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A particle is moving along the circle $x^2 + y^2 = a^2$ in anti clock wise direction. The $x-y$ plane is a rough horizontal stationary surface. At the point $(a\, cos\theta , a\, sin\theta )$, the unit vector in the direction of friction on the particle is:
$\cos \theta \,\hat i + \sin \theta \,\hat j$
$ - \left( {\cos \theta \,\hat i + \sin \theta \,\hat j} \right)$
$\sin \theta \,\hat i - \cos \theta \,\hat j$
$\cos \theta \,\hat i - \sin \theta \,\hat j$
Solution
Unit vector of point $P$ from origin is $:$
$\vec{P}=\cos \theta \hat{\imath}+\sin \theta \hat{j}$
If $\hat{\vec{F}}$ be the unit vector acting along friction. then,
$\hat{\vec{F}} \cdot \hat{\vec{P}}=0$
since friction is clockwise sense component along $\hat{i}$ direction $+ve.$
$\hat{F}=\sin \theta \hat{i}-\cos \theta \hat{j}$
