$A$ motorcyclist going around a circular track at a constant speed has:

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:

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

$A$ scooter is going round a circular road of radius $100 \,m$ at a speed of $10 \,m/s$. The angular speed of the scooter will be ......... $rad/s$.

As shown in the figure,a particle is moving with a constant speed $v = \pi \, m/s$ in a circular path. Considering its motion from point $A$ to point $B$,where the angle subtended at the center is $120^{\circ}$,the magnitude of the average velocity is:

If $\omega_1$ is the angular velocity of the hour hand of a clock and $\omega_2$ is the angular velocity of the Earth,then the ratio $\omega_1 : \omega_2$ is