3.1 (problem 2.2) First, replace the length of the bar and the two angles by symbols. Choose a basis \((\boldsymbol{\hat{\imath}},\boldsymbol{\hat{\jmath}},\boldsymbol{\hat{k}})\). Then relate \(\boldsymbol{v}_B\) to \(\boldsymbol{v}_A\) and account for the directions of these vectors (the bar is constrained by point-contacts at \(A\) and \(B\)).
You can recover \(v_B\) from \(v_A\) by using equiprojectivity.
Follow the same procedure for the accelerations of \(A\) and \(B\).
Solution:
- a. Define \(l\), \(\beta\), \(\gamma\), \((\boldsymbol{\hat{\imath}},\boldsymbol{\hat{\jmath}},\boldsymbol{\hat{k}}=\boldsymbol{\hat{\imath}}\times \boldsymbol{\hat{\jmath}})\) as shown in the Figure. To solve this problem, we apply formula \(\boldsymbol{v}_B = \boldsymbol{v}_A + \boldsymbol{\omega}\times \boldsymbol{r}_{AB}\) and account for the direction of vectors \(\boldsymbol{v}_A\) and \(\boldsymbol{v}_B\). First we write \(\boldsymbol{\omega}= \omega\boldsymbol{\hat{k}}\). Then \[ v_B (\cos\gamma\boldsymbol{\hat{\imath}}+\sin\gamma \boldsymbol{\hat{\jmath}}) =v_A (\cos\beta\boldsymbol{\hat{\imath}}-\sin\beta\boldsymbol{\hat{\jmath}}) +\omega\boldsymbol{\hat{k}}\times l\boldsymbol{\hat{\imath}} \] This yields two equations \[ v_B \cos\gamma =v_A \cos\beta, \qquad v_B \sin\gamma =-v_A \sin\beta + l \omega\qquad\text{[1-2]} \] [1-2] solve for \(v_B\) and \(\omega\): \[ \boxed{ v_B = v_A \frac{\cos\beta}{\cos\gamma}, \qquad \omega= \frac{v_A}{l} \frac{\sin(\beta+\gamma)}{\cos\gamma} } \qquad\text{[3-4]} \] Numerically: \(v_B=1.92\) m/s and \(\omega\) =3.5 rad/s (ccw).
Note: we clearly see that the result of [3] is a consequence of equiprojectivity: \(\boldsymbol{v}_A \cdot \boldsymbol{\hat{\imath}}= \boldsymbol{v}_B \cdot \boldsymbol{\hat{\jmath}}\).
- b. We cannot solve for accelerations \((a_B, \alpha)\) by taking the time-derivative of [3-4]. Instead we apply formula \(\boldsymbol{a}_B = \boldsymbol{a}_A + \boldsymbol{\alpha}\times \boldsymbol{r}_{AB}- \omega^2 \boldsymbol{r}_{AB}\) and account for \(\boldsymbol{a}_A=\boldsymbol{0}\) (since \(\boldsymbol{v}_A\) is constant in magnitude and direction) and the direction of \(\boldsymbol{a}_B\). With \(\boldsymbol{\alpha}= \alpha\boldsymbol{\hat{k}}\) we obtain \[ a_B (\cos\gamma\boldsymbol{\hat{\imath}}+\sin\gamma \boldsymbol{\hat{\jmath}})= \alpha\boldsymbol{\hat{k}}\times l\boldsymbol{\hat{\imath}}- l \omega^2 \boldsymbol{\hat{\imath}} \] This yields two equations \[ a_B \cos\gamma = -l \omega^2, \qquad a_B \sin\gamma = l \alpha\qquad\text{[5-6]} \] [5-6] solve for \(a_B\) and \(\alpha\): \[ \boxed{ a_B = - \frac{v_A^2}{l}\frac{\sin^2(\beta+\gamma)}{\cos^3\gamma}, \qquad \alpha= -\frac{v_A^2}{l^2} \frac{\sin^2(\beta+\gamma)\sin\gamma}{\cos^3\gamma} } \]
3.2. Follow the steps taken in Recitation 3. Start by finding \(\boldsymbol{v}_{A\in 1}\). Then find \(\boldsymbol{v}_{B\in 2}\) from \(\boldsymbol{v}_{A\in 2}\). Finally, relate \(\boldsymbol{v}_{B\in 3}\) to \(\boldsymbol{v}_{C\in 3} ={\bf 0}\). This should give you 2 scalar equations for the 2 unknowns of the problem (adopt a basis of unit vectors).
