- #1

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I don't know where to begin. Any advice?

- Thread starter Punchlinegirl
- Start date

- #1

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I don't know where to begin. Any advice?

- #2

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[tex]\frac{mv^2}{r}=mg[/tex]

- #3

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I don't know the velocity or acceleration

- #4

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you find that by

[tex]mgh=.5mv^2[/tex]

find v, plug that into the first equation and solve for h

[tex]mgh=.5mv^2[/tex]

find v, plug that into the first equation and solve for h

- #5

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Punchline, use Urban's first equation to determine the minimum velocity that the marble needs at the top of the loop. Then apply conservation of energy to find minimum h. You will have to use

[tex] \Delta PE = .5mv^2 + .5I\omega^2 [/tex]

- #6

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- #7

xanthym

Science Advisor

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(Note: FromSkomatth said:

Punchline, use Urban's first equation to determine the minimum velocity that the marble needs at the top of the loop. Then apply conservation of energy to find minimum h. You will have to use

[tex] \Delta PE = .5mv^2 + .5I\omega^2 [/tex]

{TOTAL Energy Required by Marble at Loop's Top} =

= {Linear Kinetic Energy @ Top} + {Rotational Kinetic Energy @ Top} + {Potential Energy @ Top} =

= (1/2)*m*v

= (1/2)*m*v

= (1/2)*m*v

= m*{(7/10)*v

The above TOTAL Energy must be obtained from the Potential Energy at an initial height "H" above Loop's bottom given by:

{Initial Required Potential Energy} = m*g*H* =

= m*{(7/10)*v

::: ⇒ H = (7/10)*v

The velocity "v" is then determined from requiring centripetal force at loop's top to equal the marble's weight "m*g":

m*v

::: ⇒ v = sqrt{g*R} ::: (Eq #2)

Place Eq #2 into Eq #1 and determine "H" to complete solution.

~~

Last edited:

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