A particle moves so that its position vector | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

$\begin{array}{l} \,\,\,\,\,\,\,\,\,\,\,\,\,Give,\,\vec r = \cos \omega t\,\hat x + \sin \,\omega t\,\hat y\ 	herefore \,\,\,\,\vec v = \frac{{d\vec

Vedclass · Accepted Answer

Velocity is perpendicular to $\overrightarrow {\;r} \;$ and acceleration is directed towards the origin.

Vedclass · Answer

$\begin{array}{l} \,\,\,\,\,\,\,\,\,\,\,\,\,Give,\,\vec r = \cos \omega t\,\hat x + \sin \,\omega t\,\hat y\ 	herefore \,\,\,\,\vec v = \frac{{d\vec r}}{{dt}} =  - \omega \,\sin \,\omega t\,\hat x + \omega \,\cos \omega t\,\hat y\ \,\,\,\,\,\,\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyleightharpoonup$}}  \over a}  = \frac{{d\vec v}}{{dt}} =  - {\omega ^2}\,\cos \,\omega t\,\hat x - {\omega ^2}\,\sin \,\omega t\,\hat y =  - {\omega ^2}\vec r\ {m{Since}}\,position\,vector\,\left( {\bar r} ight)\,is\,directed\,away\ from\,the\,origin,\,so,\,acceleration\,\left( { - {\omega ^2}\bar r} ight)\ is\,directed\,towards\,the\,origin.\ Also,\ \vec r \cdot \vec v = \left( {\cos \omega t\,\hat x + \sin \,\omega t\,\hat y} ight) \cdot \left( { - \omega \sin \omega t\,\hat x + \omega \cos \omega t\,\hat y} ight)\  =  - \omega \sin \omega t\cos \omega t + \omega \sin \omega t\cos \omega t = 0\ \,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow \bar r\, \bot \bar v \end{array}$

A particle moves so that its position vector is given by $\overrightarrow {\;r} = cos\omega t\,\hat x + sin\omega t\,\hat y$ , where $\omega$ is a constant. Which of the following is true?

Similar Questions

A stone tied to $180 cm$ long string at its end is making 28 revolutions in horizontal circle in every minute. The magnitude of acceleration of stone is $\frac{1936}{ x }\,ms ^{-2}$. The value of $x.........\left(\text { Take } \pi=\frac{22}{7}\right)$

A body is revolving with a uniform speed $v$ in a circle of radius $r$. The tangential acceleration is

A particle $P$ is moving in a circle of radius $'a'$ with a uniform speed $v$. $C$ is the centre of the circle and $AB$ is a diameter. When passing through $B$ the angular velocity of $P$ about $A$ and $C$ are in the ratio

What happens to the centripetal acceleration of a revolving body if you double the orbital speed $v$ and half the angular velocity $\omega $

If the equation for the displacement of a particle moving on a circular path is given by $(\theta) = 2t^3 + 0.5$, where $\theta$ is in radians and $t$ in seconds, then the angular velocity of the particle after $2\, sec$ from its start is ......... $rad/sec$