A) parallel to position vector
B) perpendicular to position vector
C) directed towards the origin
D) directed away from the origin
Correct Answer: B
Solution :
Position vector, \[r=(a\cos \omega t)\widehat{\mathbf{i}}+(a\sin \omega t)\widehat{\mathbf{j}}\] velocity\[v=\frac{dr}{dt}\] \[=\frac{d}{dt}[(a\cos \omega t)\widehat{\mathbf{i}}+(a\sin \omega t)\widehat{\mathbf{j}}]\] \[=(-a\sin \omega t)\widehat{\mathbf{i}}+(a\cos \omega t)\widehat{\mathbf{j}}\] \[v\cdot r=[(-a\sin \omega t)\widehat{\mathbf{i}}+(a\cos \omega t)\widehat{\mathbf{j}}]\] \[[(a\cos \omega t)\widehat{\mathbf{i}}+(a\sin \omega t)\widehat{\mathbf{j}}]\] \[v\cdot r=0\] \[i.e.,\]velocity vector is perpendicular to position vector.
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