# Homework H1.H - Sp23

 Problem statement Solution video DISCUSSION
This problem involves the joint descriptions of Cartesian and polar coordinates: Cartesian coordinates are used to describe the path of P, and the polar description is to be used since r_dot = -vA for the motion of P.

As given in the problem statement, you are to first use the Cartesian description to derive the velocity and acceleration vectors for P. Since y = y(x), you will need to use the chain rule of differentiation to find y_dot:  y_dot = (dy/dx)*x_dot = (-4x)*x_dot. Taking an additional derivative gives: y_ddot = -4*x_dot^2 - 4*x*x_ddot. Using these in the Cartesian description gives v and a in terms of x_dot and x_ddot.

Now draw in the unit vectors of er and eθ: For θ = 0, we see that i = er and j = eθ. We use this to find x_dot and θ_dot from the velocity equation, and x_ddot from the acceleration equation.

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## 47 thoughts on “Homework H1.H - Sp23”

1. Raymond Ginter Frazee says:

I'm pretty stuck on this one, and so are most people I've spoken to. I've tried using polar, cartesian, and path descriptions, but nothing seems to work quite right. The closest to an answer for finding velocity would be from using path with the formula v_bar=ve_t but this winds up telling us that the particle is only moving in the y direction, which can't be right.
Am I on the right track with any of this?

1. Abraham Pugh says:

I too am stuck on this one. I think it would be hard to use polar coordinates due to us not knowing if theta_dot is constant. Based on the diagram when the y=0 there seems like there could be two answers for velocity and two answers for acceleration. If we set y=0 into the y(x) equation, we will result in +/- 2 for x. I personally looked back at Example 1.A.1 for reference and then solved for y_dot and x_dot in terms of each other. My velocity had one term where the x_dot would be (next to i_hat, although this term did not have x_dot in it as I said it was solved in terms of y_dot) and had the y_dot term minus the Va term in the j_hat direction.

2. CMK says:

I have added some additional points of discussion above related to this problem. Please check it out, and let us know if you have questions.

3. Lily Waterman says:

Based on the updated description, we should use cartesian and polar descriptions for this problem if that helps.

2. Sharon Ni says:

Do we need to find the relation between the cartesian components and the polar components, and rewrite the information (like relate the e^r and e^theta to the cartesian x and y coordinates)?

3. Samuel Elliot Greenaway says:

Does y_P = 0 mean that the particle is on the x-axis, and that theta is zero?

1. eaub says:

I believe that's right, theta is 0 for this analysis

2. CMK says:

Yes.

4. eaub says:

I am struggling to understand how the Va component relates to the velocity vector. Would it factor into that as the y-component or add on to the dy/dx*dx/dt?

1. CMK says:

If you use a polar description for the motion of P, then r_dot = -v_A.

5. Noah Benjamin Rush says:

When solving for y_dot you will need x_dot when using the chain rule to solve. My question is, how to we obtain x_dot to be used to find y_dot?

1. Clairice Holmes says:

I think for the start, just find y dot in terms of x dot, since the equation given for the y position is only in terms of x. Once you have the equations for velocity and acceleration in cartesian in terms of x dot and x ddot, you can use the polar equations to solve for the missing values. I assume you will have to use the relation of r_dot = -V_A, but I only just started the problem so I haven’t gotten to that point yet.

6. Madeline B says:

For the rate of change, can we solve a rate of change in polar? Or only in path?

7. Lily Waterman says:

Since theta is increasing to the left, when P is in the position where y = 0, would the x-coordinate of position therefore be negative?

8. Tyler Daniel Collins says:

I keep getting an acceleration that is negative in the j direction, but this seems wrong in the context of the problem and given animation. I think it stems from the fact that I found x_ddot to be equal to 0 since x_dot was a constant but I'm not sure if this was a correct calculation. Does anybody have an explanation for why acceleration can be downward in this problem or how my logic could be off?

1. Benjamin Carl Wassgren says:

I think that it makes sense that the acceleration is downwards, because as the point approaches the top of the arch, it will need to stop moving upwards, and after it crosses the top of the arch it will start moving downwards.

2. CMK says:

TYLER: It is NOT given that x_dot = constant.

1. Sam Plumer says:

But if x dot is equal to r dot and r dot is a constant, how is x dot not constant?

9. Jacob Russell Bunton says:

How do we solve for x_dot and x_ddot?

10. ZeBang Zhou says:

I'm not sure I did correct. Here is what I thought:

->r_dot is always equal to Va=1.5m/s
->when x=2, taking cartesian and polar as 2 coordinate.
->At this moment, we have theta = 0, r=2m , x_dot = (Va)= r_dot

But should I regard theta_dot, and theta_dotdot equal to 0 at this moment?

1. Sam Plumer says:

I think you are correct in saying x_dot is equal to r_dot, but what it seems like we should be doing is then equating y_dot (-4xx_dot) to r*theta_dot because that is the corresponding term in the polar equations. Then use this theta_dot to find x_ddot and y_ddot. However, when doing this I get different answers when doing y_ddot=r*theta_ddot+2*r*theta_dot (from the polar equation) versus using y_ddot in terms of x_dot and x_ddot (from chain rule), which is -4x_ddot^2-4*x*xddot. I am not sure if I am doing the correct approach though.

11. J Cena says:

I got to the point where theta=0 and I can equate i and er. However all that does is give me the fact that -8x_dot = r_dot. (which I do not know)

1. CMK says:

Don't forget that r_dot = -v_A.

12. J Cena says:

Also to find theta_dot would I be correct in making a triangle where the legs are y_dot and x_dot?

