# Velocities at B and C on a rotating rod

by MrWarlock616
Tags: rotating, velocities
 P: 152 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 vB|A=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, vB, vB|A, I'm only left with vC vB= 5.95 m/s vB|A=7.778 m/s Now, vC|A should be (0.3+0.15)*ω, but what do I do after that? Attached Thumbnails
 P: 152 No one?
P: 296
 Quote by MrWarlock616 Now, vC|A should be (0.3+0.15)*ω, but what do I do after that?
vC|A = Vc - Va by definition.

P: 152

## Velocities at B and C on a rotating rod

But that's in the vector form! The answer is VC=9.28 m/s at an anticlockwise angle of 15.61° with the horizontal. WHA?
P: 296
 Quote by MrWarlock616 But that's in the vector form! The answer is VC=9.28 m/s at an anticlockwise angle of 15.61° with the horizontal. WHA?
Vc|a = Vc - Va

You know the direction and magnitude of Va and you know the direction and magnitude of Vc|a 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 Vc|a? 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 Vc|a must be directed perpendicular to the line joining AC.
P: 152
 Quote by aralbrec Vc|a = Vc - Va You know the direction and magnitude of Va and you know the direction and magnitude of Vc|a 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 Vc|a? 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 Vc|a must be directed perpendicular to the line joining AC.
Please check if I have this right:
letters in bold are vectors.

rC|A=0.45sin(40°)i-0.15cos(40°)j

vA=-5j

vC|A= ω x rC|A

vC=vA+vC|A

I get vC=-5j+(2.979i +(edited) 7.943j
) which is obviously wrong.
I don't get why the vC makes an angle of 15.61° with the horizontal, as given in my book!!
P: 296
 Quote by MrWarlock616 rC|A=0.45sin(40°)i-0.15cos(40°)j
That should be 0.45 on the cos as well, maybe that is it.

 I don't get why the vC makes an angle of 15.61° with the horizontal, as given in my book!!
I've attached a diagram showing a graphical solution.

Vc will be up from the horizontal because Vb is flat horizontally. Vc|a is longer than Vb|a because r is larger at C so the resultant Vc will rise above the horizontal.
Attached Thumbnails

P: 152
 Quote by aralbrec That should be 0.45 on the cos as well, maybe that is it.
Oh my god, unbelievable. Thank you so much!!

Can you also explain the geometry that goes into making the angle of vC with the horizontal?
EDIT: that angle can easily be found from VC, so nvm...
P: 296
 Quote by MrWarlock616 EDIT: that angle can easily be found from VC, so nvm...
Yes, just in case:

The magnitude of Vc is found from the law of cosines:

Vc2 = Va2 + Vc|a2 - 2VaVc|a cos 50

and then the law of sines gets you alpha:

Vc/sin 50 = Vc|a/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+Vp|a where Va is constant and pointing downward and Vp|a 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 Vc|a. The length you travel along Vc|a to find Vp|a 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 Vc|a for each point p. Vb, eg, is the horizontal vector I drew from A to the vector Vc|a in that diagram.
P: 152
 Quote by aralbrec Yes, just in case: The geometry does give some more insight though. All points on the rod AC travel at a velocity Va+Vp|a where Va is constant and pointing downward and Vp|a 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 Vc|a. The length you travel along Vc|a to find Vp|a 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 Vc|a for each point p. Vb, eg, is the horizontal vector I drew from A to the vector Vc|a in that diagram.
YEah! And had the channel not been there for the second block, the vb WOULDN'T be horizontal, right? So, it's just luck that we knew it's direction. This is exactly why I was confused at point c because I didn't know what was the angle made by vc, so I wasn't able to figure out its components!!
P: 296
 Quote by MrWarlock616 YEah! And had the channel not been there for the second block, the vb WOULDN'T be horizontal, right? So, it's just luck that we knew it's direction. This is exactly why I was confused at point c because I didn't know what was the angle made by vc, so I wasn't able to figure out its components!!
These things are machines so the movements are normally constrained like that on purpose so that the machines will perform motions that are predictable and useful.

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.
HW Helper
Thanks
P: 9,255
 Quote by MrWarlock616 YEah! And had the channel not been there for the second block, the vb WOULDN'T be horizontal, right? So, it's just luck that we knew it's direction. This is exactly why I was confused at point c because I didn't know what was the angle made by vc, so I wasn't able to figure out its components!!
The two channels were at right angles: it was not stated, but rather obvious from the picture. You can choose the most convenient coordinate system: here you say that A moves vertically, along the y axis and B moves horizontally, along the x axis. The rod connects them, the length of the rod is fixed so xb2+ya2=L2, and you get the relation between velocities by implicit derivation: xava +ybvb=0, vb=(-ya/xb)vb

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
Attached Thumbnails

 P: 152 thanKs a lot ehild and aralbrec, you guys helped a lot!

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