# Spinning ball

1. Mar 18, 2005

### stunner5000pt

have a look a the diagram
heavy particle of mass m is tied to one end of a string which is of length l. The other end is attached to the ceiling. The particle forms uniform circular motion. Angular velocity of omega. String is angle theta. Ignore friction and mass of string.

Find magnitude of string force S in terms of m, l, omega
horizontal components $$F_{x} = F sin \theta = m \omega^2 l sin \theta$$
vertical components = $$F_{y} = F cos \theta = mg$$
total components $$F = m \sqrt{g^2 + \omega^4 l^2 sin^2 \theta)$$
the text book however omits the sin theta and the g part which gives $$F = ml \omega^2$$

who is right??

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2. Mar 18, 2005

### xanthym

The book is correct:
{String Length} = L
{Horizontal Radius of Circular Orbit} = r = L*sin(θ)
{Centripetal Force} = m*v2/r =
= m*ω2*r =
= m*ω2*{L*sin(θ)} =
= m*L*ω2*sin(θ)

{String Tension} = T
{Horizontal Component of String Tension} = T*sin(θ)

At equilibrium:
{Horizontal Component of String Tension} = {Centripetal Force}
::: ⇒ T*sin(θ) = m*L*ω2*sin(θ)
::: ⇒ T= m*L*ω2

~~

3. Mar 18, 2005

### Staff: Mentor

Both answers are "correct" but the book's is the simpler and more useful one. Realize that your equation for the horizontal components ($F sin \theta = m \omega^2 l sin \theta$) immediately implies the book's answer.