# Homework H2.I - Sp23

 Problem statement Solution video Any questions??

DISCUSSION
In some sense, this is a very standard kinematics of rigid bodies problem. A rigid link connects points A and B. To relate the motion of these two points, you will need the following kinematics equations:

vA = vB + ω x rA/B
aA = aBα x rA/B - ω2 rA/B

The nuance in this problem is the acceleration of point A. From what we learned earlier in the semester, the acceleration of a point can be written in terms of its path coordinates; that is, here the acceleration of A can be resolved into its tangential and normal components. As you proceed on this problem, you first need to recognize the tangential and normal unit vectors, et and en, for the motion of A. Then write down the acceleration of A, first in its path components, and then in its Cartesian component. In the end, you will have two scalar equations coming from the rigid body acceleration equation in terms of two unknowns. ## 26 thoughts on “Homework H2.I - Sp23”

1. Sharon Ni says:

When I write my a_A equation, how do I separate the i and j components, where do we get the other equation to set the components equal to?

1. CMK says:

You need to relate the motion of A and B through the rigid body kinematics equations. Use the fact that you know the path of A. That is, you know the direction of the velocity for A, and you know the directions for the tangential and normal components of acceleration of A. Balance out the i and j components of the acceleration equation relating A and B, and using the path description for A.

2. Lorraine Nangatie Zaumu says:

Try using the other definition of acceleration provided to us at the beginning of the semester that states that acceleration is the sum of the rate of change of speed (v_dot) and velocity divided by rho.

2. Henry Francis Hellmann says:

Can we assume that v_A is in the i direction?

1. CMK says:

v_A is tangent to the semi-circular guide, with A being at the top of this guide for this instant.

2. Jack Charles Sampson says:

Yes you can assume that

3. Arshon Bozorgi says:

v_a is always tangent when it is on a circular path

3. Sam Plumer says:

How do find v dot to use in the path description for a's acceleration?

1. CMK says:

Sam: As state in the above DISCUSSION, your kinematics equation will produce two scalar equations with two unknonwns. Those two unknowns are the angular acceleration of AB and v_dot for A. Solve those two equations together for your answers.

4. Lily Waterman says:

I am confused about how to find the velocity and acceleration of link AB. Should I be using the instant center approach?

1. Elijah Allan Collins says:

No, I believe that you should use the kinematic equations given underneath the problem in the Discussion portion.
vA = vB + ω x rA/B
aA = aB + α x rA/B - ω2 rA/B

5. J Cena says:

I am getting very large numbers for a. Is this normal because the measurements given are in mm or did I make a mistake somewhere.

1. Allie says:

I'm also getting really large numbers. I'm hoping that this is just because the units are mm and the given parameters are in the hundreds because other than that I'm pretty confident in my processes so I think you're probably okay on that.

6. Michael Bradley Mayhew says:

Once we solve for the Cartesian acceleration equations, are we able to plug them straight into vectors et and en? I am thinking this because et is purely horizontal and en is purely vertical, so theoretically wouldn't those two vectors equal the i and j vectors from Cartesian?

1. Michael Bradley Mayhew says:

It feels like I should be able to, but I am not sure what I would set the equations equal to. I think both the i and j components of acceleration are non-zero, and I am not sure what other values can be put in to solve the two equations.

1. CMK says:

When working with the acceleration equation for link AB:

aA = aB + α x rA/B - ω2 rA/B

you need to specify what you know about aA and aB.

For aB, it is a known value aB in a direction that is aligned with the ramp.

For aA, you know that it has two components: one in the e_t direction (the unknown, vA_dot) and one in the e__n direction (the known, vA^2/rho).

Substitute both of those expressions for aA and aB into the acceleration equation above, and balance out your i and j components. This gives you two scalar equations in terms of two unknowns: vA_dot and alpha_AB. Solve for those two unknowns.

1. Peter Thomas Hays says:

Is aB not zero? The problem statement makes it clear that vB is constant.

1. CMK says:

Sorry, I was thinking about a different version of the problem from an earlier semester. Yes, here aB = 0.

7. Peter Thomas Hays says:

I've written out both my acceleration equations, but I'm not sure how to relate the two of them. Do I use the angle we are given to convert the path equations to cartesian with trig functions?

1. CMK says:

Peter: Not sure of the acceleration equations to which you refer here. You have really only one acceleration equation relating the motions of A and B. In that equation, you need to provide expressions for aA and aB (the latter is zero, as you have pointed out earlier). This gives you one vector equation, which becomes two scalar equations. Solve for the unknowns. Please read the detailed response that was given earlier, and repeated below.
______________________________

When working with the acceleration equation for link AB:

aA = aB + α x rA/B - ω2 rA/B

you need to specify what you know about aA and aB.

For aB, it is a known value aB in a direction that is aligned with the ramp.

For aA, you know that it has two components: one in the e_t direction (the unknown, vA_dot) and one in the e__n direction (the known, vA^2/rho).

Substitute both of those expressions for aA and aB into the acceleration equation above, and balance out your i and j components. This gives you two scalar equations in terms of two unknowns: vA_dot and alpha_AB. Solve for those two unknowns.

8. Nadra Dunston says:

At the top of the arc path that point A follows, the acceleration components of A would be exactly vertical and horizontal in Path and Cartesian coordinates. So, is finding expressions for i and j and equating them to the expressions for et and en the correct method? How should I handle v_dot and v^2 in the path expressions? If they're the same as v_A and a_A, it seems like I would end up going in equation circles.

1. Mason Fellwock says:

I am not sure if this is right but what I did was get my e_t and e_n equations set up and then I said that V^2/rho = wAB^2*R because w=v/r --> w^2=v^2/r^2. This allowed me to solve for alpha AB but i got a very small number for my angular acceleration.

1. Zachary William Delahunty says:

I did something similar to this. I set e^n equal to the j vector of acceleration so I had aAj = Va^2/P=pe^n. I believe that for this problem the j and e^n vectors are equivalent considering the position of particle A which allows us to equate these two vectors.

2. Myles Christian Kappes says:

The way I saw it, e_t is in the same direction as v_A (i_hat) because they are both tangent to the semicircular path. This made e_n point in the -j_hat direction. Using this, I broke my kinematic accel. equation ( aA = aB + α x rA/B - ω2 rA/B ) into i and j, and then replaced i and j with vA_dot and vA^2/rho, respectively. This allowed me to solve for alpha and vA_dot.

9. Anthony Michael Cialdella says:

To make this problem easier can we assume that the acceleration of B is zero since it says that Vb is constant?

1. CMK says:

It is not an assumption. It is given that v_B = constant, and therefore for a straight-line path, the acceleration of B is zero.