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Phy 231
Your 'cq_1_20.1' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** CQ_1_20.1_labelMessages **
The problem:
The metal strap used in the Angular Velocity of a Strap experiment is constrained by a vertical push pin to rotate about a hole in a die. The die is glued in place to a massive tabletop. A rubber band is attached to a point 15 cm from the axis and stretched so that it exerts a force of 3 Newtons, directed perpendicular to the rod. If this force is unopposed it will accelerate the system rapidly. You want to attach a second rubber band 5 cm from the axis to prevent the system from rotating.
How much force will that rubber band have to exert?
answer/question/discussion: ->->->->->->->->->->->-> sion:
If we imagine the point the first rubber band is attached to as a radius of 15cm on an imaginary circle, and the second rubber band at a point on a smaller circle with a radius of 5cm, we can see that a 3N force that accelerates a point on the larger circle at a certain rate would simultaneously accelerate a point the smaller circle at a lesser rate, because it is travelling a shorter distance every second.
Looking at it another way, a force exerted on the smaller circle would cause a point on it to accelerate at a certain rate, and would cause a point on the larger circle to accelerate at a greater rate.
So, to counteract the would-be acceleration of the 3N force on a point on the larger circle means counteracting a smaller would-be acceleration on a point on the smaller circle. So we would need a force smaller than 3N to cause a (theoretical) equal-but-opposite would-be acceleration at that point.
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Your argument here is very good, except that it implicitly assumes equal masses at both points.
Let's consider the effect of the 3 N force on the kinetic energy of the strap. If the 3 N force acts through a distance of, say, 1 cm it will do 3 N cm of work on the strap as the strap rotates through 1/15 radian. At the 5 cm point the same rotation would correspond only to a 1/3 cm distance. To do the same work, 3 times the force would be required.
Now if the 3 N force at the 15 cm position and an opposing 9 N force at the 5 cm position act simultaneously as the strap rotates, they 9 N force acts through only 1/3 the distance and for forces do zero work. If the system is stationary, it will therefore remain stationary.
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The force would be smaller in proportion to the difference in radii. (I don't think it's as simple as one-third the radius equals one-third the force, but maybe. In that case we'd need a force of 1N on the rubber band at the 5cm mark.)
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The intuitive idea here is leverage. The closer you are to the axis of rotation, the less leverage you have and the greater the force required.
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Once both of these forces are in place and the system is stationary, what (if anything) will happen if the glue holding the die to the tabletop comes loose?
answer/question/discussion: ->->->->->->->->->->->-> sion:
I don't think anything will happen. The forces are balanced, the system should be at rest. If the forces were not balanced, the strap would be rotating about the push pin.
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You give a good, but slightly flawed, argument from basic principles. Excellent thinking despite the flaw.
Check my note, then check out the discussion below. No revision is required:
&#See any notes I might have inserted into your document, and before looking at the link below see if you can modify your solutions. If there are no notes, this does not mean that your solution is completely correct.
Then please compare your old and new solutions with the expanded discussion at the link
Solution
Self-critique your solutions, if this is necessary, according to the usual criteria. Insert any revisions, questions, etc. into a copy of this posted document. Mark any insertions with &&&& so they can be easily identified.If your solution is completely consistent with the given solution, you need do nothing further with this problem. &#
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