$A$ person with his hand in his pocket is skating on ice at the rate of $10 \ m/s$ and describes a circle of radius $50 \ m$. What is his inclination to the vertical? (Take $g = 10 \ m/s^2$)

$A$ car is moving with velocity $v$ at the top of a semi-circular hill of radius $40 \,m$ such that the normal force on it is zero. Find the velocity $(v)$ of the car. [Use $g=10 \,ms^{-2}$] (in $\,ms^{-1}$)

If the radii of circular paths of two particles of same mass are in the ratio of $1: 2$, then to have a constant centripetal force, the ratio of their speeds should be

$A$ car is moving on an overbridge of radius $R$ with a constant speed $v$. As the car is descending on the overbridge from point $B$ to $C$,the normal reaction on it due to the bridge:

Two particles of the same mass are moving in circular orbits because of a force given by $F(r) = -\frac{16}{r} - r^3$. The first particle is at a distance $r = 1$,and the second is at $r = 4$. The best estimate for the ratio of kinetic energies of the first and the second particle is closest to:

$A$ mass of $0.5 \ kg$ is attached to a string moving in a horizontal circle with an angular velocity of $10 \ cycle/min$. Keeping the radius constant,the tension in the string is made $4$ times by increasing the angular velocity to $\omega$. The value of $\omega$ for that mass will be: