Q. The collar A in Figure slides along the rotating rod OB. The angular position
of the rod is given by ๐ =2/3 *(๐๐ก)ยฒ
rad and the distance of the collar from O varies as r =18tโด+4 m, where time t is measured in seconds. Determine the velocity and acceleration vectors of the collar at t =0.4 s.
Solution :-of the rod is given by ๐ =2/3 *(๐๐ก)ยฒ
rad and the distance of the collar from O varies as r =18tโด+4 m, where time t is measured in seconds. Determine the velocity and acceleration vectors of the collar at t =0.4 s.
Given polar co-ordinates are
r = 18tโด+4 and ๐ =2/3 ร(๐๐ก)ยฒ
r`= 72tยณ ๐` =4/3 ร(๐๐ก)
r``= 216tยฒ ๐`` =4/3 ร(๐)
At t=0.4 sec
r = 18(0.4)โด+4 = 4.46m
r`= 72(0.4)ยณ = 4.608 m/a
r``= 216(0.4)ยฒ = 34.56 m/sยฒ
๐ = 2ร(3.14ร0.4)ยฒ/3 =0.335 rad
๐`= 4ร3.14ร0.4/3 = 1.675 rad/sec
๐``= 4ร3.14/3 = 4.188 rad/secยฒ
The velocity vector is given by
V = (r`) Er + (rร๐`) E๐
= 4.608 Er +4.46ร1.675 E๐
= 4.608 Er + 7.47 E๐
Where Er - radial unit vector
E๐ - angular unit vector
Also acceleration vector is given by
A = (r``- r๐`ยฒ)Er + (2r`๐`+r๐``) E๐
= (34.56 - 4.46ร1.675ยฒ)Er + [(2ร4.608ร1.675) + 4.46ร4.188] E๐
= 22.04 Er +34.11 E๐
These are the velocity and acceleration vectors.
Please note that V and A denotes vector quantity.
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