Posted by Rachel on Friday, May 25, 2012 at 2:56pm.
frequency f = 2 rev/s
angular velocity = 2•π•f = 4•π rad/s
(a) Velocity vector at P = (1.25x4x3.14)j = 15.7 j,
where j is unit vector along positive y axis.
Also velocity vector of the particle at R = -15.7i,
where i is unit vector along positive x direction.
As the direction changes uniformly,
the average velocity over the interval PR is
(15.7/2)(j-i)= 7.85(j-i).
Magnitude = sqrt[7.85^2+7.85^2] =
=1.414x7.85 = 11.10 m/s
and the direction is 45 degrees west of north.
(b) Average velocity, V1 over interval PQ is
= 0.5{15.7 j + [15.7/sqrt(2)](j-i)} or
V1 = 0.5[26.80 j - 11.10i] = 13.40 j - 5.55 i.
Magnitude of V1 = sqrt[13.40^2 + 5.55^2] = 14.50 m/s.
due theta west of north given by theta = tan^-1 [5.55/13.40] = 22.5 degrees west of north.
(c) v= 2•π•f•R = 15.7 m/s due north
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