Abhisar
  • Abhisar
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
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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raymonde20
  • raymonde20
Is this considered math? Because I had a question like this, but it was through something as biology.
Abhisar
  • Abhisar
You can call it maths but it's basically physics. c:
raymonde20
  • raymonde20
Interesting.

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raymonde20
  • raymonde20
10/\(\sqrt{3}\) rad/s ?
raymonde20
  • raymonde20
I had professor help me on my question, he showed me how to do it and I completely blanked out. XD
anonymous
  • anonymous
can u get a detailed explanation
ParthKohli
  • ParthKohli
Pretty easy, let's do this!
ParthKohli
  • ParthKohli
|dw:1441817703074:dw|
ParthKohli
  • ParthKohli
\[\rm N\sin \theta = mg\tag{vertical}\]\[\rm N\cos \theta = m R\omega^2\tag{horizontal} \]Divide:\[\rm \tan \theta = \frac{R\omega^2 }{g}\]\[\Rightarrow \frac{3}{4}=\frac{0.4 \times \omega^2}{10 }\]\[\Rightarrow \omega^2 = \frac{30}{1.6 }=\frac{300}{16 }\]
ParthKohli
  • ParthKohli
Coming out as\[\omega = \frac{5\sqrt3}{2}\]
Abhisar
  • Abhisar
Hmm...but the answer is \(\Large \frac{10}{\sqrt{3}}\)
ParthKohli
  • ParthKohli
yw
Abhisar
  • Abhisar
You have actually taken \(\sf tan \theta\) wrong. You have written the reciprocal. c:
Abhisar
  • Abhisar
But, thanks. I get it now :)
ParthKohli
  • ParthKohli
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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