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# Net3 - Kinetics Newton's Second Law written in abstract vector form for a particle is F = ma in which m is a scalar quantity called mass

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Kinetics Newton’s Second Law written in abstract vector form for a particle is X ~ F = m~a in which m is a scalar quantity called mass. Historically, mass is called inertia or resistance to motion . Euler later uniﬁed the theory for both solid and ﬂuid by writing “Euler’s First Law” as X ~ F = d dt ~ L = d dt ( m~v ) in which ~ L = is the linear momentum of a particle or a continuous medium. Since the above laws are written in abstract vector form, they are applicable in any coordinate systems. Examples Rectlinear Motion (only one component): X F = ma Cartesian Coordinates: X F x F y F z = m a x a y a z = m ¨ x ¨ y ¨ z Cylindrical Coordinates: X F r F θ F z = m a r a θ a z = m ¨ r - r ˙ θ 2 r ¨ θ +2˙ r ˙ θ ¨ z Planar nt Coordinates: X F n F t F z = m a n a t a z = m v 2 ˙ v ¨ z –3/1–

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Example – Cylindrical Coordinates A sphere of mass m is initially attached to the wall at angle θ by a cord. If the cord is cut, the sphere will be allowed to swing as a pendulum. Find as a function of θ the ratio: T 2 T 1 = Tension after cut of cord Tension before cut of cord . T θ cord m ˆ e r ˆ e θ f r θ mg T 1 FBD: before cut T 2 FBD: after cut ˆ e θ ˆ e r θ mg θ Since pendulum problems are best described by polar coordinates, use X ± F r F θ ² = m ± ¨ r - r ˙ θ 2 r ¨ θ +2˙ r ˙ θ ² Before Cut: To obtain the sum of the forces, it is convenient to write the forces in different coordinate systems, for example, X ~ F = ± 0 - mg ² xy + ± - T 1 0 ² + ± f r 0 ² xy = ± f r - mg ² xy + ± - T 1 0 ² . To complete the sum, use the transformation matrix deﬁned as [ Q ( xy )] = ³ cos( π 2 - θ ) cos( π - θ ) cos θ cos( π 2 - θ ) ´ = ³ sin θ - cos θ cos θ sin θ ´ . x y ˆ e r ˆ e θ θ θ Therefore, X ± F r F θ ² = ³ sin θ - cos θ cos θ sin θ ´± f r - mg ² xy + ± - T 1 0 ² = ± mg cos θ + f r sin θ - T 1 - mg sin θ + f r cos θ ² = m ± 0 0 ² –3/2–
The acceleration, ~a , is set to zero because this is a static problem. Now, by equating the orthogonal components, the solution can be obtained as ± f r = mg sin θ/ cos θ = mg tan θθ component T 1 = mg cos θ + f r sin θ = mg cos θ + mg sin 2 cos θr component Simplifying further yields T 1 = mg (cos 2 θ + sin 2 θ ) cos θ = T 1 = mg cos θ After Cut: At the instant the cord is cut, the acceleration is nonzero but the velocity is still zero momentarily. For this reason, we can state that ˙ θ =0 . Also, since the string is unstretchable, it implies that ˙ r r . The equation of motion is therefore simpliﬁed to X ± F r F θ ² = ± 0 - mg ² xy + ± - T 2 0 ² = m ± 0 - r (0) 2 r ¨ θ + 2(0)(0) ² = m ± 0 r ¨ θ ² The summation of forces, using the transformation matrix, yields X ~ F = ³ sin θ - cos θ cos θ sin θ ´± 0 - mg ² xy + ± - T 2 0 ² = ± mg cos θ - T 2 - mg sin θ ² .

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## This note was uploaded on 04/16/2008 for the course CE 325 taught by Professor Docwong during the Spring '08 term at USC.

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Net3 - Kinetics Newton's Second Law written in abstract vector form for a particle is F = ma in which m is a scalar quantity called mass

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