- #1

- 2

- 0

.........m

I------------------

I........I.........θ......\

I........I..................\

I........I....................\..n

I.............................[]

I

I

m = 3 meters

n = 5 meters

θ = 30

[] = swing

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- Thread starter robdawg
- Start date

- #1

- 2

- 0

.........m

I------------------

I........I.........θ......\

I........I..................\

I........I....................\..n

I.............................[]

I

I

m = 3 meters

n = 5 meters

θ = 30

[] = swing

- #2

Integral

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Science Advisor

Gold Member

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- #3

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here's what I know so far by doing a free body diagram.

5cos(30)= y

length y = Sqrt(18.75)

5sin(30) = x

length x = 2.5

R = 3 + 2.5 = 5.5

using the pythagorean thereom we now that

sqrt((2.5)^2 + 18.75) = 5

force y = tcos(30)

force x = tsin(30)

circumference = 2pi(R)

about the forces we know that 5cos(30) = mg

and that tsin(30) = mω^2R

now how do I get rpms from this? what do I solve for?

thanks

5cos(30)= y

length y = Sqrt(18.75)

5sin(30) = x

length x = 2.5

R = 3 + 2.5 = 5.5

using the pythagorean thereom we now that

sqrt((2.5)^2 + 18.75) = 5

force y = tcos(30)

force x = tsin(30)

circumference = 2pi(R)

about the forces we know that 5cos(30) = mg

and that tsin(30) = mω^2R

now how do I get rpms from this? what do I solve for?

thanks

Last edited:

- #4

Integral

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Science Advisor

Gold Member

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