To solve this problem, we need to determine the velocity of the collar at \( A \) and the angular velocity of \( AB \). Given that the crank \( BC \) rotates with a constant angular velocity of \( \omega_{BC} = 6 \, \text{rad/s} \), we can use the principles of rotational kinematics and relative velocity. ### Step 1: Determine the velocity of point \( B \) The velocity of point \( B \) due to the rotation of crank \( BC \) can be found using the formula: \[ v_B = \omega_{BC} \times r_{BC} \] where \( r_{BC} \) is the length of the crank \( BC \). Given: \[ \omega_{BC} = 6 \, \text{rad/s} \] \[ r_{BC} = 1 \, \text{m} \] So, \[ v_B = 6 \, \text{rad/s} \times 1 \, \text{m} = 6 \, \text{m/s} \] ### Step 2: Determine the direction of \( v_B \) Since \( BC \) rotates in the \( xy \)-plane, the velocity \( v_B \) will be perpendicular to \( BC \). Given the configuration, \( v_B \) will be in the negative \( y \)-direction. ### Step 3: Determine the velocity of collar \( A \) The velocity of collar \( A \) can be found using the relative velocity equation: \[ \vec{v}_A = \vec{v}_B + \vec{\omega}_{AB} \times \vec{r}_{A/B} \] Since \( A \) and \( B \) are connected by the rod \( AB \), and \( AB \) is perpendicular to the angular velocity vector \( \omega_{AB} \), we can write: \[ \vec{v}_A = \vec{v}_B + \vec{\omega}_{AB} \times \vec{r}_{A/B} \] Given: \[ \vec{v}_B = -6 \, \hat{j} \, \text{m/s} \] \[ \vec{r}_{A/B} = 3 \, \hat{k} - 5 \, \hat{j} \, \text{m} \] Assuming \( \vec{\omega}_{AB} = \omega_{AB} \, \hat{i} \), we get: \[ \vec{\omega}_{AB} \times \vec{r}_{A/B} = \omega_{AB} \, \hat{i} \times (3 \, \hat{k} - 5 \, \hat{j}) \] \[ = \omega_{AB} (5 \, \hat{k} + 3 \, \hat{j}) \] So, \[ \vec{v}_A = -6 \, \hat{j} + \omega_{AB} (5 \, \hat{k} + 3 \, \hat{j}) \] Since \( A \) is constrained to move vertically along the \( z \)-axis, the \( \hat{j} \) component of \( \vec{v}_A \) must be zero: \[ -6 + 3 \omega_{AB} = 0 \] \[ \omega_{AB} = 2 \, \text{rad/s} \] ### Step 4: Calculate the velocity of collar \( A \) Now, substituting \( \omega_{AB} \) back into the equation for \( \vec{v}_A \): \[ \vec{v}_A = -6 \, \hat{j} + 2 (5 \, \hat{k} + 3 \, \hat{j}) \] \[ \vec{v}_A = -6 \, \hat{j} + 10 \, \hat{k} + 6 \, \hat{j} \] \[ \vec{v}_A = 10 \, \hat{k} \, \text{m/s} \] So, the velocity of the collar at \( A \) is \( 10 \, \text{m/s} \) in the \( z \)-direction, and the angular velocity of \( AB \) is \( 2 \, \text{rad/s} \).

For the position shown where ߠ = 30°, point A on the sliding collar has a constant velocity ݒ = 0.3 ݉/ݏ with corresponding lengthening of the hydraulic cylinder AC. For this same position BD is horizontal and DE is vertical. Determine the angular acceleration of DE at this instant.

As shown in the figure, a 2 kg collar C is on an inclined rod at a distanceof L as 0.4 m from the rotation axis. The coefficient of static frictionbetween the collar and the rod is μs = 0.5 and the angular velocity of therod is ω. Determine the minimum and maximum angular velocities ofthe inclined rod so that the collar does not slip. Neglect the size of thecollar.

For the single slider crank mechanism shown in figure 1, the crank OBhas a constant clockwise angular velocity of 2000 rpm. For the crankposition indicated, determine (a) the angular velocity of the connectingrod BD, and (b) the velocity of the piston P (c) the angular accelerationof the connecting rod BD and the acceleration of point D (The Piston)

To solve this problem, we need to determine the time required for the reel to obtain an angular velocity of \(5 \, \text{rad/s}\) when subjected to a force \(F = 400 \, \text{N}\). We will use the principles of rotational dynamics and kinematics. ### Given: - Mass of the reel, \( m = 50 \, \text{kg} \) - Radius of gyration, \( k_G = 2 \, \text{m} \) - Force applied, \( F = 400 \, \text{N} \) - Desired angular velocity, \( \omega = 5 \, \text{rad/s} \) - Coefficient of kinetic friction, \( \mu_k = 0.2 \) - Radius of the reel, \( R = 3 \, \text{m} \) ### Steps to solve: 1. **Determine the moment of inertia \( I \) of the reel:** The moment of inertia \( I \) about the center of mass \( G \) is given by: \[ I = m k_G^2 = 50 \, \text{kg} \times (2 \, \text{m})^2 = 200 \, \text{kg} \cdot \text{m}^2 \] 2. **Determine the torque \( \tau \) applied by the force \( F \):** The torque \( \tau \) applied by the force \( F \) is: \[ \tau = F \times R = 400 \, \text{N} \times 3 \, \text{m} = 1200 \, \text{N} \cdot \text{m} \] 3. **Determine the frictional force \( f_k \):** The normal force \( N \) is equal to the weight of the reel: \[ N = m g = 50 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 490.5 \, \text{N} \] The kinetic frictional force \( f_k \) is: \[ f_k = \mu_k N = 0.2 \times 490.5 \, \text{N} = 98.1 \, \text{N} \] 4. **Determine the net torque \( \tau_{\text{net}} \):** The net torque \( \tau_{\text{net}} \) is the applied torque minus the torque due to friction. The frictional force acts at the radius \( R = 3 \, \text{m} \): \[ \tau_{\text{friction}} = f_k \times R = 98.1 \, \text{N} \times 3 \, \text{m} = 294.3 \, \text{N} \cdot \text{m} \] Therefore, the net torque is: \[ \tau_{\text{net}} = \tau - \tau_{\text{friction}} = 1200 \, \text{N} \cdot \text{m} - 294.3 \, \text{N} \cdot \text{m} = 905.7 \, \text{N} \cdot \text{m} \] 5. **Determine the angular acceleration \( \alpha \):** Using the relation between torque and angular acceleration: \[ \tau_{\text{net}} = I \alpha \] \[ \alpha = \frac{\tau_{\text{net}}}{I} = \frac{905.7 \, \text{N} \cdot \text{m}}{200 \, \text{kg} \cdot \text{m}^2} = 4.5285 \, \text{rad/s}^2 \] 6. **Determine the time \( t \) required to reach the angular velocity \( \omega \):** Using the kinematic equation for rotational motion: \[ \omega = \omega_0 + \alpha t \] where \(\omega_0 = 0 \, \text{rad/s}\) (initial angular velocity). Solving for \( t \): \[ t = \frac{\omega}{\alpha} = \frac{5 \, \text{rad/s}}{4.5285 \, \text{rad/s}^2} \approx 1.104 \, \text{s} \]

The triangular plate rotates about a fixed axis through point O. The angular velocity is ω =9 rad/s anti-clockwise and the angular acceleration is α = 2 rad/s2 clockwise. The lengths ofthe vertical side is h = 30 cm and the length of the horizontal side is b = 45 cm.What are the magnitudes of velocity and acceleration of point A at the instant shown