For particle $P$ revolving round the centre $O$ with radius of circular path $r$ and angular velocity $\omega$, as shown in below figure, the projection of $OP$ on the $x$-axis at time $t$ is $.................$.
$x(t)=r \cos \left(\omega t+\frac{\pi}{6}\right)$
$x(t)=r \cos (\omega t)$
$x(t)=r \sin \left(\omega t+\frac{\pi}{6}\right)$
$x(t)=r \cos \left(\omega t-\frac{\pi}{6} \omega\right)$
The work done on a particle of mass $m$ by a force, $k\left[\frac{x}{\left(x^2+y^2\right)^{3 / 2}} \hat{i}+\frac{y}{\left(x^2+y^2\right)^{3 / 2}} \hat{j}\right]$ ( $K$ being a constant of appropriate dimensions), when the particle is taken from the point $(a, 0)$ to the point $(0, a )$ along a circular path of radius a about the origin in the $x$-y plane is :
A particle describes a horizontal circle of radius $r$ on the smooth surface of an inverted cone as shown. The height of plane of circle above vertex is $h$. The speed of particle should be
A particle starting from rest, moves in a circle of radius $r$. It attains a velocity of $\mathrm{V}_{0} \;\mathrm{m} / \mathrm{s}$ in the $\mathrm{n}^{\text {th }}$ round. Its angular acceleration will be
A particle is moving on a circular path of radius $r$ with uniform velocity $v$. The change in velocity when the particle moves from $P$ to $Q$ is $(\angle POQ = 40^\circ )$
If a force of constant magnitude acts in direction perpendicular to the motion of a particle, then its