Finding Radius of Circular Movement w/ Angle & Angular Velocity

[Dell]

in the following question,

http://62.90.118.184/Index.asp?CategoryID=318

i am given the following only:
angle of the cord = \alpha
angular velocity = \omega

and am asked to find the radius

what i did was :
Newtons 2nd on the radial and "y" axis

F_y=T*cos(\alpha)-mg=0 (T being tension in the cord)
T=mg/cos(\alpha)

F_r = T*sin(\alpha) = ma_r =m\omega²R

T*sin(\alpha) =m\omega²R

now i use the T i found (T=mg/cos(\alpha)) and i get
mg*tg(\alpha)=m\omega²R

R=g*tg(\alpha)/\omega²

which according to the answer sheet is the correct answer, but i cannot understand how the radius does not depend on the length of the cord or the length of the top, horizontal bar??

 

[jmb]

but i cannot understand how the radius does not depend on the length of the cord or the length of the top, horizontal bar??

The radius does depend on those quantities, but their value is encoded in the value of \alpha. If \omega is fixed and we increase the length of the cord then \alpha increases (and if we decrease the length of the cord it decreases). Similarly, for a fixed \omega, an increase in the length of the horizontal bar will increase \alpha. The physics of what is happening (spinning a mass on a string) means that the value of \alpha has a dependency on these quantities.

If you wanted, you could make this dependency explicit. From the geometry of the problem you can show that,

<br /> \operatorname{tg}{\alpha}=\frac{R-a}{\sqrt{b^2-\left(R-a\right)^2}}<br />

where a is the length of the horizontal bar and b is the length of the cord. If you substitute this into your answer then you can get an expression for R that depends explicitly on a and b instead of \alpha. However rearranging the answer involves solving a quartic equation for R which is not easy.

As an interesting aside, note that understanding the physics doesn't give us any 'information for free'. If we are given \alpha then to find R geometrically we require two more pieces of information: the length of the horizontal bar and the length of the string. To find R from \alpha using instead the physics of the situation, we still require two more pieces of information (the speed of rotation \omega and the downward acceleration the ball would experience if removed from the string, g).