Question

Radius of gyration of a thin uniform circular disc about the axis passing through its centre and perpendicular to its plane is \(K_c\). Radius of gyration of the same disc about a diamter of the disc is \(K_d\). the ratio \(K_d:K_c\) is

• A. \(1:4\)
• B. \(1:\sqrt{2}\)
• C. \(\sqrt{2}:1\)
• D. \(2:1\)

Hint:

At the highest point the net force toward the centre is mg minus the normal reaction.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
At the highest point of a convex bridge the centre of curvature is below the car. The net force toward the centre must supply the centripetal force \(\frac{mv^2}{R}\).

Step 2: Write the equation:
Weight \(mg\) acts downward (toward the centre) and the normal reaction N acts upward:
\[ mg - N = \frac{mv^2}{R} \Rightarrow N = mg - \frac{mv^2}{R} \]

Step 3: Thrust:
The thrust on the bridge equals N by Newton's third law, so it is \(mg - \frac{mv^2}{R}\). It is less than the weight, which is why we feel lighter at the top of a hump. Option (D) would apply at the bottom of a dip.

Final Answer:
The thrust is \(mg - \frac{mv^2}{R}\), option (C). \[ \boxed{mg - \frac{mv^2}{R}} \]