# Solution to Seatwork 5 ## **Given:** - Cardioid equation: \( r = .2(1 + \cos \theta) \) m - \( \theta = 30^\circ = \frac{\pi}{6} \) rad - Angular velocity: \( \dot{\theta} = 3 \) rad/s **Required:** Find the **magnitude of the velocity** of the collars at this point. --- ## **Step 1: Find the Position and its Derivative** ### **a) Compute \( r \) and \( \dot{r} \)** - \( r = .2(1 + \cos \theta) \) - \( \dot{r} = \frac{dr}{dt} = \frac{dr}{d\theta} \cdot \dot{\theta} \) \[ \frac{dr}{d\theta} = .2 \cdot (-\sin \theta) \] \[ \dot{r} = .2(-\sin \theta)\cdot \dot{\theta} \] Plug in the values: - \( \sin 30^\circ = .5 \) - \( \cos 30^\circ = .866 \) \[ r = .2 (1 + .866) = .2 \times 1.866 = .3732~\text{m} \] \[ \dot{r} = .2(-.5) \times 3 = -.1 \times 3 = -.3~\text{m/s} \] --- ## **Step 2: Find the Magnitude of Velocity** The velocity in polar coordinates is: \[ v = \sqrt{\dot{r}^2 + (r\dot{\theta})^2} \] Plug in the values: - \( \dot{r} = -.3~\text{m/s} \) - \( r = .3732~\text{m} \) - \( \dot{\theta} = 3~\text{rad/s} \) \[ v = \sqrt{(-.3)^2 + (.3732 \times 3)^2} \] \[ v = \sqrt{.09 + (1.1196)^2} \] \[ v = \sqrt{.09 + 1.2535} \] \[ v = \sqrt{1.3435}\approx 1.159~\text{m/s} \] --- ### **Final Answer** \[ \boxed{v = 1.16~\text{m/s}} \] --- ### **Graph** Below is a sketch of the cardioid and velocity vectors:  *Alt text: Cardioid guide with OA at 30 degrees, showing r and velocity vector.* --- #### **Summary Table** | Step | Calculation | Result | |------|----------------------------------------------------|-------------| | 1 | \( r = .2(1+\cos 30^\circ) \) | .3732 m | | 1 | \( \dot{r} = .2(-\sin 30^\circ)\cdot 3 \) | -.3 m/s | | 2 | \( v = \sqrt{(-.3)^2 + (.3732 \times 3)^2} \) | 1.16 m/s | --- Let me know if you want the vector decomposition or more details!