Air Track Glider and Hanging Block Dynamics

[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 v_x of the glider and the speed v_y of the hanging object are related by v_x = uv_y, where u = z(z² - h₀²)^-2.

(b) The glider is released from rest. Show that at that instant the acceleration a_x of the glider and the acceleration a_y of the hanging object are related by a_x = ua_y. (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 = 0.5a_y -Tcos(theta) = a_x = -T(z² – h₀^2)^.5/z -ua_x = T -ua_x - 0.5g = 0.5a_y

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) = v_x v_y = 2T(t) – (Th₀/z)(t) = 2T(t) - 0.5g(t) v_y/(2T-(Th₀/z)) = uv_x/(T) = v_y/(2 – h₀/z) = -uv_x v_x = v_y/(uh_o/2z – u) = 2zv_y/(uh₀ – 2zu)


(c) is easy enough. Tcos(30.0) = (1.00)a_x = ua_y (0.500)g - T = (0.500)a_y --> g - 2.00T = a_y 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) = ua_y = 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.

 

[Filip Larsen]

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.

 

[Subdot]

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 a_x. Fortunately, acceleration can easily be found from velocity and vice versa in this case. I thought about using vcos(theta) = v/u = v_x, 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.

 

[Filip Larsen]

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.

 

[Subdot]

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 v_x = uv_y 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): uv_x = v_y (or the same thing but with accelerations).

 

[Filip Larsen]

If you know that dh/dt = v_y what do you then know about dz/dt and can you relate that to dx/dt = v_x?

 

[Subdot]

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) = -uv_y. Since v_x = dx/dt = -dw/dt, -v_x = -dx/dt = -uv_y. Thus, v_x = uv_y, 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!