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Abhisar

  • one year ago

Let \(\theta\) denote angular displacement of a simple pendulum oscillating in a vertical plane. If the mass of the bob is (m), then show that the tension in string is mgCos(\(\theta\)) at the extreme positions.

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  1. Abhisar
    • one year ago
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    |dw:1441802230606:dw|

  2. Abhisar
    • one year ago
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    According to me T should be equal to \(\Large \frac{mV^2}{r}\) at extreme positions...??

  3. anonymous
    • one year ago
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    At extreme positions velocity is equal to zero but tension is not, so that formula is not correct (it is when you have a uniform rotatory movement but this is not since acceleration and velocity aren't constant). If you calculate the radial and the tangent components of mg (where the radial one is in the same direction of the string and the tangent one forms a pi/2 angulus with it), you see that, for every theta, the only forces acting on the mass are tension and weight, so that T = mgcos(theta). (whan theta equals to zero, you see that tension is mg, as expected) I hope to have been helpful and that my English will be understandable :D

  4. Abhisar
    • one year ago
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    Ahaan..Got it. Thanks c:

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