- #1

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I have determined that w = 12.7 rad/s

Now suppose that the blocks each have a mass m = 35 g. For the value of w you just found, what is the tension in the string?

For some reason, I cannot seem to get the right answer. SOMEONE PLEASE HELP!!!!

- Thread starter erogers4
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- #1

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I have determined that w = 12.7 rad/s

Now suppose that the blocks each have a mass m = 35 g. For the value of w you just found, what is the tension in the string?

For some reason, I cannot seem to get the right answer. SOMEONE PLEASE HELP!!!!

- #2

Doc Al

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The same equations used to solve for [itex]\omega[/itex] (the maximum rotational speed without slipping) will include the tension.

- #3

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1) T - umg = m R1 w^2

2) -T - umg = m R2 w^2

i was able to get w, but then i plug values back in and its not accepting my answer for T as being correct. i have no idea what im doing wrong. please help!

- #4

Doc Al

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- #5

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1) -T + umg = m r1 w^2

2) T + umg = m r2 w^2

i believe that is how i got 12.7 for w (i have so much work here and half of it is wrong, im not sure which is which anymore). I just tried solving for T though and it is still not right. ah im so confused now!

- #6

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- #7

Doc Al

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(To find [itex]\omega[/itex], start by adding those two equations.)

- #8

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T = umg - m r1 w^2

as well as

T = m r2 w^2 -umg

m=.035kg

r1=.03m

r2=.06m

w=12.7 rad/sec

u=.74

g=9.81

i plug those in and get

T=.0847

which apparently is right...i had that before i dont know y it wouldnt take it as being correct...thanks for all you help!!!!

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