- #1

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i have it so where mu_s = v^2/r but I don't know if thats right, help please! thanks :-)

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- Thread starter quickclick330
- Start date

- #1

- 83

- 0

i have it so where mu_s = v^2/r but I don't know if thats right, help please! thanks :-)

- #2

PhanthomJay

Science Advisor

Homework Helper

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I think you mean mu_s(g) = v^2/r. But in any case, if mu_s is reduced by 3, then v^2 must be reduced by what factor to keep the car from slipping?

i have it so where mu_s = v^2/r but I don't know if thats right, help please! thanks :-)

- #3

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3/r maybe?

- #4

- 486

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- #5

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then you must divide the other side of the equation to balance it, so would it be ((v^2/r)/3)?

- #6

- 486

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What should you do to v to get the whole expression to equal [tex]{\frac{v^2}{3r}[/tex]?

- #7

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I'm not sure...not having the radius measurement is throwing me off

- #8

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You have [tex]\frac{v^2}{r}[/tex]

You need to divide all of that by three, but you're NOT allowed to touch r. All you're allowed to change is v. What might you do to v so that

[tex]\frac{v^2}{3r}=\frac{(v')^2}{r}[/tex] ?

- #9

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I tried to divide by 3 again...what is v prime??

- #10

hage567

Homework Helper

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I tried to divide by 3 again...what is v prime??

The speed the driver must slow down to in order to continue along the curve safely. This is what you are trying to solve for.

- #11

- 486

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I tried to divide by 3 again...what is v prime??

v prime is your new velocity. The one that makes the origianl exprssion (v^2/r) get divided by 3.

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