Respuesta :
Answer:
[tex]\Delta x=0.002287\ m=2.287\ mm[/tex]
Explanation:
Given:
- un-stretched length of the spring, [tex]l=0.2\ m[/tex]
- speed of revolution of the ball in horizontal plane, [tex]v=3\ m.s^{-1}[/tex]
- length stretch during the motion, [tex]\Delta l=0.01\ m[/tex]
Now the radius of revolution of the ball:
[tex]r=l+\Delta l[/tex]
[tex]r=0.2+0.01[/tex]
[tex]r=0.21\ m[/tex]
Now in this case the centrifugal force is equal to the spring force:
[tex]F_c=F_s[/tex]
[tex]m.\frac{v^2}{r} =k.\Delta l[/tex]
where:
m = mass of the ball
k = spring constant
[tex]m\times \frac{3^2}{0.21} =k\times 0.01[/tex]
[tex]k=(4285.714\times m)\ N.m^{-1}[/tex]
Now the extension in the spring upon hanging the ball motionless:
[tex]m.g=k.\Delta x[/tex]
[tex]9.8\times m=(4285.714\times m)\times \Delta x[/tex]
[tex]\Delta x=0.002287\ m=2.287\ mm[/tex]
