Engineering Mechanics | polar co-ordinates problem

Q. The collar A in Figure slides along the rotating rod OB. The angular position
of the rod is given by 𝜃 =2/3 *(𝜋𝑡)²
rad and the distance of the collar from O varies as r =18t⁴+4 m, where time t is measured in seconds. Determine the velocity and acceleration vectors of the collar at t =0.4 s.

Solution :-

Given polar co-ordinates are
r = 18t⁴+4 and 𝜃 =2/3 ×(𝜋𝑡)²
r`= 72t³ 𝜃` =4/3 ×(𝜋𝑡)
r``= 216t² 𝜃`` =4/3 ×(𝜋)

At t=0.4 sec
r = 18(0.4)⁴+4 = 4.46m
r`= 72(0.4)³ = 4.608 m/a
r``= 216(0.4)² = 34.56 m/s²

𝜃 = 2×(3.14×0.4)²/3 =0.335 rad
𝜃`= 4×3.14×0.4/3 = 1.675 rad/sec
𝜃``= 4×3.14/3 = 4.188 rad/sec²

The velocity vector is given by

V = (r`) Er + (r×𝜃`) E𝜃
= 4.608 Er +4.46×1.675 E𝜃
= 4.608 Er + 7.47 E𝜃
Where Er - radial unit vector
E𝜃 - angular unit vector

Also acceleration vector is given by

A = (r``- r𝜃`²)Er + (2r`𝜃`+r𝜃``) E𝜃
= (34.56 - 4.46×1.675²)Er + [(2×4.608×1.675) + 4.46×4.188] E𝜃
= 22.04 Er +34.11 E𝜃

These are the velocity and acceleration vectors.
Please note that V and A denotes vector quantity.

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