Solution: Choose a basis \((\boldsymbol{\hat{\imath}}, \boldsymbol{\hat{\jmath}}, \boldsymbol{\hat{k}}= \boldsymbol{\hat{\imath}}\times \boldsymbol{\hat{\jmath}})\).
Part 1: Velocity Analysis
Start with the velocity \(\boldsymbol{v}_{A\in 1} = -\omega \boldsymbol{\hat{k}}\times \boldsymbol{r}_{OA}= r \omega \boldsymbol{\hat{\imath}}=\boldsymbol{v}_{A\in 2}\). Then find \(\boldsymbol{v}_{B\in 2}\): \[ \boldsymbol{v}_{B\in 2} = \boldsymbol{v}_{A\in 2} + \omega_2 \boldsymbol{\hat{k}}\times \boldsymbol{r}_{AB} = r\omega\boldsymbol{\hat{\imath}}+ \omega_2\boldsymbol{\hat{k}}\times 2r \boldsymbol{\hat{\imath}}= r\omega\boldsymbol{\hat{\imath}}+ 2r \omega_2 \boldsymbol{\hat{\jmath}} \] Finally find \(\boldsymbol{v}_{B\in 3}\) and set the equality \(\boldsymbol{v}_{B\in 2}= \boldsymbol{v}_{B\in 3}\): \[ \boldsymbol{v}_{B\in 3} = \omega_3 \boldsymbol{\hat{k}}\times \boldsymbol{r}_{CB} = \omega_3 \boldsymbol{\hat{k}}\times (-r\boldsymbol{\hat{\jmath}}) = r\omega_3 \boldsymbol{\hat{\imath}} \] This gives two equations: \[ r\omega= r\omega_3 , \qquad 2r\omega_2 = 0 \] or \[ \boxed{ \boldsymbol{\omega}_2 =\boldsymbol{0}, \qquad \boldsymbol{\omega}_3 = \omega\boldsymbol{\hat{k}} } \qquad \text{[1-2]} \]
Note: The fact that \(\boldsymbol{\omega}_2 = \boldsymbol{0}\) is zero is easily shown by the method of ICs. Since the velocities \(\boldsymbol{v}_{A\in 2}\) and \(\boldsymbol{v}_{B\in 2}\) are parallel (and not normal to line \(AB\)), the IC of body 2 is at infinity.
Part 2: Acceleration Analysis.
We follow the same steps, setting \(\boldsymbol{\alpha}_1 = \boldsymbol{0}\): \[\begin{align*} \boldsymbol{a}_{B\in 2} & = \boldsymbol{a}_{A\in 2} + \alpha_2 \boldsymbol{\hat{k}}\times \boldsymbol{r}_{AB} -\omega_2^2 \boldsymbol{r}_{AB}\\ & = -r\omega^2 \boldsymbol{\hat{\jmath}}+2r \alpha_2 \boldsymbol{\hat{\jmath}}\\ & = \boldsymbol{a}_{B\in 3} = r\alpha_3 \boldsymbol{\hat{\imath}}+r\omega_3^2 \boldsymbol{\hat{\jmath}} \end{align*}\] This gives two equations (we set \(\omega_2 =0\)): \[ -r\omega^2 +2 r\alpha_2 = r\omega_3^2 , \qquad 0 = r\alpha_3 \] Finally \[ \boxed{ \boldsymbol{\alpha}_2 = \omega^2 \boldsymbol{\hat{k}}, \qquad \boldsymbol{\alpha}_3= \boldsymbol{0} } \] From \(\boldsymbol{v}_{B\in 3} = r\omega\boldsymbol{\hat{\imath}}\), we find that the tangential unit vector associated with the motion of \(B\) is \(\boldsymbol{\hat{\imath}}\). Then the acceleration of \(B\) can be written as: \[ \boldsymbol{a}_B = \boldsymbol{a}_{B\in 3} = r\alpha_3 \boldsymbol{\hat{\imath}}+r\omega_3^2 \boldsymbol{\hat{\jmath}}= r\omega^2 \boldsymbol{\hat{\jmath}} \] which shows that \(a_{Bt}= 0\) and \(a_{Bn}= r\omega^2\).
