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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

Is this considered math? Because I had a question like this, but it was through something as biology.

2. Abhisar

You can call it maths but it's basically physics. c:

3. raymonde20

Interesting.

4. raymonde20

10/$$\sqrt{3}$$ rad/s ?

5. raymonde20

I had professor help me on my question, he showed me how to do it and I completely blanked out. XD

6. anonymous

can u get a detailed explanation

7. ParthKohli

Pretty easy, let's do this!

8. ParthKohli

|dw:1441817703074:dw|

9. 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 }$

10. ParthKohli

Coming out as$\omega = \frac{5\sqrt3}{2}$

11. Abhisar

Hmm...but the answer is $$\Large \frac{10}{\sqrt{3}}$$

12. ParthKohli

yw

13. Abhisar

You have actually taken $$\sf tan \theta$$ wrong. You have written the reciprocal. c:

14. Abhisar

But, thanks. I get it now :)

15. 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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