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a

Is there another way of calculating it??

_{com}=3g/4.I have calculated it from the radial and tangential components.Is there another way of calculating it??

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- Thread starter Tanya Sharma
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- #26

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Is there another way of calculating it??

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- #27

ehild

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You can get the components of the acceleration of COM also from the acceleration of the tip.

a

ehild

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- #29

ehild

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ehild

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Okay...i understand what you say...

Why is there a -ve sign in the expression for N and ω^{2} ? I mean what does the -ve sign signify considering ω^{2} is a positive quantity.

Why is there a -ve sign in the expression for N and ω

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- #31

ehild

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Have you got the direction of the acceleration of the COM?

ehild

- #32

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Have you got the direction of the acceleration of the COM?

ehild

Yes , I have got the direction of acceleration of COM.I worked with radial and tangential components and found their resultant i.e a

- #33

ehild

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You would get the same from the x,y components of acceleration. I like this method, as I usually get confused with the radial and tangential components.

The coordinates of the COM of a homogeneous rod with the ends at (x1,y1) and (x2,y2) are Xcom=(x1+x2)/2, Ycom=(y1+y2)/2. (x1,y1)=(0,0), and (x2,y2)=(x,y) here, so Xcom=x/2, Ycom=y/2.

You derived the x component of acceleration already, the same has to be done with the y component.

y=Lsin(θ) y'=Lcos(θ) θ',

y''= L(-sin(θ)θ'

When loosing contact, y"=L(-1/2 ω

From the condition that the x component of acceleration is zero, you have got that

ω'=-√3 ω

and Ycom"=-3g/4.

ehild

- #34

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Its a fantastic experience learning from you .You taught me Physics ,Calculus,Polar Coordinates all in one question

- #35

ehild

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Its a fantastic experience learning from you .You taught me Physics ,Calculus,Polar Coordinates all in one question

So you discovered that it was about relation between polar and Cartesian coordinates? I did not want to confuse you by speaking about polar coordinates: :tongue2:

ehild

- #36

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yes process is correct

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