3.3. Follow the steps taken in Recitation 3. Be sure to find the normal acceleration \(a_B\) of pivot \(B\). You should be able to relate all unknowns in terms of \(a_B\).
Solution:
- Part 1: Velocity Analysis
Step 1. Choose a basis \((\boldsymbol{\hat{\imath}},\boldsymbol{\hat{\jmath}},\boldsymbol{\hat{k}}=\boldsymbol{\hat{\imath}}\times \boldsymbol{\hat{\jmath}})\). Set \(\boldsymbol{\omega}_i = \omega_i \boldsymbol{\hat{k}}\), \(i=1,2,3\).
Step 2. We start with the determination of \(\boldsymbol{v}_A\) as a point of body 1: \[ \boldsymbol{v}_{A\in 1} = \omega_1 \boldsymbol{\hat{k}}\times \boldsymbol{r}_{OA}= \omega_1\boldsymbol{\hat{k}}\times r\boldsymbol{\hat{\imath}}= r\omega_1 \boldsymbol{\hat{\jmath}}= \boldsymbol{v}_{A\in 2} \] Step 3. We find \(\boldsymbol{v}_B\) (treat both \(A\) and \(B\) as two points of body 2): \[ \boldsymbol{v}_{B\in 2} =\boldsymbol{v}_{A\in 2}+ \omega_2 \boldsymbol{\hat{k}}\times \boldsymbol{r}_{AB} = r\omega_1 \boldsymbol{\hat{\jmath}}+ \omega_2 \boldsymbol{\hat{k}}\times (-r\boldsymbol{\hat{\jmath}}) = r\omega_1 \boldsymbol{\hat{\jmath}}+ r \omega_2 \boldsymbol{\hat{\imath}} \] Step 4. We find \(\boldsymbol{v}_B\) as a point of body 3 (in rotation about \(C\)) \[ \boldsymbol{v}_{B\in 3} = \omega_3 \boldsymbol{\hat{k}}\times \boldsymbol{r}_{CB} = \omega_3 \boldsymbol{\hat{k}}\times (-r\boldsymbol{\hat{\imath}}-r\boldsymbol{\hat{\jmath}}) = - r\omega_3 (\boldsymbol{\hat{\jmath}}-\boldsymbol{\hat{\imath}}) \] Step 5. Equate \(\boldsymbol{v}_{B\in 2}\) to \(\boldsymbol{v}_{B\in 3}\): \(r\omega_1 \boldsymbol{\hat{\jmath}}+ r \omega_2 \boldsymbol{\hat{\imath}}= - r\omega_3 (\boldsymbol{\hat{\jmath}}-\boldsymbol{\hat{\imath}})\) to find \[ \begin{cases} r\omega_1 = -r\omega_3\\ r\omega_2 = r\omega_3 \end{cases} \Longrightarrow \boxed{ \boldsymbol{\omega}_2 = \boldsymbol{\omega}_3 = - \omega_1 \boldsymbol{\hat{k}} } \qquad\text{[1]} \]
.
- Part 2: Acceleration Analysis:
Step 1. Set \(\boldsymbol{\alpha}_1 = \boldsymbol{0}\). We seek \(\boldsymbol{\alpha}_2 = \alpha_2 \boldsymbol{\hat{k}}\) and \(\boldsymbol{\alpha}_3 = \alpha_3 \boldsymbol{\hat{k}}\).
Step 2. We find \(\boldsymbol{a}_{A\in 1} = -r \omega_1^2 \boldsymbol{\hat{\imath}}= \boldsymbol{a}_{A\in 2}\).
