Let's break the problem into logical steps: --- ### **Given Data:** - Angular velocity, \( \omega = 8 \) rad/s (**clockwise**) - Angular acceleration, \( \alpha = 19 \) rad/s² (**counterclockwise**) - Radius of tyre, \( r = 0.25 \) m - Linear velocity of the center, \( v_o = 10.23 \) m/s The direction of axes: - \( +x \) to the right - \( +y \) into the plane - \( +z \) upward Point \( p \) is at an angle \( 67^\circ \) from the center in the 2D view. --- ## **Step 1: Find the Velocity Vector at Point \( p \)** The velocity at point \( p \) on the rim is \[ \vec{v}_p = \vec{v}_o + \vec{\omega} \times \vec{r}_{p/o} \] ### **a. Find \(\vec{v}_o\):** \[ \vec{v}_o = v_o \hat{i} = 10.23 \hat{i} \text{ m/s} \] ### **b. Find \(\vec{\omega}\):** - Clockwise about the \( z \)-axis, which is \( -\hat{k} \): \[ \vec{\omega} = -8 \hat{k} \text{ rad/s} \] ### **c. Find the position vector \(\vec{r}_{p/o}\):** Point \( p \) is at radius \( r \) and angle \( 67^\circ \) from the center, so \[ \vec{r}_{p/o} = r(\cos 67^\circ \hat{i} + \sin 67^\circ \hat{j}) = 0.25 (\cos 67^\circ \hat{i} + \sin 67^\circ \hat{j}) \] \[ \cos 67^\circ \approx 0.3907,\quad \sin 67^\circ \approx 0.9205 \] \[ \vec{r}_{p/o} = 0.25 (0.3907 \hat{i} + 0.9205 \hat{j}) = (0.0977\hat{i} + 0.2301\hat{j}) \text{ m} \] ### **d. Compute \( \vec{\omega} \times \vec{r}_{p/o} \):** \[ \vec{\omega} \times \vec{r}_{p/o} = |-8\hat{k}| \times (0.0977\hat{i} + 0.2301\hat{j}) \] - \( \hat{k} \times \hat{i} = \hat{j} \) - \( \hat{k} \times \hat{j} = -\hat{i} \) So, \[ (-8\hat{k}) \times (0.0977\hat{i} + 0.2301\hat{j}) = -8\left(0.0977 \hat{k} \times \hat{i} + 0.2301 \hat{k} \times \hat{j}\right) \] \[ = -8(0.0977\hat{j} - 0.2301\hat{i}) \] \[ = (-8 \times 0.0977) \hat{j} + (8 \times 0.2301) \hat{i} \] \[ = -0.7816 \hat{j} + 1.8408 \hat{i} \] ### **e. Add to \( \vec{v}_o \):** \[ \vec{v}_p = 10.23 \hat{i} + 1.8408 \hat{i} - 0.7816 \hat{j} \] \[ = (12.0708 \hat{i} - 0.7816 \hat{j}) \text{ m/s} \] --- ### **Final Answer for Velocity at Point \( p \):** \[ \boxed{ \vec{v}_p = 12.07\,\hat{i} - 0.78\,\hat{j} \ \text{m/s} } \] --- ## **Step 2: Find the Acceleration of Point \( p \) in Vector Form** The acceleration at \( p \) is: \[ \vec{a}_p = \vec{a}_o + \vec{\alpha} \times \vec{r}_{p/o} + \vec{\omega} \times (\vec{\omega} \times \vec{r}_{p/o}) \] ### **a. Center acceleration, \( \vec{a}_o \):** Since the center is moving at constant velocity, \( \vec{a}_o = 0 \). ### **b. Angular acceleration vector, \( \vec{\alpha} \):** - Counterclockwise about \( z \)-axis → \( +\hat{k} \): \[ \vec{\alpha} = 19 \hat{k} \text{ rad/s}^2 \] ### **c. Compute \( \vec{\alpha} \times \vec{r}_{p/o} \):** \[ \vec{\alpha} \times \vec{r}_{p/o} = 19 \hat{k} \times (0.0977\hat{i} + 0.2301\hat{j}) = 19 (0.0977\hat{j} - 0.2301\hat{i}) \] \[ = (19 \times 0.0977) \hat{j} - (19 \times 0.2301) \hat{i} \] \[ = 1.8563\hat{j} - 4.372\hat{i} \] ### **d. Compute \( \vec{\omega} \times (\vec{\omega} \times \vec{r}_{p/o}) \):** First, find \( \vec{\omega} \times \vec{r}_{p/o} \) (already computed above): \[ \vec{\omega} \times \vec{r}_{p/o} = 1.8408 \hat{i} - 0.7816 \hat{j} \] Now, take cross product with \( \vec{\omega} \): \[ \vec{\omega} \times (\vec{\omega} \times \vec{r}_{p/o}) = (-8\hat{k}) \times (1.8408 \hat{i} - 0.7816 \hat{j}) \] \[ = -8\hat{k} \times 1.8408\hat{i} + 8\hat{k} \times 0.7816\hat{j} \] \[ = -8(1.8408\hat{k} \times \hat{i}) + 8(0.7816\hat{k} \times \hat{j}) \] \[ = -8(1.8408\hat{j}) + 8(-0.7816\hat{i}) \] \[ = -14.7264 \hat{j} - 6.2528 \hat{i} \] ### **e. Add all terms:** \[ \vec{a}_p = (1.8563\hat{j} - 4.372\hat{i}) + (-14.7264 \hat{j} - 6.2528 \hat{i}) \] \[ = (1.8563 - 14.7264)\hat{j} + (-4.372 - 6.2528)\hat{i} \] \[ = -12.8701 \hat{j} - 10.6248 \hat{i} \] --- ### **Final Answer for Acceleration at Point \( p \):** \[ \boxed{ \vec{a}_p = -10.62\,\hat{i} - 12.87\,\hat{j} \ \text{m/s}^2 } \] --- ## **Summary of Answers:** ### **Velocity at Point \( p \):** \[ \boxed{ \vec{v}_p = 12.07\,\hat{i} - 0.78\,\hat{j} \ \text{(m/s)} } \] ### **Acceleration at Point \( p \):** \[ \boxed{ \vec{a}_p = -10.62\,\hat{i} - 12.87\,\hat{j} \ \text{(m/s}^2\text{)} } \]