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.

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 \( 2 \, \text{rad/s} \).

In the mechanism shown on the left, at an instant in time whenOB is vertical its angular velocity is 𝜔 = 1.2 rad/s in thecounter-clockwise direction, and its angular acceleration is 𝛼 =2.4 rad/s2 in the clockwise direction. What is the acceleration ofpoint A at this instant? The length r is 2 m.Useful information: 𝛼⃗ 𝐵/𝐴 = 𝛼⃗𝐴𝐵 × 𝑟⃗𝐵/𝐴 − 𝜔2𝑟⃗𝐵/𝐴Hint: This problem is much easier if you find the instantaneouscentre of zero velocity for AB, in order to calculate 𝜔𝐴𝐵

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

A string is wrapped around a uniform solid cylinder of radius r𝑟, as shown in  (Figure 1). The cylinder can rotate freely about its axis. The loose end of the string is attached to a block. The block and cylinder each have mass m𝑚.Figure1 of 1Part AFind the magnitude α𝛼 of the angular acceleration of the cylinder as the block descends.Express your answer in terms of the cylinder's radius r𝑟 and the magnitude of the free-fall acceleration g𝑔.View Available Hint(s)for Part AHint 1for Part A. How to approach the problemHint 2for Part A. Find the net force on the blockHint 3for Part A. Find the net torque on the pulleyHint 4for Part A. Relate linear and angular accelerationHint 5for Part A. Putting it togetherActivate to select the appropriates template from the following choices. Operate up and down arrow for selection and press enter to choose the input value typeActivate to select the appropriates symbol from the following choices. Operate up and down arrow for selection and press enter to choose the input value typeα𝛼 =2(g+a)r