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Given the following kinematic problem, how would one calculate the velocity $v_C$ of point C given a certain horizontal velocity $v_{E}$ in point E? Given is that the disk does not slip and cannot move in the vertical direction. Point E can also only move horizontally.

Kinematic problem

My train of thought is as follows: Calculate $v_D=v_E+\omega_{DE} \times r_{D/E}$. Then use $v_D=v_C+\omega_{BD}\times r_{D/B}$. With these two equations I could find $v_C$. However, I cannot find the angular velocities. What am I missing here?

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  • $\begingroup$ What is the significance of $R_1$?. $\endgroup$
    – AJN
    Oct 3, 2022 at 15:16
  • $\begingroup$ @AJN For this question in specific it's not relevant. The full dynamics problem continuous about the kinematics/kinetics of a particle A moving in the slot $R_1$. $\endgroup$ Oct 3, 2022 at 18:00

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You need one more equation $v_B=v_C+\omega_{BD}\times r_{C/B}$ and the fact that $\omega_{DE}=\dot{\theta }$.

Now we have $v_B=v_C+\omega_{BD}\ \hat{k} \times R_3\ \hat{j}=v_C-\omega_{BD}\ R_3\hat{i}$ and since $v_B=0$, we get $v_C=\omega_{BD}\ R_3\hat{i}$.

For D we have, $v_D=v_C-\omega_{BD}\ \hat{k} \times R_3\ \hat{j}=v_C+\omega_{BD}\ R_3\hat{i} = 2\ \omega_{BD}\ R_3\hat{i}$

Thus $$v_C=v_D/2$$

Using $v_D=v_E+\omega_{DE} \times r_{D/E}$, we get $v_D=v_E\hat{i}-\dot{\theta} \hat{k} \times \{-L_{DE} \cos\theta\hat{i}+L_{DE} \sin\theta\hat{j}\} = v_E\hat{i}+\dot{\theta}L_{DE}\cos\theta\hat{j}+\dot{\theta}L_{DE}\sin\theta\hat{i}$

Thus $$v_C=\frac{1}{2}(v_E\hat{i}+\dot{\theta}L_{DE}\sin\theta\hat{i}+\dot{\theta}L_{DE}\cos\theta\hat{j})$$

(For the points C and D to remain at the constant height with no vertical motion, either $\theta=90 {}^{\circ}$ or $\dot{\theta}=0$.)

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At time zero, I would expect $\vec{v}_C$ to be half of $\vec{v}_E$ and the angular velocity should be $\vec{v}_E \times \vec{BD}$. because you can think of point $B$ as stationary. This is of course not true after some movement.

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