform solid cylinder, \(M = 20 \text{ kg}\)
• Radius of the solid cylinder, \(R = 10 \text{ cm} = 0.10 \text{ m}\)
• Parallel offset distance between the axes, \(d = 5.0 \text{ cm} = 0.05 \text{ m}\)

Step 2: Calculating the moment of inertia about the center-of-mass axis (\(I_{\text{cm}}\)).
For a uniform solid cylinder rotating about its central longitudinal axis, the formula is: \[ I_{\text{cm}} = \frac{1}{2}MR^2 \] Substitute the given values: \[ I_{\text{cm}} = \frac{1}{2} \times 20 \times (0.10)^2 \] \[ I_{\text{cm}} = 10 \times 0.01 = 0.10 \text{ kg}\cdot\text{m}^2 \]

Step 3: Calculating the additional parallel displacement term (\(Md^2\)).
Calculate the shift term from the parallel displacement: \[ Md^2 = 20 \times (0.05)^2 \] \[ Md^2 = 20 \times 0.0025 = 0.05 \text{ kg}\cdot\text{m}^2 \]

Step 4: Applying the Parallel Axis Theorem.
Sum the center-of-mass moment of inertia and the parallel displacement term together: \[ I = I_{\text{cm}} + Md^2 \] \[ I = 0.10 + 0.05 = 0.15 \text{ kg}\cdot\text{m}^2 \] This matches option (1) exactly.