A particle at a distance of 1 m from the origin starts moving such that \frac{dr}{d\theta }= r, where (r, θ) are polar coordinates. Then the angle between resultant velocity and tangential velocity component is

Dependent on where the particle is

KVPY SX SB Exam 2016 Question Paper

Question : 29 of 120

Marks: +1, -0

A particle at a distance of 1 m from the origin starts moving such that

{dr}/{dθ }= r

, where (r, θ) are polar coordinates. Then the angle between resultant velocity and tangential velocity component is

Solution:

Given, the particle is at

1 m

from the origin

⇒ r =1

v _{\;\text "tangential "\;}=\;{ d θ}/{ d t }

v _{\;\text "radial "\;}=\;{ dr }/{ dt }

v _{\;\text "resultant "\;}= v _{{\;\text "radial "\;}↖{→}}+ v _{{\;\text "tangential "\;}↖{→}}

Let

φ

be the angle between the resultant velocity and tangential velocity componment
Then

\tan φ=\;{ v _{\;\text "radial "\;}}/{v_{\;\text "tangential "\;}}

⇒ \tan φ=\;{({ \dr }/{ \dt })}/{(\;{ \d θ}/{ \dt })}

⇒ \tan φ=\;{ \dr }/{ \d θ}

Given

\;{ \dr }/{ \d θ}= r

⇒ \tan φ= r =1

⇒ φ=\tan ^{-1}(1)=45^{∘}

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