- #1

whatdoido

- 48

- 2

1. Homework Statement

1. Homework Statement

The diameter of a carousel is 15 m and it rotates 8,0 rounds per minute. Calculate how large of an angle is formed with respect to the vertical plane by the swing that hangs by the ropes fixed to the edge of carousel. The length of ropes is 2,5 m.

So I mark:

[itex]n=8,0 r/min=0,1333... r/s[/itex]

[itex]r=(15/2) m=7,5 m[/itex]

[itex]x=2,5 m[/itex]

Angle between the vertical plane at the edge of carousel and swing is [itex]\theta[/itex]

## Homework Equations

[itex] F=m\frac {v^2}{r}[/itex]

[itex] F=mg [/itex]

[itex] \omega=n2\pi[/itex]

[itex][/itex]

## The Attempt at a Solution

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The angular velocity:

[itex]\omega=n2\pi=0,1333... r/s*2\pi=0,8377... 1/s[/itex]

Then I thought that I need the distance of swing from the edge of carousel. I mark this distance as R

Thus, [itex]R=x*sin\theta[/itex]

Then the total distance of swing from the middle is [itex]r+R[/itex], when tangential velocity [itex]v=\omega(r+R)[/itex]

I figured that the angle should be solvable from vertical and horizontal forces which act on the swing:

[itex]\left\{\begin{array}{cc} Fsin\theta=m\frac{v^2}{r+R}\\Fcos\theta=mg\end{array}\right.[/itex]

Those divided gives:

[itex]\frac{sin\theta}{cos\theta}=\frac{v^2}{g(r+R)}=\frac{\omega^2(r+R)^2}{g(r+R)}[/itex]

[itex]tan\theta=\frac{\omega^2r+\omega^2x*sin\theta}{g}[/itex]

[itex]g*tan\theta-\omega^2x*sin\theta-\omega^2r=0[/itex]

and this is where I got stuck.. I tried solving this equation with substituting [itex]tan\theta=\frac{sin\theta}{\pm\surd(1-sin\theta)}[/itex] and [itex]sin\theta=u[/itex] which did not help. So maybe I got some idea wrong and my equation is wrong or I am just missing some trick by how to solve that equation. On the other hand I checked that does the correct angle (32) work on that equation and it gives about 0=0, so my equation should be correct (I think?).