# Velocity of a slider connect to a wheel

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1. Sep 14, 2016

### Jonski

1. The problem statement, all variables and given/known data

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.

2. Relevant equations
va = vb + va/b
velocity at edge of circle in line with O = sqrt(2)*v
Vc = 0
3. The attempt at a solution

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.

2. Sep 15, 2016

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.

3. Sep 15, 2016

### Jonski

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

4. Sep 15, 2016

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

5. Sep 15, 2016

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.

6. Sep 16, 2016

### Jonski

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.

7. Sep 16, 2016

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.

Last edited: Sep 16, 2016
8. Sep 17, 2016

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.