# Homework H3.A - Sp23

 Problem statement Solution video Discussion and hints Although the motion of B is made up of simple components (constant rotation rate for link OA, constant rotation rate for AB and constant extension of the telescoping link), the motion of B is quite complicated, as evidenced by the path of B and the velocity and acceleration of B shown above. However, for an observer on link AD, the path of B is rather simple: that observer sees a straight-line path for B, moving only in the x-direction. This is shown in the above animation giving the view of this moving observer. Are you able to visualize this observed motion of B?

Clarification
The angles theta and phi are BOTH measured from fixed, horizontal lines. Therefore, the rotation rate of OA is theta_dot, and the rotation of AD is phi_dot; that is, the rotation rate of AD is independent of theta_dot.

## 14 thoughts on “Homework H3.A - Sp23”

1. William Grant Dierking says:

For anyone struggling to find acceleration of a with e theta and e r, you can also do it using a rigid body kinematic equation.

1. CMK says:

This is not intended to be a polar coordinates problem in kinematics. Instead, you should apply the moving reference frame kinematics equations to the problem. Place the observer on link AD. As shown in the animation above, the motion of P is a straight line in the x-direction.

2. Allie says:

Is this assignment going to be due before or after the exam? Asking because it's not in Gradescope yet and it's not exam 1 content.

1. Emily says:

My gradescope says its due Wednesday, Feb 8th

3. Henry Francis Hellmann says:

Is the angular velocity just theta_dot plus phi_dot?

1. CMK says:

Place the observer of link AD. For the angular velocity of the observer, you should use phi_dot since that is the angular velocity of link AD.

4. Owen says:

I am assuming we use length b for r B/A when solving for velocity and acceleration as we are not shown if the length has changed in the right diagram?

1. Sharon Ni says:

Yes, you can use length b. The right diagram is the position that you are supposed to solve at, while the left diagram I assume is there to give you a better understanding of what the given measurements are, such as which angles are theta and which angles are phi.

5. Won Jun Lee says:

For the Velocity of B in respect to A, would it be positive B dot or negative B dot?

6. eaub says:

Where does the angular velocity of link OA come in for the equation? Should the first term of the equation be : a_B = a_A...? and how would you find a_A? or would it be a_B = a_O .... where a_O is = ? I'm just having trouble conceptualizing the relationships into an equation.

7. Madeline B says:

I want to make sure I am on the right path with this. Segment OA and AD are rigid links, meaning we can use the equations from chapter two. We can use a_A = a_O... where a_O = 0 and we can solve for a_A and ang. accel. Then, we can do a_B = a_A (with the solved a_A and angular acceleration = 0) using the equation from chapter 3. Is this somewhat on the right track?

1. Madeline B says:

Or are we treating point A as a pin with zero movement, so A does not have an acceleration?

1. CMK says:

Point A is on a rigid body connected to ground at O. Use the rigid body kinematics equations from Chapter 2. Don’t forget what we just learned.

8. Drew Thomas Casterton says:

Is it okay to substitute -ω^2(r_B/A) for the last (double cross product) term in the acceleration formula as was done in chapter 2?