# Air Track Glider and Hanging Block Dynamics

• Subdot
In summary, the conversation discusses a problem involving a glider on a horizontal air track being pulled by a string at an angle theta, with the string running over a pulley and attached to a hanging object of mass 0.500 kg. The conversation shows that the speed of the glider, vx, and the speed of the hanging object, vy, are related by vx = uvy, where u = z(z2 - h02)-1/2. It also shows that at the instant the glider is released from rest, the acceleration of the glider, ax, and the acceleration of the hanging object, ay, are related by ax = uay. The conversation concludes by finding the tension in the string at the instant
Subdot

## 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.

## 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!

existentialist_us

## What is an air glider?

An air glider is a type of aircraft that uses air currents to stay aloft and move forward. It does not have an engine and relies on the force of gravity and wind to fly.

## How does an air glider work?

An air glider works by using its wings to generate lift as it moves through the air. The glider's weight is balanced by the lift force, allowing it to stay airborne. By adjusting the angle of its wings, the glider can control its direction and speed.

## What is the purpose of a hanging block on an air glider?

A hanging block on an air glider is used to adjust the weight distribution of the glider. By shifting the position of the hanging block, the pilot can change the center of gravity of the glider, which affects its stability and maneuverability.

## Can an air glider fly without air currents?

No, an air glider requires air currents to stay aloft and move forward. Without air currents, the glider will lose lift and eventually descend to the ground.

## What is the difference between an air glider and a traditional airplane?

The main difference between an air glider and a traditional airplane is that an airplane has an engine to generate thrust and move through the air, while an air glider relies on gravity and wind to stay aloft and move forward.