Question

small ball is attached to one end of a spring that has an unstrained length of 0.245 m. The spring is held by the other end, and the ball is whirled around in a horizontal circle at a speed of 3.55 m/s. The spring remains nearly parallel to the ground during the motion and is observed to stretch by 0.0194 m. By how much would the spring stretch if it were attached to the ceiling and the ball allowed to hang straight down, motionless? Number Enter your answer in accordance to the question statement Units Choose the answer from the menu in accordance to the question statement

Answer 1

Answer:

$x = 4 \times 10^{-3} m$

Explanation:

As we know that the force required to move the mass in circle with uniform speed is known as centripetal force

This centripetal force is given as

$F_c = \frac{mv^2}{R}$

while mass is revolving in horizontal circle the force is due to spring so it is given as

$F_c = kx$

$kx = \frac{mv^2}{R}$

$k(0.0194) = \frac{m\times 3.55^2}{(0.245 + 0.0194)}$

$k(0.0194) = 47.66 m$

now we have

$\frac{k}{m} = 2457$

now when mass is suspended at the end of spring then we have

$mg = kx$

$x = \frac{mg}{k}$

$x = \frac{9.81}{2457}$

$x = 4 \times 10^{-3} m$