Step 3. Find the acceleration of \(B\) \[\begin{align*} \boldsymbol{a}_{B\in 2} & = \boldsymbol{a}_{A\in 2} +\boldsymbol{\alpha}_2 \times \boldsymbol{r}_{AB} - \omega_2^2 \boldsymbol{r}_{AB} &=-r \omega_1^2 \boldsymbol{\hat{\imath}}+ r\alpha_2 \boldsymbol{\hat{\imath}}+r\omega_2^2 \boldsymbol{\hat{\jmath}} \end{align*}\]
Step 4. Find \(\boldsymbol{a}_B\) as a point of body 3 (in rotation about \(C\)) \[\begin{align*} \boldsymbol{a}_{B\in 3}& = \alpha_3 \boldsymbol{\hat{k}}\times \boldsymbol{r}_{CB} -\omega_3^2 \boldsymbol{r}_{CB} \\ & = r\alpha_3 (\boldsymbol{\hat{\imath}}-\boldsymbol{\hat{\jmath}}) - r\omega_3^2 (-\boldsymbol{\hat{\imath}}-\boldsymbol{\hat{\jmath}}) \end{align*}\]
Step 5. Equate \(\boldsymbol{a}_{B\in 2}\) to \(\boldsymbol{a}_{B\in 3}\) to find \[ \boxed{ \boldsymbol{\alpha}_3 = \boldsymbol{0}, \qquad \boldsymbol{\alpha}_2 = 2 \omega_1^2 \boldsymbol{\hat{k}} } \qquad\text{[2]} \]
- Conclusion:
With \(\boldsymbol{\alpha}_3 =\boldsymbol{0}\), we now find that \(\boldsymbol{a}_{B\in 3} =r\omega_3^2 (\boldsymbol{\hat{\imath}}+\boldsymbol{\hat{\jmath}})\) is directed along the normal \(\boldsymbol{\hat{e}}_N = (\boldsymbol{\hat{\imath}}+\boldsymbol{\hat{\jmath}})/\sqrt{2}\) to the circular path of \(B\). At the instant shown, the tangential acceleration of \(B\) is zero.
We conclude that \[ a_B = \sqrt{2} r\omega_3^2 \] Since \(\omega_3 = -\omega_1\), we find \[ \boxed{\omega_1 = \Big(\frac{a_B}{r\sqrt{2}} \Big)^{1/2} } \] This result is only valid for the given configuration. At any other time, it is likely that \(\boldsymbol{\alpha}_3 \neq 0\). In fact all the results found in [1-2] are only valid at the time of the configurartion.3.4. (problem 2.5) Define a basis \((\boldsymbol{\hat{\imath}},\boldsymbol{\hat{\jmath}},\boldsymbol{\hat{k}}=\boldsymbol{\hat{\imath}}\times \boldsymbol{\hat{\jmath}})\) and an oriented angle \(\theta_2\) to define the orientation of body 2. You will need a relationship between \(\theta_1\) and \(\theta_2\).
Relate the velocities of \(A\) and \(B\) as two points of body 2. Impose the direction of \(\boldsymbol{v}_{B \in 3}\). The velocity \(\boldsymbol{v}_{A\in 1}\) is easily found as a point rotating about \(O\) with body 1.
Repeat this analysis to find \(\boldsymbol{a}_{B\in 3/0}\).
Solution:
- a. Define basis \((\boldsymbol{\hat{\imath}},\boldsymbol{\hat{\jmath}},\boldsymbol{\hat{k}}=\boldsymbol{\hat{\imath}}\times \boldsymbol{\hat{\jmath}})\) as shown in the Figure. Define oriented angle \(\theta_2\) to define the orientation of body 2.
