For a particle in uniform circular motion, the acceleration $\vec a$ at a point $P(R,\theta)$ on the circle of radius $R$ is (Here $\theta$ is measured from the $x-$ axis)
$\frac{{{V^2}}}{R}\widehat i + \frac{{{V^2}}}{R}\widehat j$
$ - \frac{{{V^2}}}{R}\cos \theta \widehat i + \frac{{{V^2}}}{R}\sin \theta \widehat j$
$ - \frac{{{V^2}}}{R}\sin \theta \widehat i + \frac{{{V^2}}}{R}\cos \theta \widehat j$
$ - \frac{{{V^2}}}{R}\cos \theta \widehat i - \frac{{{V^2}}}{R}\sin \theta \widehat j$
If a body moving in circular path maintains constant speed of $10\,ms^{-1},$ then which of the following correctly describes relation between acceleration and radius ?
particle is moving in a circular path. The acceleration and momentum of the particle at a certain moment are $\vec a\, = \,(4\hat i + 3\hat j)\,\,m/{s^2}$ and $\vec P\, = \,(8\hat i\, - \,6\hat j)\,kg\, - \,m/s$ . The motion of the particle is
The driver of a car travelling at velocity $v$ suddenly see a broad wall in front of him at a distance $d$. He should
A particle of mass $m$ describes a circle of radius $r$. The centripetal acceleration of the particle is $4/r^2$. What will be the momentum of the particle?
What is uniform circular motion ? By using proper figure, obtain equation of acceleration ${a_c}\, = \,\frac{{{v^2}}}{r}$ for uniform circular motion. Show that its direction is towards centre.