# Kinematics of actuated disk with no slip

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.

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?

• What is the significance of $R_1$?.
– AJN
Commented Oct 3, 2022 at 15:16
• @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 $R_1$. Commented Oct 3, 2022 at 18:00

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 $$\dot{\theta}=0$$.)

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.