The ball moves around in circular motion. The centripetal force keeping it in circular motion is given by:
F = mv²/r
F = centripetal force, m = mass of ball, v = velocity of ball, r = radius of motion
The force the spring exerts on the ball is given by:
F = kΔx
F = spring force, k = spring constant, Δx = change of spring length
The spring provides the centripetal force that keeps the ball in circular motion, so set the spring force equal to the centripetal force:
kΔx = mv²/r
Let's do some algebra:
m/k = rΔx/v²
Now if we attach the same spring to the ceiling with the ball still fixed to one end and let the ball hang in static equilibrium, the spring force would balance the ball's weight:
kΔx' = mg
k = spring constant, Δx' = new change in spring length, m = mass, g = gravitational acceleration
Let's do some more algebra:
Δx' = gm/k
Substitute m/k with our previous result, rΔx/v²:
Δx' = grΔx/v²
Given values:
g = 9.81m/s²
r = 0.169m + 0.0131m = 0.1821m
Δx = 0.0131m
v = 3.15m/s
Plug in the values and solve for Δx':
Δx' = 9.81(0.1821)(0.0131)/3.15²
Δx' = 0.0024m
We have that for the Question "By how much would the spring stretch if it were attached to the ceiling and the ball allowed to hang straight down, motionless?" it can be said that how much would the spring stretch is
From the question we are told
Generally the equation for the maximum stretch position is mathematically given as
[tex]\frac{mv^2}{r}=k(2r)\\\\\frac{m}{k}=\frac{(0.0191/2)^2}{2.81^2}\\\\\frac{m}{k}=2.31*10^{-5}\\\\[/tex]
[tex]mg=ky\\\\y=\frac{m}{k}g\\\\y=2.31*10^{-5}*9,81\\\\[/tex]
[tex]y=2.27*10^{-4}m[/tex]
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