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Please help

- Thread starter Torater
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Please help

- #2

tiny-tim

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(btw,

Instead of a rolling condition, you'll have an

Please help

show us what you get.

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Giving me:

∑τ=Iα = Fr

∑Fy=-m2a = T-m2g

I= 1/2mr²

OR if I had an X equation it would be:

∑Fx=m1a = T-F

Therefore 1/2m1r²(a/r)= (m2(g-a) - m1a) (r)

1/2(8)(.12)²(a/.12) = (5(9.81-a)) (.12)

2.48a=5.886

a= 2.34m/s²

T= (5(9.81-2.34))

T= 37.4N

Is this the right approach?

- #4

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I'm beginning to think we are in the same school Torater.. (Macewan?)

BTW, yeah I don't believe there would be an x direction because that would include friction which arises from a rolling sphere and in this case it's an unrolling condition.

I'm happy we've made it through these problems together hahah

BTW, yeah I don't believe there would be an x direction because that would include friction which arises from a rolling sphere and in this case it's an unrolling condition.

I'm happy we've made it through these problems together hahah

Last edited:

- #5

tiny-tim

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That's right … the pulley doesn't move, it only rotates, so you don't have an F = ma for the pulley.wouldn't there NOT be a X direction for this problem?

First, you could save yourself a lot of trouble by noticing that every term in the first equation is a multiple of r, so you can just ignore r, instead of remembering which power of .12 to multiply by.Therefore 1/2m1r²(a/r)= (m2(g-a) - m1a) (r)

1/2(8)(.12)²(a/.12) = (5(9.81-a)) (.12)

(this isn't a conicidence, it happens in

Second, I can't quite follow what you've done, but I think you didn't move the a (from g-a) over to the LHS before solving for a.

- #6

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Ok lets try this again:

∑τ=Iα = Tr

∑Fy=-m2a = T-m2g

I= 1/2mr²

Therefore 1/2m1r²(a/r)= m2(g-a) (r)

1/2m1 a = m2(9.81-a)

1/2 (8) (a) = 5(9.81-a)

a= 5.45m/s²

T= (5(9.81-5.45))

T= 21.8N

???

I hope your not in my class CHOD!! lol

∑τ=Iα = Tr

∑Fy=-m2a = T-m2g

I= 1/2mr²

Therefore 1/2m1r²(a/r)= m2(g-a) (r)

1/2m1 a = m2(9.81-a)

1/2 (8) (a) = 5(9.81-a)

a= 5.45m/s²

T= (5(9.81-5.45))

T= 21.8N

???

I hope your not in my class CHOD!! lol

Last edited:

- #7

tiny-tim

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Hi Torater!

Yes that's fine.

(note that, using**gneill**'s trick in the https://www.physicsforums.com/showthread.php?p=3033838&posted=1#post3033838", you can say that the effective mass of the pulley is its "rolling mass", 8/2, so the whole thing has mass 5 + 8/2 = 9, and force 5g, so using F = ma you get a = F/m = 5g/9 = 5.45 … *which is how I checked it! *… though perhaps the question doesn't allow that short-cut)

Yes that's fine.

(note that, using

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