Absolutely! Let's break down **Pb. 4.4** step by step. --- ## **Given Data** - Frame is rotating about the **y-axis**: - Angular velocity: \(\omega_1 = 6\, \text{rad/s}\) - Angular acceleration: \(\dot{\omega}_1 = 2\, \text{rad/s}^2\) - Rod **FG** is rotating (relative to the frame) about the **x-axis**: - \(\omega_2 = 4\, \text{rad/s}\) - \(\dot{\omega}_2 = 3\, \text{rad/s}^2\) - Collar **C** moves along rod **FG** toward **F**: - Relative velocity: \(v_{C/F} = .5\, \text{m/s}\) - Relative acceleration: \(a_{C/F} = .7\, \text{m/s}^2\) - Distance of C from O (origin): \(.4\, \text{m}\) - Distance CF: \(.2\, \text{m}\) --- ## **Step 1: Define Reference Frames** - **Frame **: Inertial frame (ground) - **Frame 1**: Rotating frame (rotates about y-axis) - **Frame 2**: Rod FG (rotates about x-axis of Frame 1) --- ## **Step 2: Express Angular Velocities** - \(\vec{\omega}_1 = 6\, \hat{j}\) (about y-axis) - \(\vec{\omega}_2 = 4\, \hat{i}'\) (about x'-axis, which moves with frame 1) **Total angular velocity (of FG relative to ground):** \[ \vec{\omega}_{FG} = \vec{\omega}_1 + \vec{\omega}_2 \] --- ## **Step 3: Collar's Position Vector** Let’s define the position of C relative to G (center of FG): \[ \vec{r}_{C/G} = .2\, \hat{i}' \] But G is .2 m from O along the x'-axis, so \[ \vec{r}_{G/O} = .2\, \hat{i}' \] \[ \vec{r}_{C/O} = .4\, \hat{i}' \] --- ## **Step 4: Velocity of Collar (Absolute)** The general velocity formula for a point C moving along a rotating rod FG: \[ \vec{v}_C = \vec{v}_G + \vec{\omega}_{FG} \times \vec{r}_{C/G} + \left( \frac{d\vec{r}_{C/G}}{dt} \right)_{FG} \] - \(\vec{v}_G\): velocity of point G - The last term: velocity of C w.r.t. FG (given as \(.5\, \text{m/s}\) toward F, i.e., along \(\hat{i}'\)) Let’s break it down: ### **A. Velocity of G (\(\vec{v}_G\))** Since G is on the frame rotating about the y-axis: \[ \vec{v}_G = \vec{\omega}_1 \times \vec{r}_{G/O} = (6\,\hat{j}) \times (.2\,\hat{i}') = 6 \cdot .2 \cdot (\hat{j} \times \hat{i}') = 1.2\,(-\hat{k}') \] So, \(\vec{v}_G = -1.2\,\hat{k}'\,\text{m/s}\) ### **B. \(\vec{\omega}_{FG} \times \vec{r}_{C/G}\)** \[ \vec{\omega}_{FG} = 6\,\hat{j} + 4\,\hat{i}' \] \[ \vec{r}_{C/G} = .2\,\hat{i}' \] Only the \(6\,\hat{j}\) part will contribute: \[ (6\,\hat{j}) \times (.2\,\hat{i}') = 1.2\,(-\hat{k}') \] (Again, \(\hat{j} \times \hat{i}' = -\hat{k}'\)) \[ (4\,\hat{i}') \times (.2\,\hat{i}') = \] So, this term is \(1.2\,(-\hat{k}')\) ### **C. Relative Velocity along FG** \[ \left( \frac{d\vec{r}_{C/G}}{dt} \right)_{FG} = v_{C/F}\,\hat{i}' = .5\,\hat{i}' \] --- ## **Step 5: Add Up the Velocity Terms** \[ \vec{v}_C = (-1.2\,\hat{k}') + (-1.2\,\hat{k}') + (.5\,\hat{i}') \] \[ \vec{v}_C = .5\,\hat{i}' - 2.4\,\hat{k}' \quad \text{(units: m/s)} \] --- ## **Step 6: Acceleration of Collar (Absolute)** General formula: \[ \vec{a}_C = \vec{a}_G + \dot{\vec{\omega}}_{FG} \times \vec{r}_{C/G} + \vec{\omega}_{FG} \times \left( \vec{\omega}_{FG} \times \vec{r}_{C/G} \right) + 2\vec{\omega}_{FG} \times \left( \frac{d\vec{r}_{C/G}}{dt} \right)_{FG} + \left( \frac{d^2\vec{r}_{C/G}}{dt^2} \right)_{FG} \] Let’s summarize the terms (using given values): ### **A. Acceleration of G (\(\vec{a}_G\))** \[ \vec{a}_G = \dot{\vec{\omega}}_1 \times \vec{r}_{G/O} + \vec{\omega}_1 \times (\vec{\omega}_1 \times \vec{r}_{G/O}) \] \[ \dot{\vec{\omega}}_1 = 2\,\hat{j} \] \[ (2\,\hat{j}) \times (.2\,\hat{i}') = .4\,(-\hat{k}') \] \[ \vec{\omega}_1 \times (\vec{\omega}_1 \times \vec{r}_{G/O}) = 6\,\hat{j} \times (6\,\hat{j} \times .2\,\hat{i}') = 6\,\hat{j} \times (-1.2\,\hat{k}') = 6 \times (-1.2) (\hat{j} \times \hat{k}') = -7.2\,\hat{i}' \] So: \[ \vec{a}_G = -.4\,\hat{k}' - 7.2\,\hat{i}' \] --- ### **B. \(\dot{\vec{\omega}}_{FG} \times \vec{r}_{C/G}\)** \[ \dot{\vec{\omega}}_{FG} = \dot{\vec{\omega}}_1 + \dot{\vec{\omega}}_2 = 2\,\hat{j} + 3\,\hat{i}' \] \[ \vec{r}_{C/G} = .2\,\hat{i}' \] \[ (2\,\hat{j}) \times (.2\,\hat{i}') = .4\,(-\hat{k}') \] \[ (3\,\hat{i}') \times (.2\,\hat{i}') = \] So, \(-.4\,\hat{k}'\) --- ### **C. Centripetal Term: \(\vec{\omega}_{FG} \times (\vec{\omega}_{FG} \times \vec{r}_{C/G})\)** First, find \(\vec{\omega}_{FG} \times \vec{r}_{C/G}\): Already found above: \(6\,\hat{j} \times .2\,\hat{i}' = -1.2\,\hat{k}'\) So, \[ \vec{\omega}_{FG} \times (-1.2\,\hat{k}') \] Now, \[ (6\,\hat{j}) \times (-1.2\,\hat{k}') = 6 \cdot -1.2\, (\hat{j} \times \hat{k}') = -7.2\,\hat{i}' \] \[ (4\,\hat{i}') \times (-1.2\,\hat{k}') = 4 \cdot -1.2\, (\hat{i}' \times \hat{k}') = -4.8\,(-\hat{j}) = 4.8\,\hat{j} \] So, total: \( -7.2\,\hat{i}' + 4.8\,\hat{j} \) --- ### **D. Coriolis Term: \(2\vec{\omega}_{FG} \times \left( \frac{d\vec{r}_{C/G}}{dt} \right)_{FG}\)** \[ 2 \times (6\,\hat{j} + 4\,\hat{i}') \times (.5\,\hat{i}') \] \[ 6\,\hat{j} \times .5\,\hat{i}' = 3\,(-\hat{k}') \] \[ 4\,\hat{i}' \times .5\,\hat{i}' = \] So, total: \(2 \times -3\,\hat{k}' = -6\,\hat{k}'\) --- ### **E. Relative Acceleration along FG** \[ \left( \frac{d^2\vec{r}_{C/G}}{dt^2} \right)_{FG} = a_{C/F}\,\hat{i}' = .7\,\hat{i}' \] --- ## **Step 7: Add All Acceleration Terms** Sum all terms in \(\hat{i}'\), \(\hat{j}\), and \(\hat{k}'\): - \(\hat{i}': -7.2 + (-7.2) + .7 = -13.7\) - \(\hat{j}: 4.8\) - \(\hat{k}': -.4 + (-.4) + (-6) = -6.8\) So, \[ \vec{a}_C = -13.7\,\hat{i}' + 4.8\,\hat{j} - 6.8\,\hat{k}' \quad \text{(units: m/s}^2\text{)} \] --- ## **Summary** ### **Velocity of Collar C**: \[ \boxed{\vec{v}_C = .5\,\hat{i}' - 2.4\,\hat{k}' \quad [\text{m/s}]} \] ### **Acceleration of Collar C**: \[ \boxed{\vec{a}_C = -13.7\,\hat{i}' + 4.8\,\hat{j} - 6.8\,\hat{k}' \quad [\text{m/s}^2]} \] --- ### **Let Me Know if You Want Step-by-Step Vector Calculations or Coordinate Conversions!**