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Abhisar

  • one year ago

A small mass of 10 gm lies in a hemispherical bowl of radius 0.5 m at a height of 0.2 m from the bottom of the bowl. The mass will be in equilibrium if the bowl rotates at an angular speed of? The answer should be 10/\(\sqrt{3}\) rad/s

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  1. raymonde20
    • one year ago
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    Is this considered math? Because I had a question like this, but it was through something as biology.

  2. Abhisar
    • one year ago
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    You can call it maths but it's basically physics. c:

  3. raymonde20
    • one year ago
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    Interesting.

  4. raymonde20
    • one year ago
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    10/\(\sqrt{3}\) rad/s ?

  5. raymonde20
    • one year ago
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    I had professor help me on my question, he showed me how to do it and I completely blanked out. XD

  6. anonymous
    • one year ago
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    can u get a detailed explanation

  7. ParthKohli
    • one year ago
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    Pretty easy, let's do this!

  8. ParthKohli
    • one year ago
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    |dw:1441817703074:dw|

  9. ParthKohli
    • one year ago
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    \[\rm N\sin \theta = mg\tag{vertical}\]\[\rm N\cos \theta = m R\omega^2\tag{horizontal} \]Divide:\[\rm \tan \theta = \frac{R\omega^2 }{g}\]\[\Rightarrow \frac{3}{4}=\frac{0.4 \times \omega^2}{10 }\]\[\Rightarrow \omega^2 = \frac{30}{1.6 }=\frac{300}{16 }\]

  10. ParthKohli
    • one year ago
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    Coming out as\[\omega = \frac{5\sqrt3}{2}\]

  11. Abhisar
    • one year ago
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    Hmm...but the answer is \(\Large \frac{10}{\sqrt{3}}\)

  12. ParthKohli
    • one year ago
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    yw

  13. Abhisar
    • one year ago
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    You have actually taken \(\sf tan \theta\) wrong. You have written the reciprocal. c:

  14. Abhisar
    • one year ago
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    But, thanks. I get it now :)

  15. ParthKohli
    • one year ago
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    Yes, goddammit. I kept refreshing the page again and again just to correct this small mistake that I immediately realised. Why does OS always hang at these seriously vital moments?

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spraguer (Moderator)
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