Hoop Point B Acceleration 78E207

1. **Problem statement:** We have a hoop rolling on a rough surface with angular velocity $\omega = 4$ rad/s and angular acceleration $\alpha = 5$ rad/s$^2$. The center $O$ has velocity $v_O = 5$ m/s and acceleration $a_O = -2$ m/s$^2$ (deceleration). We want to find the acceleration of point $B$ located at radius $r = 0.3$ m from $O$ at an angle of $45^\circ$ below the horizontal. 2. **Known values:** - $\omega = 4$ rad/s - $\alpha = 5$ rad/s$^2$ - $v_O = 5$ m/s (rightward) - $a_O = -2$ m/s$^2$ (leftward, so acceleration vector points left) - $r = 0.3$ m - Angle of $OB$ with horizontal: $45^\circ$ downward 3. **Formula for acceleration of a point on a rotating body:** $$\vec{a}_B = \vec{a}_O + \vec{\alpha} \times \vec{r} - \omega^2 \vec{r}$$ where: - $\vec{a}_O$ is acceleration of center $O$ - $\vec{\alpha}$ is angular acceleration vector - $\vec{r}$ is position vector from $O$ to $B$ - $\omega^2 \vec{r}$ term is centripetal acceleration directed toward center 4. **Set coordinate system:** - Positive $x$-axis to the right - Positive $y$-axis upward 5. **Vectors:** - $\vec{a}_O = -2 \hat{i}$ m/s$^2$ (leftward) - $\vec{\alpha} = 5 \hat{k}$ rad/s$^2$ (out of plane, counterclockwise) - $\vec{r} = r(\cos 45^\circ \hat{i} - \sin 45^\circ \hat{j}) = 0.3(0.7071 \hat{i} - 0.7071 \hat{j}) = 0.2121 \hat{i} - 0.2121 \hat{j}$ m 6. **Calculate $\vec{\alpha} \times \vec{r}$:** $$\vec{\alpha} \times \vec{r} = 5 \hat{k} \times (0.2121 \hat{i} - 0.2121 \hat{j}) = 5(0.2121 \hat{k} \times \hat{i} - 0.2121 \hat{k} \times \hat{j})$$ Using right-hand rule: - $\hat{k} \times \hat{i} = \hat{j}$ - $\hat{k} \times \hat{j} = -\hat{i}$ So: $$= 5(0.2121 \hat{j} + 0.2121 \hat{i}) = 5(0.2121 \hat{i} + 0.2121 \hat{j}) = 1.0605 \hat{i} + 1.0605 \hat{j}$$ 7. **Calculate centripetal acceleration $-\omega^2 \vec{r}$:** $$-\omega^2 \vec{r} = -16 (0.2121 \hat{i} - 0.2121 \hat{j}) = -3.3936 \hat{i} + 3.3936 \hat{j}$$ 8. **Sum all components:** $$\vec{a}_B = \vec{a}_O + \vec{\alpha} \times \vec{r} - \omega^2 \vec{r} = (-2 \hat{i}) + (1.0605 \hat{i} + 1.0605 \hat{j}) + (-3.3936 \hat{i} + 3.3936 \hat{j})$$ Combine $\hat{i}$ components: $$-2 + 1.0605 - 3.3936 = -4.3331$$ Combine $\hat{j}$ components: $$0 + 1.0605 + 3.3936 = 4.4541$$ So: $$\vec{a}_B = -4.3331 \hat{i} + 4.4541 \hat{j}$$ 9. **Calculate magnitude of acceleration:** $$|\vec{a}_B| = \sqrt{(-4.3331)^2 + (4.4541)^2} = \sqrt{18.77 + 19.84} = \sqrt{38.61} = 6.21 \text{ m/s}^2$$ 10. **Calculate direction (angle $\theta$ above horizontal to left):** $$\theta = \tan^{-1} \left( \frac{4.4541}{4.3331} \right) = \tan^{-1}(1.028) = 45.6^\circ$$ Since $\hat{i}$ component is negative and $\hat{j}$ positive, acceleration points $45.6^\circ$ above the negative $x$-axis (to the left and upward). **Final answer:** The acceleration of point $B$ is approximately $$6.21 \text{ m/s}^2$$ directed $45.6^\circ$ above the horizontal to the left.