mecanica-para-ingenieria-dinamica-anthony-bedford-5-edicion-solucionario

Problem 13.182 The constant velocity v = 2 m/s. ay
What is the acceleration of point P in terms of polar
coordinates when x = 0.25 m? aθ

Solution: See the solution of Problem 13.192. The polar angle θ is ar

θ = arctan y = arctan 0.141 = 29.5◦. Then θ
x 0.25 ax

ar = ax cos θ + ay sin θ = 0 + (−5.58) sin 29.5◦ = −2.75 m/s2,

aθ = −ax sin θ + ay cos θ = 0 + (−5.58) cos 29.5◦ = −4.86 m/s2.

Problem 13.183 A point P moves along the spiral P

path r = (0.1)θ m, where θ is in radians. The angular r
position θ = 2t rad, where t is in seconds, and r = 0 θ
at t = 0. Determine the magnitudes of the velocity and
acceleration of P at t = 1 s.

Solution: The path r = 0.2t m, θ = 2t rad. The velocity compo-

nents are

dr dθ
vr = dt = 0.2 m/s, vθ = r dt = (0.2t)2 = 0.4t.
At t = 1 seconds the magnitude of the velocity is

√
|v| = vr2 + vθ2 = 0.22 + 0.42 = 0.447 m/s

The acceleration components are:

ar = d2r −r dθ 2
dt2 dt
= −(0.2t)(22) m/s2,

aθ = r d2θ +2 dr dθ = 2(0.2)(2) = 0.8 m/s2.
dt2 dt dt

The magnitude of the acceleration is |a| = ar2 + aθ2 = 1.13 m/s2

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