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Rodmechanism relative movements problem

  • Thread starter wouterbeke
  • Start date
1. Homework Statement
g73rnWh.png

The rods are hengedly connected. O is fixed, C is free to move in the x-direction and AB is parallel to OC. OA turns around O with an angular velocity of ##\omega##.

further given: $$\theta = 30^{\circ}$$ $$|OA|=|BC|=l_1=0.30m$$ $$|AB|=l_2=0.40m$$ $$\omega=15rad/s$$ $$v_c = 6 m/s$$
Question: Find ##v_{B}##.

2. Homework Equations
$$v_{B,\text{abs}} = v_{B,\text{sleep}} + v_{B,\text{rel}}$$ $$v_{B,\text{sleep}} = v_{0'} + \omega '\cdot |OB| = v_A = \omega\cdot l_1$$

3. The Attempt at a Solution

I define a non-rotating coordinate system o'x'y' that is fixed in point A.
k5eCsg7.png


$$v_{B,\text{abs,x}} = -v_{B,\text{tan}}\cos{\beta} + v_c = v_{B,\text{sleep}}\cos{\beta}$$ $$v_{B,\text{abs,y}} = v_{B,\text{tan}}\sin{\beta} = -v_{B,\text{rel}} - v_{B,\text{sleep}}\sin{\beta}$$ $$v_{B,\text{sleep}}=\frac{-\omega l_1\cos{\beta}+v_c}{\cos{\beta}}=7.5m/s$$ I think this is incorrect. $$v_{B,\text{rel}}=-v_{B,\text{tan}}\sin{\beta}-v_{B,\text{sleep}}\sin{\beta}$$ This can't be correct, because this is a negative number and I know from the solution that ##\omega_{BC} = \omega_{OA}## so, I believe, the velocity of B relative to o'x'y' should work out to zero.

What am I doing wrong?

(extra) written version: http://i.imgur.com/WAeCf8L.jpg
 
Last edited:

TSny

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Hello, wouterbeke. Welcome to the forum!

It appears to me that your ##\vec{v}_{B,tan}## is the velocity of B relative to C. So, ##\vec{v}_{B,tan}## is perpendicular to rod BC (rather than "tangent").

Shouldn't the y-component of ##\vec{v}_{B,tan}## be negative? See your drawing of the vectors at B. You have written the y-component as positive.
 

haruspex

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The rods are hengedly connected. O is fixed, C is free to move in the x-direction and AB is parallel to OC. OA turns around O with an angular velocity of ##\omega##.

further given: $$\theta = 30^{\circ}$$ $$|OA|=|BC|=l_1=0.30m$$ $$|AB|=l_2=0.40m$$ $$\omega=15rad/s$$ $$v_c = 6 m/s$$
Question: Find ##v_{B}##.
I must be misunderstanding something - it seems to be overspecified (and contradictory).
If AB is held parallel to OC then the length of AB is irrelevant for most purposes. A moves at speed 15*0.3 = 4.5m/s. Its horizontal component should be half that, and C's speed should be double that, so also 4.5m/s. Should it say OA=BC=0.40m?
 

TSny

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I must be misunderstanding something - it seems to be overspecified (and contradictory).
If AB is held parallel to OC then the length of AB is irrelevant for most purposes. A moves at speed 15*0.3 = 4.5m/s. Its horizontal component should be half that, and C's speed should be double that, so also 4.5m/s.
I agree that the horizontal component of velocity of A is 2.25 m/s, but I don't see how to get that C's speed should be 4.5 m/s. The horizontal rod AB is rotating.

I get an answer that I believe is consistent with the given data, but I don't get that rods OA and BC have the same angular speed as stated by the OP.
 
Last edited:

haruspex

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The horizontal rod AB is rotating.
OK, I see - it's only transiently parallel to OC. Thought I must have had something wrong.
 

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