Velocity of a slider connect to a wheel

[Jonski]

Homework Statement

A wheel of radius 42 mm is rolling without slipping and its centre O has a velocity 7.2 m/s in the direction shown.

The slider B is driven by the link AB and A is connected to the wheel.

If AC= 28.9 mm, AB= 73.9 mm and angle ACB= 73°, what is the velocity of the slider B? Take the right-hand side direction positive.

Homework Equations

va = vb + va/b
velocity at edge of circle in line with O = sqrt(2)*v
Vc = 0

The Attempt at a Solution

[/B]
My initial thoughts are that B first has to move to the left as A is going to pull it rather than push. (So the answer will be negative)
I first constructed the triangle ABC, but without angle CAB or side CB, I don't know how to complete it.Any help would be greatly appreciated, Thanks.

 

[Charles Link]

Suggestion is to try to determine the velocity vector for the point A. Do you see that the point A is going to be instantaneously moving in a circle centered at point C? Once you figure out how point A moves, go to point B and try relating the coordinates of B (to first order in $\Delta t$ ) to the coordinates of A.

 

[Jonski]

Would A be moving straight down as that would be perpendicular to OA?

 

[Charles Link]

Would A be moving straight down as that would be perpendicular to OA?

The pivot point is at C and the radius is CA. The direction of $v_A$ will be perpendicular to CA.

 

[Charles Link]

The pivot point is at C and the radius is CA. The direction of $v_A$ will be perpendicular to CA.

Just an an additional comment is you could also treat the problem as the center at O translating at constant speed with the wheel rotating about O at a rate that will make it roll without slipping. Considering it as pivoting about point C is a shortcut that should get you the same answer.

 

[Jonski]

Just an an additional comment is you could also treat the problem as the center at O translating at constant speed with the wheel rotating about O at a rate that will make it roll without slipping. Considering it as pivoting about point C is a shortcut that should get you the same answer.

So could I say that the instantaneous centre of OA is at C:
7.2/0.042 = 171.429 rad/s = Woa
Va = 171.429*0.0289 = 4.954 m/s
Then we can construct a right angle trigangle of Va, Vb and Va/b with angle 73 degrees. Then for Vb it would be Va/sin(73) = 5.18m/s => -5.18m/s (right is positive)
However, this is still wrong any tips on what I did incorrect.

 

[Charles Link]

First I recommend you express $v_A$ as a vector with x and y components. Use C as your origin. Find the location of $A$ coordinates (x,y) at time t=0. Then at time $\Delta t$ , $A'=A_0 +v_A \Delta t$. Also write location $B$ (at t=0) in terms of $A$, and then $B'$ (at $t=\Delta t$ ) in terms of $A'$. (Both of these steps are simple=uses Pythagorean theorem.) First get the exact expression for $B'$ and then expand and simply keep first order term in $\Delta t$. Finally, $B'-B=v_B \Delta t$. $\\$ One helpful hint for the direction of $v_A$ : CA is perpendicular to $v_A$ and points in direction (unit vector) $\hat{t}=\cos(\theta)\ \hat{i} +\sin(\theta) \hat{j}$. The perpendicular to this is (unit vector) $\hat{n}=\sin(\theta) \hat{i}-\cos(\theta) \hat{j}$ which will be your direction for $v_A$. (Notice that $\hat{t} \cdot \hat{n}=0$ ). ... editing... I agree with your result for the amplitude of $v_A=4.954 \, m/sec$, but unless the direction of $v_A$ is horizontal, the computation for $v_B$ is non-trivial. The method I presented above for finding $v_B$ should work... editing... I did get an answer that I think is correct and it (edited=first time I didn't compute it precisely) just slightly greater than +4.954 m/s and in the "+" direction=to the right.

 

[Charles Link]

First I recommend you express $v_A$ as a vector with x and y components. Use C as your origin. Find the location of $A$ coordinates (x,y) at time t=0. Then at time $\Delta t$ , $A'=A_0 +v_A \Delta t$. Also write location $B$ (at t=0) in terms of $A$, and then $B'$ (at $t=\Delta t$ ) in terms of $A'$. (Both of these steps are simple=uses Pythagorean theorem.) First get the exact expression for $B'$ and then expand and simply keep first order term in $\Delta t$. Finally, $B'-B=v_B \Delta t$. $\\$ One helpful hint for the direction of $v_A$ : CA is perpendicular to $v_A$ and points in direction (unit vector) $\hat{t}=\cos(\theta)\ \hat{i} +\sin(\theta) \hat{j}$. The perpendicular to this is (unit vector) $\hat{n}=\sin(\theta) \hat{i}-\cos(\theta) \hat{j}$ which will be your direction for $v_A$. (Notice that $\hat{t} \cdot \hat{n}=0$ ). ... editing... I agree with your result for the amplitude of $v_A=4.954 \, m/sec$, but unless the direction of $v_A$ is horizontal, the computation for $v_B$ is non-trivial. The method I presented above for finding $v_B$ should work... editing... I did get an answer that I think is correct and it (edited=first time I didn't compute it precisely) just slightly greater than +4.954 m/s and in the "+" direction=to the right.

Additional comment=your answer $v_B=+5.18 \, m/sec$ is close to the numerical answer I got, but your method of computing it from the $v_A$ I believe is incorrect.