Roller Velocity Acceleration 0Daed4

1. **Problem Statement:** We have a bar $AB$ of length $0.6$ m inclined at $30^\circ$ with the vertical at roller $A$. Roller $A$ has a velocity of $4$ m/s downward and acceleration of $6$ m/s$^2$ downward. The track at $B$ is inclined $30^\circ$ downward to the right. We need to find the velocity and acceleration of roller $B$, and the angular velocity $\omega$ and angular acceleration $\alpha$ of the bar. 2. **Known Data:** - Length $AB = 0.6$ m - Angle of bar with vertical at $A$: $30^\circ$ - Velocity of $A$, $v_A = 4$ m/s downward - Acceleration of $A$, $a_A = 6$ m/s$^2$ downward - Track at $B$ inclined $30^\circ$ downward to the right 3. **Formulas and Concepts:** - Velocity relation for rigid body rotation: $$\vec{v}_B = \vec{v}_A + \vec{\omega} \times \vec{r}_{B/A}$$ - Acceleration relation: $$\vec{a}_B = \vec{a}_A + \vec{\alpha} \times \vec{r}_{B/A} - \omega^2 \vec{r}_{B/A}$$ - $\vec{r}_{B/A}$ is the position vector from $A$ to $B$. - The bar length and angle give components of $\vec{r}_{B/A}$. - Directions must be resolved into components. 4. **Step 1: Define coordinate system and vectors** - Let $x$ axis be horizontal to the right, $y$ axis vertical upward. - Since $A$ velocity and acceleration are downward, $v_A = -4 \hat{j}$ m/s, $a_A = -6 \hat{j}$ m/s$^2$. - Position vector $\vec{r}_{B/A}$: length $0.6$ m, angle $30^\circ$ with vertical means: $$r_{B/A,x} = 0.6 \sin 30^\circ = 0.6 \times 0.5 = 0.3 \text{ m}$$ $$r_{B/A,y} = -0.6 \cos 30^\circ = -0.6 \times 0.866 = -0.5196 \text{ m}$$ (negative because $B$ is below $A$) 5. **Step 2: Velocity of $B$** - Velocity of $B$ along the track inclined $30^\circ$ downward to the right. - Unit vector along track at $B$: $$\hat{u}_B = \cos 30^\circ \hat{i} - \sin 30^\circ \hat{j} = 0.866 \hat{i} - 0.5 \hat{j}$$ - Let velocity of $B$ be $v_B$ along $\hat{u}_B$: $$\vec{v}_B = v_B (0.866 \hat{i} - 0.5 \hat{j})$$ - Angular velocity $\omega$ is scalar about axis perpendicular to plane (out of page), direction given by right-hand rule. - Cross product: $$\vec{\omega} \times \vec{r}_{B/A} = \omega \hat{k} \times (0.3 \hat{i} - 0.5196 \hat{j}) = \omega (0.5196 \hat{i} + 0.3 \hat{j})$$ - Velocity relation: $$\vec{v}_B = \vec{v}_A + \vec{\omega} \times \vec{r}_{B/A}$$ $$v_B (0.866 \hat{i} - 0.5 \hat{j}) = -4 \hat{j} + \omega (0.5196 \hat{i} + 0.3 \hat{j})$$ 6. **Step 3: Equate components for velocity** - $x$ component: $$v_B \times 0.866 = \omega \times 0.5196$$ - $y$ component: $$-v_B \times 0.5 = -4 + \omega \times 0.3$$ 7. **Step 4: Solve velocity equations** From $x$ component: $$v_B = \frac{0.5196}{0.866} \omega = 0.6 \omega$$ Substitute into $y$ component: $$-0.5 (0.6 \omega) = -4 + 0.3 \omega$$ $$-0.3 \omega = -4 + 0.3 \omega$$ $$-0.3 \omega - 0.3 \omega = -4$$ $$-0.6 \omega = -4$$ $$\omega = \frac{4}{0.6} = 6.67 \text{ rad/s}$$ Then velocity of $B$: $$v_B = 0.6 \times 6.67 = 4.0 \text{ m/s}$$ 8. **Step 5: Acceleration of $B$** - Let acceleration of $B$ be $a_B$ along track: $$\vec{a}_B = a_B (0.866 \hat{i} - 0.5 \hat{j})$$ - Angular acceleration $\alpha$ about $\hat{k}$. - Compute terms: $$\vec{\alpha} \times \vec{r}_{B/A} = \alpha \hat{k} \times (0.3 \hat{i} - 0.5196 \hat{j}) = \alpha (0.5196 \hat{i} + 0.3 \hat{j})$$ $$-\omega^2 \vec{r}_{B/A} = - (6.67)^2 (0.3 \hat{i} - 0.5196 \hat{j}) = -44.49 (0.3 \hat{i} - 0.5196 \hat{j}) = -13.35 \hat{i} + 23.1 \hat{j}$$ - Acceleration relation: $$a_B (0.866 \hat{i} - 0.5 \hat{j}) = -6 \hat{j} + \alpha (0.5196 \hat{i} + 0.3 \hat{j}) -13.35 \hat{i} + 23.1 \hat{j}$$ $$a_B (0.866 \hat{i} - 0.5 \hat{j}) = (-6 + 23.1) \hat{j} + \alpha (0.5196 \hat{i} + 0.3 \hat{j}) - 13.35 \hat{i}$$ $$a_B (0.866 \hat{i} - 0.5 \hat{j}) = 17.1 \hat{j} + \alpha (0.5196 \hat{i} + 0.3 \hat{j}) - 13.35 \hat{i}$$ 9. **Step 6: Equate components for acceleration** - $x$ component: $$0.866 a_B = 0.5196 \alpha - 13.35$$ - $y$ component: $$-0.5 a_B = 17.1 + 0.3 \alpha$$ 10. **Step 7: Solve acceleration equations** From $x$: $$0.866 a_B - 0.5196 \alpha = -13.35$$ From $y$: $$-0.5 a_B - 0.3 \alpha = 17.1$$ Multiply second equation by $1.732$ (approx $\frac{0.866}{0.5}$) to align $a_B$ coefficients: $$-0.866 a_B - 0.5196 \alpha = 29.66$$ Add to first equation: $$(0.866 a_B - 0.5196 \alpha) + (-0.866 a_B - 0.5196 \alpha) = -13.35 + 29.66$$ $$-1.0392 \alpha = 16.31$$ $$\alpha = -15.69 \text{ rad/s}^2$$ Substitute $\alpha$ into first equation: $$0.866 a_B - 0.5196 (-15.69) = -13.35$$ $$0.866 a_B + 8.15 = -13.35$$ $$0.866 a_B = -21.5$$ $$a_B = \frac{-21.5}{0.866} = -24.82 \text{ m/s}^2$$ 11. **Final answers:** - Angular velocity: $$\boxed{\omega = 6.67 \text{ rad/s}}$$ - Angular acceleration: $$\boxed{\alpha = -15.69 \text{ rad/s}^2}$$ - Velocity of roller $B$ along track: $$\boxed{v_B = 4.0 \text{ m/s}}$$ - Acceleration of roller $B$ along track: $$\boxed{a_B = -24.82 \text{ m/s}^2}$$ The negative acceleration means roller $B$ is decelerating along the track direction.