# Circular movement

1. Sep 9, 2009

### 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 = $$\omega$$

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

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

Fr = T*sin($$\alpha$$) = mar =m$$\omega$$2R

T*sin($$\alpha$$) =m$$\omega$$2R

now i use the T i found (T=mg/cos($$\alpha$$)) and i get
mg*tg($$\alpha$$)=m$$\omega$$2R

R=g*tg($$\alpha$$)/$$\omega$$2

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??

2. Sep 9, 2009

### jmb

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,

$$\operatorname{tg}{\alpha}=\frac{R-a}{\sqrt{b^2-\left(R-a\right)^2}}$$

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, $g$).