Question

A 20-ton truck is traveling along a curved path of radius 240 m. If the center of gravity of the truck above the ground is 2 m and the distance between its wheels is 1.5 m, the maximum speed of the truck with which it can travel without toppling over is:
(Acceleration due to gravity \( g = 10 \) ms\(^{-2}\))

• A. \( 43 \) ms\(^{-1}\)
• B. \( 40 \) ms\(^{-1}\)
• C. \( 38 \) ms\(^{-1}\)
• D. \( 30 \) ms\(^{-1}\)

Hint:

Toppling occurs when centrifugal force overcomes restoring torque. - Use torque balance to find the critical speed.

Answer 1

The Correct Option is D

Step 1: Use toppling condition
The toppling condition occurs when the moment due to centrifugal force equals the moment due to weight: \[ \frac{m v^2}{r} \times h = mg \times \frac{d}{2}. \] Step 2: Solve for \( v \)
\[ \frac{v^2}{240} \times 2 = 10 \times \frac{1.5}{2}. \] \[ \frac{2v^2}{240} = 7.5. \] \[ v^2 = 900. \] \[ v = 30 \text{ m/s}. \] Thus, the correct answer is \( \boxed{30} \) ms\(^{-1}\).