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.
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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?
The work on the .15m side is .15m*3N=.45J, for them to equal out the force on the .05m side will need to equal .45J. Therefore, .05m*W=.45J, divide by .05m, W = 9N.
Good.
However note that work is equal to force * displacement along the line of the force. In this case nothing is moving so there is no displacement, so no work is involved here.
However the units are identical to those of work, so it isn't implausible to call this work.
In fact what this product gives you it torque. 3 N is the force, .15 m is called the 'moment-arm' of the force, and the torque is the product of moment-arm and force: .15 m * 3 N = .45 m * N. Note that I've expressed this as m * N to indicated that this is a torque and not work; in fact m * N and N * m are mathematically identical units with two different physical interpretations, the choice of m * N for torque or N * m for work is more or less a grammatical convention.
However when used to refer to torques, m * N (or N * m) does not equal Joules. Joules measure work, not torque.
Here we say that since the system is in equilibrium (not accelerating) the torques must be equal and opposite, so that | .15 m * 3 N | = | .05 m * F |, where F stands for the unknown force. The solution F = .45 m N / (.05 m) = 9 N, as you indicate.
That's enough for the time being. There are also conventions for signs of forces and torques, which will be developed within the next few assignments.
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?
The Strap will launch toward you. It will be under 12N of force, I think.
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5mins
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I guessed on the amount of force for the second question. I wasn't for sure how to figure it out, or if I know how to figure it out.
It was a little more than a guess, it was a sensible conclusion, with the added benefit that it turns out to be pretty much correct. The unequal forces will in fact cause some of the energy of the rubber bands to go into the rotation of the system, and the forces exerted by the rubber bands begin to change as they constract. However in the stationary system, the pin through the axis of rotation would exert a force of 12 N.
&#This looks good. See my notes. Let me know if you have any questions. &#