Air Track Glider and Hanging Block Dynamics

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Homework Help Overview

The discussion revolves around the dynamics of a system involving a glider on an air track and a hanging block, focusing on the relationship between their speeds and accelerations. The problem includes deriving equations that relate the speeds and accelerations of the glider and the hanging object, as well as calculating the tension in the string under specific conditions.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore geometric relationships and attempt to derive equations relating the speeds and accelerations of the glider and hanging block. Some question the assumptions made in the original poster's equations and suggest examining the geometry of the problem more closely.

Discussion Status

The discussion has progressed with participants providing hints and exploring various approaches to the problem. Some have successfully derived relationships and clarified definitions, while others continue to seek guidance on specific parts of the problem.

Contextual Notes

Participants note potential typographical errors in the original equations and discuss the implications of these errors on their reasoning. There is also mention of a broken attachment link that may affect the clarity of the problem setup.

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Homework Statement


1 1.00-kg glider on a horizontal air track is pulled by a string at an angle theta. The taut string runs over a pulley and is attached to a hanging object of mass 0.500 kg as in the below picture. (a) Show that the speed vx of the glider and the speed vy of the hanging object are related by vx = uvy, where u = z(z2 - h02)-2.

(b) The glider is released from rest. Show that at that instant the acceleration ax of the glider and the acceleration ay of the hanging object are related by ax = uay. (c) Find the tension in the string at the instant the glider is realeased for h0 = 80.0 cm and theta = 30.0 degrees.

Picture.jpg



Homework Equations



The usual laws of motion at an introductory level.

The Attempt at a Solution


Eh, I'm stymied. I've been so for a good year or so (like that other problem -__-). Here's my scratch work for (a) and (b). I found (c) fine enough.
T - 0.5g = 0.5ay -Tcos(theta) = ax = -T(z2 – h0^2)^.5/z -uax = T -uax - 0.5g = 0.5ay

Since nothing seemed to work, I decided to try to introduce a new force, "Ry" to represent the force the air pushed upwards on the glider. I also tried using the old kinematics equations. Note that "g(t)" and so forth is actually g * t--not a function of g at time t.

Ry +Tsin(theta) – g = 0 = Ry + Th0/z – g -T(z2-h02)1/2/z(t) = vx vy = 2T(t) – (Th0/z)(t) = 2T(t) - 0.5g(t) vy/(2T-(Th0/z)) = uvx/(T) = vy/(2 – h0/z) = -uvx vx = vy/(uho/2z – u) = 2zvy/(uh0 – 2zu)


(c) is easy enough. Tcos(30.0) = (1.00)ax = uay (0.500)g - T = (0.500)ay --> g - 2.00T = ay Now (in meters), z = h0/sin(30.0) = 1.6, so u = 1.6(1.62 - 0.82)-0.5 = 1.15 (approximately).

Thus, Tcos(30.0) = uay = 1.15(g - 2.00T) = 1.15g - 2.30T --> Tcos(30.0) + 2.30T = (cos(30.0) + 2.30)T = 1.15g. Solving for T then gives 3.56.


So, any pointers or hints for the first two parts? It was from an easy introductory course, so I'm probably missing something obvious.
 
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Its probably easier than you think. Try look at the geometry of the problem.

By the way, you're attachment link seems to be broken. I was able to open it directly if I removed "www." first.
 
Typo in the OP. u should = z(z2 - h02)-1/2.

Anyway, I've looked at the geometry of the problem and can see that cos(theta) = 1/u, but that's about as far as I can get besides using that knowledge with forces as can be seen in my scratch work. It's weird how when I do that I come up with u being multiplied by ax. Fortunately, acceleration can easily be found from velocity and vice versa in this case. I thought about using vcos(theta) = v/u = vx, but that doesn't work since the air glider's velocity (presumably) is only horizontal.

That's strange that the attachment did that. It's opening alright for me, so I'm not sure what to change. Thanks for catching it though! Sorry for the late response. I was busy studying for a math test.
 
You have a right triangle with side z, h0 and what we could call x (the horizontal side). These sides have a well-known relationship.

You also know the angle between z and x, so you can express x as a function of z and this angle, and then again as a function of z and u.

Finally, these those two equations should allow you to express u as a function of z and h0.
 
x = (z2 - h02)1/2

x/z = cos(theta) --> x = zcos(theta) = z/u

(z2 - h02)1/2 = z/u --> u = z(z2 - h02)-1/2


That gives me u, but I still don't see how to get the relationship vx = uvy from this information. =( The closest I can get is the reverse relationship, though with lots of other terms in there too (as seen in the above scratch work): uvx = vy (or the same thing but with accelerations).
 
If you know that dh/dt = vy what do you then know about dz/dt and can you relate that to dx/dt = vx?
 
Ohhhhhhh! I see it now!

Let w = x0 - x = the distance from the glider to the edge, where x0 is the original distance of the glider from the edge. w = (z2 - h02)1/2, so dw/dt = z(z2 - h02)-1/2(dz/dt) = u(dz/dt). Since h = h0 + (z0 - z), where z0 is the original value of z, dh/dt = -dz/dt. Thus, dw/dt = -u(dh/dt) = -uvy. Since vx = dx/dt = -dw/dt, -vx = -dx/dt = -uvy. Thus, vx = uvy, and the acceleration can be found by differentiating this with respect to t.

Finally done with this problem!


Thank you very much for your patience and your help!
 
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