Engineering Mechanics - Planar Kinematics of a Rigid Body (PKRB) - Discussion

Discussion :: Planar Kinematics of a Rigid Body (PKRB) - General Questions (Q.No.2)

2. 

The 2-m-long bar is confined to move in the horizontal and vertical slots A and B. If the velocity of the slider block at A is 6 m/s, determine the bar's angular velocity and the velocity of block B at the instant = 60°.

[A]. AB = 3.46 rad/s , vB = 3.46 m/s 9
[B]. AB = 3.00 rad/s , vB = 3.00 m/s 9
[C]. AB = 3.00 rad/s , vB = 6.00 m/s 9
[D]. AB = 6.00 rad/s , vB = 10.39 m/s 9

Answer: Option A

Explanation:

No answer description available for this question.

Arpit Lasod said: (Jan 4, 2011)  
consider rod AB, and A as I.C.R.(instantenous center of rotation)

Va=6m/s
r=2m(length of the rod)
Va=rw
w=3rad/s

consider rod AB, and B as I.C.R.(instantenous center of rotation)

r=2m
w=3rad/s

Vb=rw
Vb=6m/s

Rios said: (Jan 31, 2018)  
Doing a triangle, the displacement is equal to 2 sin(60).
VA= wr.
6= w(2sin(60)).
w=3.46.

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.
Now,
W=(Va/Ia)=(Vb/Ib).
(Va=6m/s).
(Ia=√3 and Ib=1 from instantaneous centre method).

Therefore wb=wa=w=3.46.

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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