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Torsional Pendulum Preliminary experiment

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Introduction

A2 Physics Coursework

Aim: To investigate a Torsional Pendulum.

Research and equations:

As we are working in circular motion, rather than linear motion, the equations that will help me investigate the Torsional pendulum will have to be derived. Here is how it is derived.

Using Force= Mass x Acceleration which is what you use for linear motion, this becomes Torque=Moment of Inertia x Angular acceleration.  Using Force= -kx from a simple pendulum, this becomes Force=- Torsional Constant x Angular displacement

Therefore image00.png    This can definitely be compared to a=-ω2x and becomes image01.pngHowever image09.png  therefore image05.png I then found out the exact expression which allowed me to directly work out I and K. The moment of inertia was simply image15.pngmL2  However for the Torsional constant I first found the formula for the polar moment of inertia which was Ipimage16.pngimage16.png=πd4/32 and the angle of twist φ=TL/GIp this was rearranged to T= GIp/L where T is the Torsional constant, then substituting in Ip I got Torsional constant= image17.pngimage18.pngUsing the equation  image05.png I can now substitute in expressions for I and K to get an overall equation which came out to be:  T=2πimage02.pngimage03.png   T=Time Period I=Moment of Inertia of the bar L=Length of wire G= Shear Modulus of material d= diameter of wire

The following web pages were used to help me derive these equations:

http://www.engin.umich.

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Middle

Using 0.1 meters as the starting point, make the length 0.1m using a meter rule, measuring from the base of the bung to the top of the bar at the knot.Turn the bar 90 degrees anticlockwise and release it, start the stopwatch at the same time of release.The time period for one complete oscillation is; for the end of the bar to go around clockwise once and changes direction then anticlockwise until it changes again, the moment it stops just before changing direction for a second time is one oscillation. Allow 5 complete oscillations for once length and divide the end time by five. Record the time period on a suitable table.Loosen the clamp and increase the length by 0.1m and repeat above steps until approximately 8 results are complete.Now measure the length of the bar using a meter ruler, and the diameter of the bar using a micrometer. Also measure the length of the wire using a meter ruler and its diameter using a micrometer. Record all these results.

To ensure that the experiment is carried out in safe environment I will make sure that I have plenty of space around me, with any obstacles removed to ensure the experiment can run smoothly.

Theory:

If simple harmonic motion applies, which I am

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Conclusion

±0.05seconds, therefore maximum reading error was (0.005/6.162)x100=0.081%, this is however a lot less significant than experimental error.The scale is accurate to ±0.05 grams. Therefore maximum error is (0.05/201.1) x 100 = 0.0249%, therefore this error was not so significant.The micrometer is accurate to ±0.005mm, as smallest division is 0.01mm, therefore error for my reading was (0.005/0.41) x 100 = 1.219%, this error was quite significant and a lot larger than I expected.

The value for the gradient I obtained was 0.4375, however I was expecting 0.5, therefore there is clearly errors in the time period and length, which is what determined the gradient, with reasons for these errors stated above. The error for the gradient will be the total error of the time and length, therefore approximately 6% error, when adding average most significant error of the time period and length.

Using the Equation T=2πimage02.pngimage03.png      I can work out the overall error of my experiment. As  2πimage03.pngimage10.png   x image11.pngimage11.png =T  and as I found out that T=16.788 x l0.4375

Therefore 2πimage10.pngimage12.png  should be equal to 16.788 if my experiment had no errors. I will now work out how close to this value I actually got.

     =2πimage13.pngimage14.png  = 14.12

Therefore the total error from what the true value should be is [(16.788-14.12)/16.788] x 100= 15.89%

From all the percentage errors above I can see that there are clearly issues with this preliminary experiment and that changes will have to be made for the final experiment to increase accuracy and reduce errors.

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