Respuesta :
Answer:
By calculation, it can be shown that;
K = [tex]\frac{F_{angular}}{2\times x\times c}[/tex]
Whereby for constant K, as [tex]{F_{angular}}[/tex] increases, x also increases.
Explanation:
The experiment set up consisted of the use of a spring force to maintain the object in circular path.
The energy in the spring is given by
[tex]\frac{1}{2}\cdot k\cdot x^2[/tex].
Rotational kinetic energy = [tex]\frac{1}{2}[/tex]·I·ω²
Inertia, I = [tex]\frac{1}{2}[/tex]·m·r²
ω = [tex]\frac{v}{r}[/tex]
Substituting gives
Rotational kinetic energy = [tex]\frac{1}{2}[/tex]·
= [tex]\frac{1}{4}[/tex]·m· v²
Equating both equations gives
K = [tex]\frac{2\times m\times v^2}{4\times x^{2} }[/tex] = [tex]\frac{1\times m\times v^2}{2\times x^{2} }[/tex]
Within the proportionality limit, x ∝ r
therefore we can write x = c·r which gives
[tex]\frac{ m\times v^2}{2\times x\times c\times r }[/tex] = [tex]\frac{v^{2} }{r} \times\frac{m}{2\times x\times c}[/tex]
Since [tex]\frac{v^{2} }{r}[/tex] = angular acceleration, α, then
m× [tex]\frac{v^{2} }{r}[/tex] = Angular force
Therefore K = [tex]\frac{F_{angular}}{2\times x\times c}[/tex]
Therefore as Force, F increases, x also increases and the size of a spring force should be proportional to the amount of stretch in the spring.
