- #1

wmrunner24

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## Homework Statement

An amusement park ride consists of a rotating circular platform 8.07m in diameter from which 10kg seats are suspended at the end of 2.87m massless chains. When the system rotates, it makes an angle of 37.3° with the vertical.

a) What is the speed of each seat?

b) If a child of mass 45.3kg sits in a seat what is the tension in the chain for the same angle?

m=55.3kg (mass of the seat + the mass of the child)

V

_{t}=5.56565m/s (calculated from part A, confirmed as correct)

Ѳ=37.3° (listed)

g=9.8m/s

^{2}(listed)

d=8.07m (listed)

l=2.87m (listed)

r=5.77419m (calculated from the length of the chain and the diameter of the circular platform, d/2+lsinѲ, which is its horizontal distance from the center of rotation, where d is the diameter of the platform and l is the length of the chain)

## Homework Equations

ƩF

_{x}=0

ƩF

_{y}=0

F

_{c}=[tex]\frac{Vt^2m}{r}[/tex]

F

_{g}=mg

## The Attempt at a Solution

I have the answer to A, and that part is verified as correct already. B is what I'm wondering about.

So, this swing is rotating in the z dimension, but in the x and y directions it is in equilibrium which means that the sum of the forces in the x direction and the y direction must be equal to 0.

ƩF

_{x}=0

ƩF

_{y}=0

Then the forces in the x are the horizontal tension in the cable in the x and the centrifugal (equal and opposite to the centripetal), and the forces in the y are the vertical tension in the cable and the force of gravity.

ƩF

_{x}=T

_{x}-F

_{c}

ƩF

_{y}=T

_{y}-F

_{g}

Since they're both equal to 0, they can be set equal to each other.

T

_{x}-F

_{c}=T

_{y}-F

_{g}

I rewrote these based on their definitions.

Tsin(Ѳ)-[tex]\frac{Vt^2m}{r}[/tex]=Tcos(Ѳ)-mg

Solve symbolically:

T=[tex]\frac{m}{sin\Theta-cos\Theta}[/tex][tex]\frac{Vt^2}{r}[/tex] -g

When I put all that together, I get T=681.28N. I have one more guess left for this problem so I need to be sure it's right beforehand. Can someone verify if this logic is correct please?