The relative angular speed of the hour hand and the second hand of a clock is (in $rad/s$):

$A$ wheel has a diameter of $1 \ m$. If it makes $30$ revolutions per second,then the linear speed of a point on its circumference will be:

$A$ wheel of radius $0.20 \ m$ starts from rest and rotates with an angular acceleration of $1 \ rad/s^2$. What will be the centripetal acceleration of a point on its circumference when it has rotated through an angle of $90^o$?

$A$ particle moves along a circle of radius $\left( \frac{20}{\pi} \right) \, m$ with constant tangential acceleration. If the velocity of the particle is $80 \, m/s$ at the end of the second revolution after motion has begun,the tangential acceleration is

The length of the second's hand in a watch is $1 \, cm$. The change in velocity of its tip in $15 \, seconds$ is:

The velocity $\vec{v}$ of a particle of mass $m$ acted upon by a constant force is given by $\vec{v}(t) = A[\cos(kt) \hat{i} - \sin(kt) \hat{j}]$. Then the angle between the force and the velocity of the particle is (Here $A$ and $k$ are constants). (in $^{\circ}$)