Q.Semi-vertical angle of the conical section of a funnel is 37°. There is a small ball kept inside the funnel. On rotating the funnel, the maximum speed that the ball can have in order to remain in the funnel is 2 m/s. Calculate inner radius of the brim of the funnel. Is there any limit upon the frequency of rotation? How much is it? Is it lower or upper limit? Give a logical reasoning. (Use g = 10 m/s² and sin 37° = 0.6)

For a ball on the funnel's smooth inner wall, $\tan\theta = \dfrac{rg}{v^2}$, so the brim (largest r) corresponds to the maximum speed. …

A conical funnel of semi-vertical angle $37^\circ$ holds a small ball that can rotate along the inside surface of the funnel; the MAXIMUM speed the ball can have while still remaining inside the funnel is given as 2 m/s (use g = 10 m/s^2, $\sin37^\circ=0.6$). Since a rotating ball inside a funnel behaves exactly like a conical pendulum with the normal reaction N playing the role of the string tension, $\tan\theta=v^2/(rg)$ gives the inner radius r at the brim (where the maximum speed is reached) as $r=\frac{v_{max}^2}{g\tan\theta}=\frac{4}{10\times0.75}\approx0.3$ m. The example's discussion then notes that as the ball approaches the (narrower) base of the funnel its linear speed decreases but its angular speed/frequenc …

Think of the wall-normal as the conical-pendulum tension: the ball's steady circle needs $n=\dfrac{1}{2\pi}\sqrt{\dfrac{g}{r}\tan\theta}$ — since $n\propto1/\sqrt{r}$, smaller circles spin FASTER, so the large …

Calling the frequency limit an upper limit: as the ball nears the funnel's narrow base, $r\to0$ and $n$ grows without bound, so no upper …