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Hinged Stick Experiment

by Shoelace Thm.
Tags: experiment, hinged, stick
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Shoelace Thm.
#1
Dec3-12, 09:50 PM
P: 60
Suppose I have a hinged meter stick of length [itex] L [/itex] and mass [itex] M [/itex] angled at angle [itex] \theta [/itex], where one end is free to rotate while the other end serves as a rotation axis. There is a ball sitting on a golf tee at the edge and a cup sitting [itex] L \cos \theta [/itex] up the ramp. As the stick falls, its free end starts to accelerate above gravity (though that is not necessarily the case when it starts!)

I have that [itex] \ddot{\theta} =\frac{3g}{2l} \cos \theta [/itex] but how can I get [itex] \theta [/itex] as a function of [itex] t [/itex]? The goal is to eventually find the critical angle where the cup reaches the ground right before the ball.

Of course, if there is an easier way to find this critical angle, please let me know.
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haruspex
#2
Dec3-12, 10:44 PM
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I don't understand the set-up (e.g. edge of what?), but just looking at this equation:
[itex] \ddot{\theta} =\frac{3g}{2l} \cos \theta [/itex]
I don't think you can solve that in closed form. If θ is not too large, you can expand cos θ as ≈ 1 - θ2/2, if that helps.
Shoelace Thm.
#3
Dec3-12, 11:03 PM
P: 60
Well the stick is hinged at one end attached to the ground. The other end is lifted at a certain angle to the horizontal. The ball is sitting at the very end of the of the lifted end (distance [itex] L [itex/] from the hinge) and the cup is attached pretty close to that point. Upon release, under certain conditions the ball will land in the cup although the cup must travel farther.

If the setup is still unclear, this is a rather well-known experiment. Just google the title. However, I have yet to find a rigorous treatment of the physics behind this experiment. Most assume that the acceleration does not increase as the stick falls. Thus the minimal angle found in this way is too large.

haruspex
#4
Dec4-12, 12:33 AM
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Hinged Stick Experiment

OK, I found a good demonstration at http://video.mit.edu/watch/hinged-st...ing-ball-6092/.
Unless you consider g to be negative, your ODE has a sign wrong. [itex]\ddot{\vartheta}[/itex] will be negative.
You can do one integration step, or equivalently write down the energy equation, but it looks even worse:
[itex]\int_{t=0}\sqrt{\frac{L}{3g(sin(\vartheta_0)-sin(\vartheta))}}d \vartheta[/itex]
The initial vertically downward component of acceleration of the tip of the plank will exceed g if cos2θ0 > 2/3.
Shoelace Thm.
#5
Dec4-12, 01:16 AM
P: 60
Why must [itex] \ddot{\theta } [/itex] be negative?

How did you obtain that energy equation and that integral? Also how would you perform the integration step or go about evaluating that integral?

And that last statement is true but it will not yield the minimum angle. The cup need not initially have a downward acceleration greater than g because its downward acceleration increases as theta decreases (a is proportional to cos theta). This is why I am introducing any of these equations in the first place.
haruspex
#6
Dec4-12, 03:01 AM
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Quote Quote by Shoelace Thm. View Post
Why must [itex] \ddot{\theta } [/itex] be negative?
Because θ is measured up from the horizontal. Initially it is > 0, and when released from rest it reduces. Therefore the second derivative is negative.
How did you obtain that energy equation and that integral?
In terms of energy, -ΔPE = ΔKE.
Working with the 2nd order ODE, multiply each side by [itex] \dot{\theta } [/itex] and each side becomes integrable.
Also how would you perform the integration step or go about evaluating that integral?
I doubt there is a closed form.
And that last statement is true but it will not yield the minimum angle.
No, I didn't mean to suggest it would. But how do you define minimum angle? Don't you first have to specify the horizontal and vertical distances (when the plank is level) from the tee to the lip of the cup? The smaller you can make those, the smaller the angle of release can be, zero being the limit.
The cup need not initially have a downward acceleration greater than g because its downward acceleration increases as theta decreases (a is proportional to cos theta).
There's a stronger reason than that. The ball is physically unable to overtake the plank, so above that critical angle the ball stays on the tee (or simply falls off if the tee is too narrow) until that angle is reached.
Shoelace Thm.
#7
Dec4-12, 08:18 AM
P: 60
The horizontal distance between ball and cup is L - Lcosθ. Assume that the vertical distance is 0, i.e. the cup has no height. This will make things simpler, the only condition being that the cup hits ground before ball does rather than by a certain interval of time.

Now we should find the minimal angle. None of your expressions have closed forms you say so how to go about this? I would like to find a number so I can confirm that it is indeed smaller than [itex] \cos ^{-1} (\sqrt{\frac{2}{3}}) [/itex]
mfb
#8
Dec4-12, 11:43 AM
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I don't think there is a minimal angle. To get a purely vertical movement of the ball, the angle is limited, however: The angle to the vertical axis has some minimal value and the angle to the table has a maximal value.
This is given by the condition that the point below the ball accelerates with >g. The acceleration in direction of rotation is ##-\frac{3}{2}g\cos(\theta)##, with a component of ##-\frac{3}{2}g\cos(\theta) \cos(\theta)## downwards. This gives [edit] ##\cos^2(\theta) = \frac{2}{3}## as critical angle.
Shoelace Thm.
#9
Dec4-12, 12:57 PM
P: 60
The movement of the ball does not have to be purely vertical. In fact above the critical angle it certainly won't be.

I think its intuitively clear that there is a minimal angle. You have given an angle at which the acceleration of the tip (which is strictly increasing as the stick falls) is always greater than g. As I keep increasing this angle the cup takes longer to reach the critical angle. Eventually there should be a point where it will not reach the ground before the ball.

EDIT: I understand what you are saying; the angle to the table has some maximal value. But that angle should be larger than the critical angle everyone is giving because of what I stated in an earlier post.
mfb
#10
Dec4-12, 01:00 PM
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P: 11,574
If you increase/decrease the angle (I am still confused by that definition), the ball will aquire a horizontal velocity as well as it sticks to the tee initially - but it will always land after the stick, and you can place the cup at the right position.
Shoelace Thm.
#11
Dec4-12, 02:46 PM
P: 60
haruspex has posted a video. By increasing the angle I mean lifting the stick higher at the start.

Why will the ball always land after the cup? Is this because above the critical angle the presence of the stick keeps the ball from accelerating downwards at g and below the critical angle the downwards acceleration of the cup exceeds g? I believe this covers all cases.
haruspex
#12
Dec4-12, 02:55 PM
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Quote Quote by mfb View Post
I don't think there is a minimal angle.
I agree. If the release angle is θ0 to the horizontal, the cup position needs to be at L(1-cos(θ0)) from the ball, and that's always > 0. Below the critical angle, the tee accelerates faster than the ball. If the top of the cup is no higher than the top of the tee, that should be enough.
To get a purely vertical movement of the ball, the angle is limited, however: The angle to the vertical axis has some minimal value and the angle to the table has a maximal value.
This is given by the condition that the point below the ball accelerates with >g. The acceleration in direction of rotation is ##-\frac{3}{2}g\cos(\theta)##, with a component of ##-\frac{3}{2}g\cos(\theta) \cos(\theta)## downwards. This gives ##\cos^2(\theta) = \frac{3}{2}## as critical angle.
2/3, not 3/2.
Quote Quote by Shoelace Thm.
Why will the ball always land after the cup? Is this because above the critical angle the presence of the stick keeps the ball from accelerating downwards at g and below the critical angle the downwards acceleration of the cup exceeds g? I believe this covers all cases.
Yes.


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