A particle at a distance of 1 m from the orig | vedclass

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Question

$(b)$ Velocity $v$ of a particle moving around a curved path can be resolved into $\operatorname{radial}\left(v_{r}=\frac{d r}{ d t}\right)$ and tange

Vedclass · Accepted Answer

$45^{\circ}$

Vedclass · Answer

$(b)$ Velocity $v$ of a particle moving around a curved path can be resolved into $\operatorname{radial}\left(v_{r}=\frac{d r}{ d t}ight)$ and tangential (or transverse, $\left.v_{	heta}=r \frac{d 	heta}{d t}ight)$ velocity components.

Angle $\alpha$ between velocity $v$ and tangential velocity $v _{r}$ is given by

$	an \alpha=\frac{v_{	heta}}{v_{r}}$

$\Rightarrow \quad 	an \alpha=\frac{\left(r \frac{d 	heta}{d t}ight)}{\left(\frac{d r}{d t}ight)}=r \cdot\left(\frac{d 	heta}{d r}ight)$

$=\frac{r}{(d r / d 	heta)}$

Here, given $\frac{d r}{d 	heta}=r$

$	herefore 	an \alpha=\frac{r}{r}=1$

$\Rightarrow \alpha=	an ^{-1}(1)=45^{\circ}$

A particle at a distance of $1 m$ from the origin starts moving, such that $d r / d \theta=r$, where $r$ and $\theta$ are polar co-ordinates. Then, the angle between resultant velocity and tangential velocity is

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