Question

The maximum velocity to avoid overturning on a circular track of radius \(r = 100\) m, with coefficient of friction \(\mu = 0.2\), is:

• A. 0.14 m/s
• B. 140 m/s
• C. 1.4 m/s
• D. 14 m/s

Hint:

Use v equals the square root of mu times g times r, since friction supplies the centripetal force.

Answer 1

The Correct Option is D

Check units and rough size first, then confirm with the exact formula.

• The formula for maximum speed on a flat circular track limited by friction is $v = \sqrt{\mu g r}$. Checking units: $\mu$ has no units, $g$ is in m/s^2, $r$ is in m, so the product is in m^2/s^2, and its square root is in m/s, a proper speed unit.
• For a quick estimate, take $g \approx 10$ m/s^2. Then $\mu g r = 0.2 \times 10 \times 100 = 200$, and $\sqrt{200} \approx 14.1$ m/s. This immediately rules out 0.14 m/s and 1.4 m/s, which are smaller by factors of 100 and 10, and rules out 140 m/s, which is ten times too large.
• Only 14 m/s is in the right range. Now do the exact calculation with $g = 9.8$ m/s^2: $\mu g r = 0.2 \times 9.8 \times 100 = 196$, and $\sqrt{196} = 14$ m/s exactly.

This confirms the maximum velocity before the vehicle overturns or skids is 14 m/s, option D.