Question

Inthisexperiment,aspringforcewasusedtokeep moving object travelingin a circularpath.The size of a spring force should be proportional to the amount of stretch in the spring. Doesthisclaimagree qualitatively with the data in your five trials? Why or why not?

Answer 1

Answer:

By calculation, it can be shown that;

K = $\frac{F_{angular}}{2\times x\times c}$

Whereby for constant K, as  ${F_{angular}}$ 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

$\frac{1}{2}\cdot k\cdot x^2$.

Rotational kinetic energy = $\frac{1}{2}$·I·ω²

Inertia,  I =  $\frac{1}{2}$·m·r²

ω = $\frac{v}{r}$

Substituting gives

Rotational kinetic energy =  $\frac{1}{2}$·

=  $\frac{1}{4}$·m· v²

Equating both equations gives

K = $\frac{2\times m\times v^2}{4\times x^{2} }$  = $\frac{1\times m\times v^2}{2\times x^{2} }$  

Within the proportionality limit, x ∝ r

therefore we can write x = c·r which gives

$\frac{ m\times v^2}{2\times x\times c\times r }$ = $\frac{v^{2} }{r} \times\frac{m}{2\times x\times c}$

Since $\frac{v^{2} }{r}$ = angular acceleration, α, then

m× $\frac{v^{2} }{r}$ = Angular force

Therefore K = $\frac{F_{angular}}{2\times x\times c}$

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.