- #1

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V(tangential) = 2pi R / T

i'm not sure if 33.33 rev/min is the V, but i tried plugging in the respective numbers and solve for R, didnt really make much sense.

- Thread starter huskydc
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- #1

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V(tangential) = 2pi R / T

i'm not sure if 33.33 rev/min is the V, but i tried plugging in the respective numbers and solve for R, didnt really make much sense.

- #2

Pyrrhus

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- #3

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and i have :

f = ma

N=mg

from that i solve for a, which = .8829

but i'm still confused as to what that 33.333 rev/min

- #4

Pyrrhus

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That's just the speed.

It's simply:

[tex] F_{friction} = m \frac{v^2}{R} [/tex]

It's simply:

[tex] F_{friction} = m \frac{v^2}{R} [/tex]

- #5

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i tried:

F = m (v^2/R)

like you said, V = 33.333

mu g = (v^2 / R) and solve for R this way, didn't work. i've been told that 33.333 is not just V. and if it is, then what's the T for?

- #6

Pyrrhus

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You need to have common units!, convert rev/min to m/s!

Last edited:

- #7

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converted rev/min to rad/s.

i did this:

33.333 rev / min x (1 min/60 sec) x (2pi rad/ 1 rev) = 3.49 rad/s

with mu g = (v^2 / R) , R= (v^2)/mu g

am i doing something wrong that i just don't see?

- #8

Pyrrhus

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Is that how the book states it, it looks to me you would need to use.

[tex] F_{frictionn} = m \frac{4 \pi^{2} R}{T^{2}} [/tex]

where T is the time for one revolution or the period T, but what you did above, should have worked just the same.

- #9

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Thanks so much!!! (physics is so hard....well for me it is...)

- #10

Pyrrhus

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it came from

[tex] v = \frac{2 \pi R}{T} [/tex]

substitute in

[tex] a_{c} = \frac{v^2}{R} = \frac{4 \pi^{2} R}{T^2} [/tex]

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