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Cyrus

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- Thread starter FZX Student
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- #36

Cyrus

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- #37

Sirus

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ZapperZ said:Sigh... maybe we ALL should wait till the original poster comes back and explain if this is all in a horizontal flat plane, or if *I* was the one who simply interpreted the question wrong...

Zz. <smacks himself silly with a baseball bat>

Edit: PS: I apologize for going off the bend with you two. I should have waited for the explanation.

I agree that the ball is moving in a flat horizontal plane. See my previous posts.

The original poster will come back to two pages of argument and regret what he started... :tongue:

- #38

Cyrus

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So the Magnitude of the tension in the string, (for the spinning mass), must equal Mg of the hanging string.

When the mass is spinning, there must be a component of tension in the Y direction to support the spinning mass from falling down. And it must equal the weight of the spinning mass.

Are we ok so far? Or do you disagree.

When the mass is spinning, there must be a component of tension in the Y direction to support the spinning mass from falling down. And it must equal the weight of the spinning mass.

Are we ok so far? Or do you disagree.

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- #39

Cyrus

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Now because the mass is spinning it has a ceneripital force along the plane of the circle. But this will ONLY contribute to the X component of force. Because as you stated, it acts along the radial direction. And the radius of the circle it traces is NOT along the string, but it is horizonal along the axis of the rod to the mass. So this force must be balanced by a tension in the X-direction of the string, for the part that is spinning.

But the two masses are equal. Any force in the x direction will make the magnitude of the tension in the spinning string more than the stationary one. So we can conclude that the stationary mass has to be heavier than the spinning one in order to allow for a component of tension in the x direction.

Did I goof off somewhere?

But the two masses are equal. Any force in the x direction will make the magnitude of the tension in the spinning string more than the stationary one. So we can conclude that the stationary mass has to be heavier than the spinning one in order to allow for a component of tension in the x direction.

Did I goof off somewhere?

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- #40

Sirus

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- #41

Cyrus

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- #42

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cyrusabdollahi said:Now because the mass is spinning it has a ceneripital force along the plane of the circle. But this will ONLY contribute to the X component of force. Because as you stated, it acts along the radial direction. And the radius of the circle it traces is NOT along the string, but it is horizonal along the axis of the rod to the mass. So this force must be balanced by a tension in the X-direction of the string, for the part that is spinning.

But the two masses are equal. Any force in the x direction will make the magnitude of the tension in the spinning string more than the stationary one. So we can conclude that the stationary mass has to be heavier than the spinning one in order to allow for a component of tension in the x direction.

Did I goof off somewhere?

So obviously we want to jump the gun and run with this scenario.

Well, OK, if you want to do that... What you THEN have, assuming your picture is the correct one, is that you have a conical pendulum at the top. However, now the calculation of the tension in the string isn't as straightforward, since now, BOTH masses contribute to the tension since the swinging mass now will have a component of its weight along the string.

If this is true, then we cannot solve this problem till we first solve the correct angle that the conical pendulum makes with the vertical for the given mass.

Zz.

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Sirus

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- #44

Cyrus

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- #45

Tide

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If [itex]m_2[/itex] is the hanging mass and [itex]m_1[/itex] is the revolving mass then the tension on the string is

[tex]T = m_2 g = \sqrt { \left(m_1 g\right)^2 + \left( \frac {m_1 v^2}{r^2}\right)^2}[/tex]

so that

[tex]v = \left(\frac {m_2^2 - m_1^2}{m_1^2}\right)^{1/4} \sqrt {gr}[/tex]

[tex]T = m_2 g = \sqrt { \left(m_1 g\right)^2 + \left( \frac {m_1 v^2}{r^2}\right)^2}[/tex]

so that

[tex]v = \left(\frac {m_2^2 - m_1^2}{m_1^2}\right)^{1/4} \sqrt {gr}[/tex]

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- #46

Cyrus

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- #47

Cyrus

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- #48

Cyrus

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- #49

Sirus

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Cyrus, Sirus is the nickname I use on PF, and I pronounce it like sigh-rus. Good to see another sirus/cyrus on the forums.

- #50

Cyrus

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- #51

fomenkoa

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Are you crazy? Clearly the masses are in a state of pure equilibrium with the net conservation of mass modeled by the equation F= GvC, where C is a constant! Chaos theory would predict that the flux of friction dissipation would be inversely proportional to its mass in a vaccuum!

Just kidding

- #52

Cyrus

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- #53

FZX Student

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2Fg = 2Fc

2mv^2/r = 2mg

*cancel 2m*

v^2/r = g

v^2 = gr (which is the same as the original weights)

How I got a completely different answer, is a complete mystery on its own.

- #54

Cyrus

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