Discussion :: Planar Kinematics of a Rigid Body (PKRB) - General Questions (Q.No.2)
|Arpit Lasod said: (Jan 4, 2011)|
|consider rod AB, and A as I.C.R.(instantenous center of rotation)
r=2m(length of the rod)
consider rod AB, and B as I.C.R.(instantenous center of rotation)
|Rios said: (Jan 31, 2018)|
|Doing a triangle, the displacement is equal to 2 sin(60).
|Jani Dhruvkumar said: (Aug 3, 2019)|
|No, the Correct one is;
Hence both points are on single link AB, therefore, the value of w=wa=wb.
(Ia=√3 and Ib=1 from instantaneous centre method).
|Pankaj said: (Apr 26, 2021)|
|Vb - Va=Vba.
Vba = wrba.
Vb - va = w rba.
Vb = 6i + w*2(-cos60i +sin60j).
Vb is in y direction so we only take y component and it's x component should be zero
Vb = 2wsin60, 6-2wcos60=0.
w = 6/(2cos60).
w = 6,
Vb = 2 * 6 * √3/2.
Vb = 6√3,
Vb = 10.39.
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