13. Elijah Allan Collins says:

I am still a little confused about how to use the unit vectors of er and eθ: For θ = 0, to find x_dot and θ_dot from the velocity equation, and x_ddot from the acceleration equation. I have my y_dot and y_dot_dot in terms of x, x_dot, and x_dot_dot but am confused after that. I would appreciate any help

1. Branson Dao says:

When you draw the particle at y_p (or look at the animation they give you), you can see that e_r becomes your i_hat unit vector and e_theta becomes your j_hat unit vector. From there you can use the polar velocity equation to identify the value that would be for x_dot.

The HW discussion gives it to you (x_dot = -V_a).

You can also isolate for x_dot/theta_dot (theta_dot becomes important in a moment) by using that polar velocity equation and substitution of the e_r/e_theta then solving for the necessary values.

From there I used the acceleration polar equation and solved for x_dot_dot in order to plug it into my y_dot_dot equation that I derived from the equation they originally give us.

Hopefully, this helps you out and I was clear enough in my explanation.

14. Melissa Jewel Brock says:

How do you find that rdot = -vA? And if the problem is asking for Cartesian coordinates, why do you need to know thetadot?

1. Branson Dao says:

If you use the polar velocity equation and make the necessary substitutions in regard to the unit vectors (for the e_theta and e_r), you are able to see that r_dot = -V_a.

If you also were to draw out the pulley, assuming the cable is perfectly straight at X = 0, you will see that V_a pulls in the negative x direction.

15. Meagan Theresa Figler says:

I have found the equations for y dot and y dot dot. However for y dot dot I cannot figure out how to solve for x dot dot using the equations given. I have that x dot dot = rdotdot - r theta dot ^2 however I dont understand how you are supposed to derive any of these values. I think you have to use the equation for y dot which is rtheta dot but I am not sure.

16. Lila Johnson says:

Since theta = 0, does that mean theta_dot would also = 0?

1. Lila Johnson says:

Also, since er=i when theta=0, does that mean x_dot=r_dot? And if I solve for x by setting y=0 in the y(x) equation, can I use x_dot and x in the y_dot=-4x*x_dot equation?

1. Liam Quentin Johnson says:

Yes that's how I did it, and we can say x_dot = -Va because r_dot = -Va

2. Branson Dao says:

No, I did not theta_dot = 0; if you use the polar velocity equation and make the necessary substitutions, you will find that the theta_dot can be isolated and will not be 0.

For example, if the velocity is pulling the ball up the curve (look at the animation provided), the theta is increasing which means theta_dot cannot be 0.

17. Sandra Edilia Bern says:

I'm still kinda confused about whether x_ddot is 0 or not. If it isn't 0 how would I solve for it?

1. Alexander M Toth says:

You can equate y_dot to r*theta_dot and then use theta_dot to solve for x_ddot

1. CMK says:

Agreed. x_ddot is not zero. You need to solve for this as Alexander describes above.

18. Peter Thomas Hays says:

Similar to how some people used theta=0 to state that e^r only has a component in the i^ direction, the same is also true for e^t. And because V=v*e^t, this would seem to suggest that there is zero velocity in the y direction at this point. Logically that doesn't make any sense, but the math seems to add up. Can anyone tell me why this could be wrong?

1. Justin James Paluch says:

Hey Peter, you are right with setting e^r = to i^ and e^t = j^ due to theta being zero. However, in order to find the velocity in the y direction at this point, you have to find y dot which is equal to dy/dx * dx/dt. Keep y dot in terms of x and x dot until you find their values and then once you plug them in you should get a positive y component for the velocity. I hope this helps!

19. Michael Lie Setiawan says:

When you set yp = 0, and you get ±2, why are we supposed to use +2 and not -2?

1. Matthew Elias Brunton says:

The diagram implies the assumption that you should use x = +2.

20. Matthew Elias Brunton says:

Alright, so I am stuck. I have solved all the way out to the acceleration vector. My problem here is that I have 3 unknowns...

X_doubledot
r_doubledot
theta_doubledot

Currently, I have solved for theta_dot and r_dot. Which of these am I to assume are constants to set their derivative to zero to be able to solve for y_doubledot and x_doubledot?

21. Elijah Allan Collins says:

In the explanation video, it says that the acceleration is really high. Does anyone know why that is? I got relatively low numbers but I guess I did it wrong.

1. Elijah Allan Collins says:

It.s just intuitively, the acceleration to be that high while the velocity is much lower doesn't make much sense to me.

2. CMK says:

Elijah: At what time marker value was "the acceleration is really high"? Let me go back to that point for the context, and I will get back with a response.

1. Elijah Allan Collins says:

At 10:02. It says the Acceleration is 260 Va^2

1. CMK says:

ELIJAH: You ask a good question here.

A little background first. Consider the position for P where the cable is perpendicular to the parabolic guide. When the cable is perpendicular to the guide, point A on the free end of the cable cannot move; that is, the ratio of speeds between A and P at that position goes as: v_A/v_P = 0. Taking the inverse of this, we see that v_P/v_A = infinity at that position. In words, if v_A ≠ 0, the v_P must be infinity.

In our problem at hand, we are specifying v_A as a constant value throughout. The mathematics says that the only way that this can be true when the cable is perpendicular to the guide is for P to be traveling infinitely fast. Practically speaking, this says that it is not possible for v_A to be a non-zero value at the point where the cable is perpendicular to the guide.

For the position that you are asked to analyze, P is near the position of where the cable is perpendicular to the guide. For this position, expect the acceleration and speed both to be very large at that position.

Note that the animation above stops near this point of where the cable is perpendicular to the guide. The reason for this is that the numerical calculations are not able to handle the infinite value for the speed of P. You can see how the speed of P is growing at a fast rate near this point.

Does this help?

1. Elijah Allan Collins says:

This does help, thank you!