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Velocities at B and C on a rotating rod 
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#1
Nov3012, 12:35 AM

P: 154

1. The problem statement, all variables and given/known data
In the attached image, find angular velocity of the rod and the velocities at points B and C. 2. Relevant equations v_{BA}=0.3ω 3. The attempt at a solution How do I know what angle the absolute velocity makes with the other vectors when put in a triangle? How would the velocity vectors look like at point C in the attached image? It's hard for me since I only have one angle. I could calculate ω, v_{B}, v_{BA}, I'm only left with v_{C} v_{B}= 5.95 m/s v_{BA}=7.778 m/s Now, v_{CA} should be (0.3+0.15)*ω, but what do I do after that? 


#2
Nov3012, 01:21 AM

P: 154

No one?



#3
Nov3012, 03:02 AM

P: 296




#4
Nov3012, 03:13 AM

P: 154

Velocities at B and C on a rotating rod
But that's in the vector form! The answer is V_{C}=9.28 m/s at an anticlockwise angle of 15.61° with the horizontal. WHA?



#5
Nov3012, 10:18 AM

P: 296

You know the direction and magnitude of Va and you know the direction and magnitude of Vca so you can solve for both magnitude and direction of Vc using this one equation. You have two unknowns and this vector equation represents two equations in two dimensions. Graphically you can draw a vector diagram which will end up being a closed triangle with Vc one of the sides. I think we're not sure what difficulty you are having... maybe you are unsure of the direction of Vca? AC is a rigid body which means the two points A and C cannot get closer together or further apart. So any velocity one part of the rigid body has with respect to another must be a rotation, ie directed perpendicular to the line connecting the two points. So Vca must be directed perpendicular to the line joining AC. 


#6
Nov3012, 11:52 AM

P: 154

letters in bold are vectors. r_{CA}=0.45sin(40°)i0.15cos(40°)j v_{A}=5j v_{CA}= ω x r_{CA} v_{C}=v_{A}+v_{CA} I get v_{C}=5j+(2.979i +(edited) 7.943j) which is obviously wrong. I don't get why the v_{C} makes an angle of 15.61° with the horizontal, as given in my book!! 


#7
Nov3012, 12:49 PM

P: 296

Vc will be up from the horizontal because Vb is flat horizontally. Vca is longer than Vba because r is larger at C so the resultant Vc will rise above the horizontal. 


#8
Nov3012, 01:13 PM

P: 154

Can you also explain the geometry that goes into making the angle of v_{C} with the horizontal? EDIT: that angle can easily be found from V_{C}, so nvm... 


#9
Nov3012, 01:56 PM

P: 296

The magnitude of Vc is found from the law of cosines: Vc^{2} = Va^{2} + Vca^{2}  2VaVca cos 50 and then the law of sines gets you alpha: Vc/sin 50 = Vca/sin alpha and θc = alpha  90 I think it's probably equally difficult to do it this way or the cross product way; maybe a bit easier using your method because there are fewer things to think about. The geometry does give some more insight though. All points on the rod AC travel at a velocity Va+Vpa where Va is constant and pointing downward and Vpa is always perpendicular to the rod. This means the one vector velocity diagram shows all velocities at every point of the rod :: any point's velocity is simply the sum of Va and a vector along Vca. The length you travel along Vca to find Vpa depends on the distance r of the point p from a (the length is rw remember). So the tail of the absolute velocity of p is anchored at A and the head sweeps along the vector Vca for each point p. Vb, eg, is the horizontal vector I drew from A to the vector Vca in that diagram. 


#10
Nov3012, 08:52 PM

P: 154




#11
Nov3012, 10:02 PM

P: 296

Had the constraint not been there, you'd have to invoke Netwon's laws (ie it would be a kinetics problem) to find out exactly how the rod moved. But from that diagram with no forces acting (including friction and gravity) and assuming no existing rotational motion, the entire rod would just move downward at 5 m/s. 


#12
Nov3012, 11:23 PM

HW Helper
Thanks
P: 10,780

C moves along a circle of radius r about B. Its velocity is perpendicular to the radius r. You can find that the shaded angles are all equal, so the velocity encloses angle θ with the horizontal. ehild 


#13
Dec112, 01:17 AM

P: 154

thanKs a lot ehild and aralbrec, you guys helped a lot!



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