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: Free body view:
Take the downward direction (toward the centre) as positive. Net downward force = $mg - N$.

Step 2: Newton's second law:
$mg - N = ma_c = \frac{mv^2}{R}$, so $N = mg - \frac{mv^2}{R}$ (C). If the speed reached $\sqrt{gR}$, N would drop to zero and the car would leave the road.

Final Answer:
$mg - \frac{mv^2}{R}$. \[ \boxed{mg - \frac{mv^2}{R}} \]