Question

A cylindrical rod of length 1 m and radius 4 cm is mounted vertically. It is subjected to a shear force of $ 10^5 $ N at the top. Considering infinitesimally small displacement in the upper edge, the angular displacement $ \theta $ of the rod axis from its original position would be: (shear moduli $ G = 10^{10} $ N/m$^2$)

• A. \( \frac{1}{160\pi} \)
• B. \( \frac{1}{4\pi} \)
• C. \( \frac{1}{40\pi} \)
• D. \( \frac{1}{2\pi} \)

Hint:

The angular displacement due to shear force can be found using the formula \( \theta = \frac{F L}{G A} \), where \( A \) is the cross-sectional area and \( G \) is the shear modulus.

Answer 1

The Correct Option is A

To find the angular displacement \( \theta \) of the rod axis from its initial position, utilize the formula for angular displacement in a cylindrical body subjected to shear force:

\(\theta = \frac{F \cdot L}{G \cdot A}\)

Where:

• \(F\) represents the shear force applied at the rod's top.
• \(L\) is the rod's length.
• \(G\) denotes the material's shear modulus.
• \(A\) is the rod's cross-sectional area.

The provided values are:

• \(F = 10^5\) N
• \(L = 1\) m
• \(G = 10^{10}\) N/m²
• The rod's radius is 4 cm, equivalent to \(0.04\) m.

Begin by calculating the cross-sectional area \(A\) of the cylindrical rod:

\(\pi r^2 = \pi (0.04)^2 = 0.0016\pi\) m²

Substitute these values into the formula:

\(\theta = \frac{10^5 \cdot 1}{10^{10} \cdot 0.0016\pi}\)

Simplify the expression:

\(\theta = \frac{10^5}{10^{10} \cdot 0.0016\pi} = \frac{10^5}{1.6 \times 10^4 \pi}\)

\(\theta = \frac{10^5}{1.6 \times 10^4 \pi} = \frac{10^5}{1.6 \times 10^4 \pi} = \frac{10^5}{1.6 \times 10^4 \pi} = \frac{1}{160\pi}\)

The calculated angular displacement is \( \theta = \frac{1}{160\pi} \). The final answer is:

\( \frac{1}{160\pi} \)