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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
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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raymonde20
 one year ago
Best ResponseYou've already chosen the best response.0Is this considered math? Because I had a question like this, but it was through something as biology.

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.0You can call it maths but it's basically physics. c:

raymonde20
 one year ago
Best ResponseYou've already chosen the best response.010/\(\sqrt{3}\) rad/s ?

raymonde20
 one year ago
Best ResponseYou've already chosen the best response.0I had professor help me on my question, he showed me how to do it and I completely blanked out. XD

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0can u get a detailed explanation

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2Pretty easy, let's do this!

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2dw:1441817703074:dw

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2\[\rm N\sin \theta = mg\tag{vertical}\]\[\rm N\cos \theta = m R\omega^2\tag{horizontal} \]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
 one year ago
Best ResponseYou've already chosen the best response.2Coming out as\[\omega = \frac{5\sqrt3}{2}\]

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.0Hmm...but the answer is \(\Large \frac{10}{\sqrt{3}}\)

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.0You have actually taken \(\sf tan \theta\) wrong. You have written the reciprocal. c:

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.0But, thanks. I get it now :)

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2Yes, 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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