Question

A small ball is attached to one end of a spring that has an un- strained length of 0.200 m. The spring is held by the other end, and the ball is whirled around in a horizontal circle at a speed of 3.00 m/s. The spring remains nearly parallel to the ground during the motion and is observed to stretch by 0.010 m. By how much would the spring stretch if it were attached to the ceiling and the ball allowed to hang straight down, motionless?

Answer 1

Answer:

$\Delta x=0.002287\ m=2.287\ mm$

Explanation:

Given:

• un-stretched length of the spring, $l=0.2\ m$

• speed of revolution of the ball in horizontal plane, $v=3\ m.s^{-1}$

• length stretch during the motion, $\Delta l=0.01\ m$

Now the radius of revolution of the ball:

$r=l+\Delta l$

$r=0.2+0.01$

$r=0.21\ m$

Now in this case the centrifugal force is equal to the spring force:

$F_c=F_s$

$m.\frac{v^2}{r} =k.\Delta l$

where:

m = mass of the ball

k = spring constant

$m\times \frac{3^2}{0.21} =k\times 0.01$

$k=(4285.714\times m)\ N.m^{-1}$

Now the extension in the spring upon hanging the ball motionless:

$m.g=k.\Delta x$

$9.8\times m=(4285.714\times m)\times \Delta x$

$\Delta x=0.002287\ m=2.287\ mm$