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A particle moves in the XY plane under the actionof force F vector such that the value of its linear moment P vector at any time t is P vector = 2costi^+2sintj^. Find the angle between F vector And P vector at = 4 sec. |
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Answer» n vector r = x i + y j in the x y plane. force vector F = m a = m dv/dt = d P / d t momentum vector P = m V P = 2 COS t i + 2 Sin t j F = - 2 sin t i + 2 Cos t jAngle of vector P with x AXIS: Tan Ф1 = 2 Sin t / (2 Cos t) = Tan tAngle of vector F with x axis : Tan Ф2 = 2 Cos t / (- 2 sin t) = - Cot t = - tan (π/2 - t) = tan [π - (π/2 - t) ] = tan (π/2 + t) or tan (3π/2 - t)Angle between P and F = Ф2 - Ф1Tan (Ф₂ - Ф₁) = - [tan t + Cot t] / [ 1 - tan t * Cot t ] = infinitySo angle Ф₂ - Ф₁ = π/2The momentum and force vectors are perpendicular to each other.This is the case of UNIFORM circular motion. F is the centripetal force. |
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