Question

A particle P moves along a path given by r = f (θ), which is symmetric around the line θ = 0. When the particle passes through the position θ = 0, where the radius of curvature is ρ , the velocity of P is v. Obtain an expression for in terms of v, r, and ρ for the motion of the particle at this point.

Answer 1

Explanation:

So, given the Constant Angular Velocity at some time "t" , we have the Expression for the Velocity Vector of the Particle as :

\vec{v} = ( - \omega r \sin \theta) \hat{i} + ( \omega r \cos \theta) \hat{j}

v

=(−ωrsinθ)

i

^

+(ωrcosθ)

j

^

Answer 2

Answer:

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Explanation:

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