The angular velocities are given by \[ \boldsymbol{\omega}_1 = \dot{\theta}_1 \boldsymbol{\hat{k}}, \qquad \boldsymbol{\omega}_2 = -\dot{\theta}_2 \boldsymbol{\hat{k}} \] We relate \(\boldsymbol{v}_B\) to \(\boldsymbol{v}_A\): \[ \boldsymbol{v}_{B\in 3/0}= \boldsymbol{v}_{B\in 2/0} = \boldsymbol{v}_{A\in 2/0} - \dot{\theta}_2 \boldsymbol{\hat{k}}\times L(\cos\theta_2 \boldsymbol{\hat{\imath}}- \sin\theta_2 \boldsymbol{\hat{\jmath}}) = \dot{\theta}_1 \boldsymbol{\hat{k}}\times r (\cos\theta_1 \boldsymbol{\hat{\imath}}+ \sin\theta_1 \boldsymbol{\hat{\jmath}}) - \dot{\theta}_2 \boldsymbol{\hat{k}}\times L(\cos\theta_2 \boldsymbol{\hat{\imath}}- \sin\theta_2 \boldsymbol{\hat{\jmath}}) \]
We impose \(\boldsymbol{v}_{B\in 3/0}= v_B \boldsymbol{\hat{\imath}}\) and find 2 equations \[ v_B = -r \dot{\theta}_1 \sin\theta_1 -L \dot{\theta}_2 \sin\theta_2, \qquad 0= r\dot{\theta}_1 \cos\theta_1 - L \dot{\theta}_2 \cos\theta_2 \] yielding \[ \boxed{ \dot{\theta}_2 = \frac{r}{L} \dot{\theta}_1 \frac{\cos\theta_1}{\cos\theta_2} } \qquad\text{[1]} \] \[ \boxed{ \boldsymbol{v}_B = v_B \boldsymbol{\hat{\imath}}= -r \dot{\theta}_1 \frac{\sin(\theta_1+\theta_2)}{\cos\theta_2} \boldsymbol{\hat{\imath}} } \qquad\text{[2]} \] Note that [2] is also a consequence of equiprojectivity. In [1-2], the angle \(\theta_2\) is a function of \(\theta_1\): \[ \sin\theta_2 = \frac{r}{L} \sin\theta_1 \]
- b. We can repeat this analysis to find \(\boldsymbol{a}_{B\in 3/0}\) (with \(\dot{\theta}_1 = constant\)) \[\begin{align} \boldsymbol{a}_{B\in 3/0} = \boldsymbol{a}_{B\in 2/0} &=& \boldsymbol{a}_{A\in 2/0} - \ddot{\theta}_2 \boldsymbol{\hat{k}}\times L(\cos\theta_2 \boldsymbol{\hat{\imath}}- \sin\theta_2 \boldsymbol{\hat{\jmath}}) -L\dot{\theta}_2^2 (\cos\theta_2 \boldsymbol{\hat{\imath}}- \sin\theta_2 \boldsymbol{\hat{\jmath}}) \\ & = & \underbrace{- r\dot{\theta}_1^2 (\cos\theta_1 \boldsymbol{\hat{\imath}}+ \sin\theta_1 \boldsymbol{\hat{\jmath}})}_{\boldsymbol{a}_{A\in 1/0}} - L\ddot{\theta}_2 (\cos\theta_2 \boldsymbol{\hat{\jmath}}+ \sin\theta_2 \boldsymbol{\hat{\imath}})-L\dot{\theta}_2^2 (\cos\theta_2 \boldsymbol{\hat{\imath}}- \sin\theta_2 \boldsymbol{\hat{\jmath}}) \end{align}\] We impose \(\boldsymbol{a}_{B\in 3/0}= a_B \boldsymbol{\hat{\imath}}\) and find 2 equations \[ a_B = -r \dot{\theta}_1^2 \cos\theta_1 -L \ddot{\theta}_2 \sin\theta_2-L\dot{\theta}_2^2 \cos\theta_2 \] \[ 0= -r\dot{\theta}_1^2 \sin\theta_1 - L \ddot{\theta}_2 \cos\theta_2 +L\dot{\theta}_2^2 \sin\theta_2 \] These 2 equations solve for \((\ddot{\theta}_2 , a_B)\). Acceleration \(a_B\) is found by eliminating \(\ddot{\theta}_2\): \[ \boxed{ \boldsymbol{a}_B =-\frac{r \dot{\theta}_1^2}{\cos\theta_2} \Big(\cos(\theta_1+\theta_2) + \frac{r}{L} \frac{\cos^2 \theta_1}{\cos^2 \theta_2} \Big) \boldsymbol{\hat{\imath}} }\qquad\text{[3]} \] Note: [3] can also be obtained by taking the time-derivative of [2].
.