If the angular velocity of a merry-go-round is $60^o/s$ and you are $3.5\,m$ from the centre of rotation,your linear velocity will be

Two objects of masses $m_1$ and $m_2$ are moving in circles of radii $r_1$ and $r_2$ respectively. Their respective angular speeds $\omega_1$ and $\omega_2$ are such that they both complete one revolution in the same time $t$. The ratio of the linear speed of $m_2$ to that of $m_1$ is:

What is the angular velocity of the second hand of a clock?

$A$ particle moving with uniform speed in a circular path maintains:

$A$ particle is revolving in an anticlockwise sense along the circumference of a circle of radius $r$ with linear velocity $v$. The angle between the linear velocity $v$ and the angular velocity $\omega$ is: (in $^{\circ}$)

Two bodies of equal masses revolve in circular orbits of radii $R_1$ and $R_2$ with the same period. Their centripetal forces are in the ratio: