How to determine the linear and angular velocity of one end of a rod that is pinned on a track on each side?

Question

I am working on trying to determine the velocity of one end of a rod that is pinned on both ends into two different tracks. One track is linear and the pinned end has known values. The other side is pinned to a circular track in which no values are known.

I have attempted to obtain the angular velocity first as I am using $$ V_B=V_A+V_{B/A} $$

by using $$ \omega = v/r $$

I have also noted to obtain $ V_{B/A} $ I need to find $ \dot\theta $ which equals $ \omega $

$ V_{B/A} $ has the equation of $$ (\dot\theta\hat k) \times I*(cos\theta\hat i - sin\theta\hat j) $$

I am stuck at finding $\theta$ or $ V_{B/A}$ because it seems both need the other in order to get an answer.

Does anyone have a suggestion of equations to look at to obtain one of those two values with the information known?

Could I treat point A is being the center of the circle and use the formula of $ V_A = r\omega $ to obtain the value for $\omega$

Answer 1

Calculate $\varphi_0 = \arctan(s)$. Then the position of point $A$ is \begin{align} &X_A = X_A\\ &Y_A = s\,X_A + b \end{align} The position of point $B$ can be expressed in terms of the angle $\theta$ as \begin{align} &X_B = X_A + l\cos(\theta + \varphi_0)\\ &Y_B = b + s\,X_A + l\sin(\theta + \varphi_0) \end{align} However, we have a restricition for point $B$, called a holonomic constrant, which is that $B$ always moves along the circle $(X-c)^2 + Y^2 = r^2$, Therefore $(X_B-c)^2 + Y_B^2 = r^2$ which explicitly is $$\big( \, X_A + l\cos(\theta + \varphi_0) - c \,\big)^2 \, + \, \big(\,b + s\,X_A + l\sin(\theta + \varphi_0)\,\big)^2 \, = \, r^2$$ Thus, the position of the bar, moving so that point $A$ is always on the line $Y = sX + b$ and point $B$ is always on the circle $(X-c)^2 + Y^2 = r^2$ can be described by the three equations \begin{align} &X_B = X_A + l\cos(\theta + \varphi_0)\\ &Y_B = b + s\,X_A + l\sin(\theta + \varphi_0)\\ &\big( \, X_A + l\cos(\theta + \varphi_0) - c \,\big)^2 \, + \, \big(\,b + s\,X_A + l\sin(\theta + \varphi_0)\,\big)^2 \, = \, r^2 \end{align} Hence, if you know the way $X_A = X_A(t)$ changes with respect to time $t$, then you can blug it in the third equation and solve it for $\theta = \theta(t)$. After you have found $\theta$ you can plug it alongisde $X_A$ in the first two equations to find the coordinates $(X_B, \, Y_B)$ of $B$.

To find the angular velocity $\frac{d\theta}{dt}$ of the bar, you simply differentiate the third equation with respect to $t$ and add the new differentiated equation to the system, as a forth equation: \begin{align} &X_B = X_A + l\cos(\theta + \varphi_0)\\ &Y_B = b + s\,X_A + l\sin(\theta + \varphi_0)\\ &\big( \, X_A + l\cos(\theta + \varphi_0) - c \,\big)^2 \, + \, \big(\,b + s\,X_A + l\sin(\theta + \varphi_0)\,\big)^2 \, = \, r^2\\ &\big( \, X_A + l\cos(\theta + \varphi_0) - c \,\big) \left(\,\frac{dX_A}{dt} - l\sin(\theta + \varphi_0) \frac{d\theta}{dt}\,\right) = \, \\ &+ \,\big(\,b + s\,X_A + l\sin(\theta + \varphi_0)\,\big) \left(\,s\,\frac{dX_A}{dt} + l\cos(\theta + \varphi_0) \frac{d\theta}{dt}\,\right) \, = \, 0 \end{align} To find $\frac{d\theta}{dt}$ you need only the last two equations: \begin{align} &\big( \, X_A + l\cos(\theta + \varphi_0) - c \,\big)^2 \, + \, \big(\,b + s\,X_A + l\sin(\theta + \varphi_0)\,\big)^2 \, = \, r^2\\ &\big( \, X_A + l\cos(\theta + \varphi_0) - c \,\big) \left(\,l\sin(\theta + \varphi_0) \frac{d\theta}{dt} - \frac{dX_A}{dt}\,\right) \, = \,\big(\,b + s\,X_A + l\sin(\theta + \varphi_0)\,\big) \left(\,s\,\frac{dX_A}{dt} + l\cos(\theta + \varphi_0) \frac{d\theta}{dt}\,\right) \end{align} Given $X_A = X_A(t)$ and $V_A = V_A(t) = \frac{dX_A}{dt}$, you can take the first equation from the latter system of two equations, plug $X_A$ in it and solve for $\theta = \theta(t)$. This equation is the hardest to solve. After that, plug in the second equation $X_A, \, \theta,\, \frac{dX_A}{dt}$ and solve for the angular speed $\frac{d\theta}{dt}$.

Finally, to find the velocity of $B$, you simply take the first two equations of the four equation system above and differentiate them with respect to $t$: \begin{align} &V_{x,B} = \frac{dX_B}{dt} = \frac{dX_A}{dt} - l\sin(\theta + \varphi_0)\frac{d\theta}{dt}\\ &V_{y,B} = \frac{dY_B}{dt} = s\,\frac{dX_A}{dt} + l\cos(\theta + \varphi_0)\frac{d\theta}{dt} \end{align} So, you just have to plug in this equation the already calculated $\theta, \,\frac{d\theta}{dt}$ and $\frac{dX_A}{dt} = V_A$.

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OLD VERSION. Let's simplify things a bit. First, perform the translation: \begin{align} &X = \tilde{x} + c \\ &Y = \tilde{y} \end{align} Then the equation of the circle becomes $$r^2 = (X - c)^2 + Y^2 = \tilde{x}^2 + \tilde{y}^2$$ Then find the angle between the line $Y = sX + b$, which in new coordinates is $\tilde{y} = s\,\tilde{x} + (sc + b)$, and the horizontal axis: the slope is the tangent of that angle, i.e. $$\varphi_0 = \arctan(s)$$ Next, perform a clock-wise rotation of angle $\varphi_0$ so that the line $\tilde{y} = s\,\tilde{x} + (sc + b)$ becomes a line $\tilde{y} = h$ (one can calculate the distance $h$ between the center of the circle (the origin) and the line in question) parallel to the horizontal $x-$axis: \begin{align} \tilde{x} = \cos(\varphi_0)\, x \, - \, \sin(\varphi_0)\, y\\ \tilde{y} = \sin(\varphi_0)\, x \, + \, \cos(\varphi_0)\, y \end{align} Denote by $x_A$ the $x-$coordinate of the point $A$ moving along the line. The $y-$coordinate is $h$ and is fixed. The equation of the upper half of the circle in these new rotated and translated coordinates can be written as $$y = \sqrt{r^2 - x^2}$$
If $\theta$ is the angle between the rod $AB$ and the line $y = h$, which is parallel to the $x-$axis, then the equations for the position of the other end of the rod, point $B$, are \begin{align} &{x}_B = x_A + l\, \cos(\theta)\\ &{y}_B = \sqrt{r^2 - \big(x_A + l\, \cos(\theta)\big)^2} \end{align} Observe, there are two free parameters for the position of $B$ on the circle, namely $x_A$ and $\theta$. However, there is another restriction - the distance between $A$ and $B$ is always $l$. Hence: $$\big(x_B - x_A\big)^2 + \big(y_B - y_A\big)^2 = l^2$$ or after substitutions $$l^2 \cos^2(\theta) \, + \, \Big(\sqrt{r^2 - \big(x_A + l\, \cos(\theta)\big)^2\,} - h\Big)^2 \, = \, l^2$$ which establishes a link between the coordinates $x_A$ and $\theta$. You can move the first term from the lefthand side to the right one, then apply a central trigonometric identity to the right hand side, after which you can take square root on both sides, and finally obtain the simplified equation $$\sqrt{r^2 - \big(x_A + l\, \cos(\theta)\big)^2\,} - h \, = \, \pm \, l\sin(\theta)$$ where you should have in mind the sign $\pm$ depends on the sign of the right hand side. On your picture, $\theta \in [0, \pi/2)$ so you can choose a plus sign and the equation is $$\sqrt{r^2 - \big(x_A + l\, \cos(\theta)\big)^2\,} - h \, = \, l\sin(\theta)$$
Now, in this latter equation $x_A = x_A(t)$ and $\theta = \theta(t)$ are function of the time $t$, so we can differentiate the equation with respect to $t$ and pair it with the latter equation above: \begin{align} &\sqrt{r^2 - \big(x_A + l\, \cos(\theta)\big)^2\,} - h \, = \, l\sin(\theta)\\ &\frac{\big(x_A + l\cos(\theta)\big)}{\sqrt{r^2 - \big(x_A + l\, \cos(\theta)\big)^2}}\left(l\sin(\theta)\frac{d\theta}{dt} \, - \, \frac{dx_A}{dt}\right) \, = \, l\cos(\theta) \frac{d\theta}{dt} \end{align} You can simplify the second equation, using the first one, by solving for the square root $\sqrt{r^2 - \big(x_A + l\, \cos(\theta)\big)^2}$ and write the system as follows: \begin{align} &\sqrt{r^2 - \big(x_A + l\, \cos(\theta)\big)^2\,} - h \, = \, l\sin(\theta)\\ &\frac{\, x_A + l\cos(\theta)\,}{h \, + \, l\sin(\theta)}\left(l\sin(\theta)\frac{d\theta}{dt} \, - \, \frac{dx_A}{dt}\right) \, = \, l\cos(\theta) \frac{d\theta}{dt} \end{align} This system of equations features four variables: $$x_A, \, \theta, \, \frac{dx_A}{dt}, \, \frac{d\theta}{dt}$$ So, if you are given any two of these, you can solve the system and find the other two. For example, if you know the position and velocity of $A$, then you know $x_A$ and $ \frac{dx_A}{dt}$. Then, you can plug $x_A$ in the first equation and solve that same first equation for $\theta$. Then, knowing already $x_A, \, \theta, \, \frac{dx_A}{dt}$, you can plug these three values in the second equation and solve it for the angular velocity $\frac{d\theta}{dt}$. This second equation is easier to solve with respect to $\frac{d\theta}{dt}$ because it is linear with respect to $\frac{d\theta}{dt}$.

The next step is to find the linear velocity of $B$, which should be tangent to the circle. If you take the equations \begin{align} &{x}_B = x_A + l\, \cos(\theta)\\ &{y}_B = \sqrt{r^2 - \big(x_A + l\, \cos(\theta)\big)^2} \end{align} By the first equation from the system of equations discussed above, you can express $\sqrt{r^2 - \big(x_A + l\, \cos(\theta)\big)^2} = l\sin(\theta) + h$ and re-write the latter parametrization as follows: \begin{align} &{x}_B = x_A + l\, \cos(\theta)\\ &{y}_B = l\sin(\theta) + h \end{align} To find the linear velocity of $B$, you just have to differentiate the latter parametrization with respect to $t$ \begin{align} &\frac{dx_B}{dt} = \frac{dx_A}{dt} - l\, \sin(\theta)\frac{d\theta}{dt}\\ &\frac{dy_B}{dt} = l \, \cos(\theta)\frac{d\theta}{dt} \end{align}
plug the already determined values of $\frac{dx_A}{dt}, \, \theta, \, \frac{d\theta}{dt}$.

Answer 2

You are very close to the solution. What you need to consider is that the velocity of each point on the curved path, has to be tangent to the curve. I.e. the velocity of point A plus the $V_{A|B}$, needs to have the same direction with the tangent at point B.

So what you need to do is:

• calculate the tangent to the curved path at point B
• express all velocities in the problems in the same frame of reference (XY, or xy). Preferably the latter xy.
• solve the system of the velocities:

e.g. if you select xy system then you should have $$\begin{bmatrix} V_{Bx}\\V_{By} \\0 \end{bmatrix} =\begin{bmatrix} V_{Ax}\\ 0\\ 0 \end{bmatrix} + \begin{bmatrix} 0\\ 0\\ \dot{\theta} \end{bmatrix} \times \begin{bmatrix} I \cos\theta\\ I\sin\theta\\ 0 \end{bmatrix} $$

This reduces to:

$$\begin{bmatrix} V_{Bx}\\V_{By} \end{bmatrix} =\begin{bmatrix} V_{Ax}\\ 0 \end{bmatrix} + \begin{bmatrix} -\dot{\theta}I\sin\theta \\ \dot{\theta}I \cos\theta \end{bmatrix} $$

The above can be solved if you keep in mind the constraint about the direction of the tangent to the curved path (lets denote it $\phi$), i.e. $\tan\phi = \frac{V_{By}}{V_{Bx}}$.