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A particle moves on a circular path of radius x of centre O(0,0) with constant speed v=20m/s from a point A(0,-x) to a point B such that OB makes an angle of 30° with x-axis. Calculate the change in velocity from A to B( Please illustrate it using Parallelogram law of vector addition)

A particle moves on a circular path of radius x of centre O(0,0) with constant speed v=20m/s from a point A(0,-x) to a point B such that OB makes an angle of 30° with x-axis. Calculate the change in velocity from A to B( Please illustrate it using Parallelogram law of vector addition)

Grade:12th pass

1 Answers

Yuvraj
21 Points
9 years ago
The question has not specified whether the motion is clockwise or anticlockwise. Well, for clockwise the answer would be as follows. Intially, when particle is at (0,-x) the velocity vector is perpendicular to x axis and along the positive y-axis. So v1=0 i+v j When the radius vector makes 30 degree angle with x axis, the velocity vector makes 150 degrees angle with the initial vector. SInce vectors are free you can write the final vector as: v2=v(cos(-60)i+sin(-60)j) (Since angle made is -60 with the x axis. So v2=0.5v i - 0.8666v j So delta v=v2-v1 =0.5v i -1.8666v j If u use parallelogram law, then delta v =v2+(-v1) -v1 makes an angle 30 degrees with v2 so delta v =v2v²+2v²cos(30) =v(v(2+v3))

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