$A$ cyclist riding a bicycle at a speed of $14\sqrt{3} \text{ m/s}$ takes a turn around a circular road of radius $20\sqrt{3} \text{ m}$ without skidding. Given $g = 9.8 \text{ m/s}^2$,what is his inclination to the vertical in degrees?

$A$ body of mass $M = 100 \; g$ is rotating in a horizontal circle with an angular frequency of $n = 2/\pi \; \text{rev/sec}$ as shown in the figure. If the radius of the circular path is $R = 1 \; m$, what is the angle $\theta$ made by the string with the vertical? $(g = 10 \; m/s^2)$

Assume a proton is rotating along a circular path of radius $1 \,m$ under a centrifugal force of $4 \times 10^{-12} \,N$. If the mass of the proton is $1.6 \times 10^{-27} \,kg$, then its angular velocity of rotation is

$A$ body of mass $10 \,g$ is tied to a string of length $0.4 \,m$ and it is whirled in a horizontal plane with a speed of $6 \,m/s$. Then the tension in the string is (in $\,N$)

$A$ particle of mass $M$ moves with constant speed along a circular path of radius $r$ under the action of a force $F$. Its speed is

$A$ string of length $L$ is fixed at one end and carries a mass $M$ at the other end. The string makes $\frac{\pi}{2}$ revolutions per second around the vertical axis through the fixed end as shown in the figure. The tension in the string is: