A small ball is attached to one end of a spring that has an unstrained length of 0.202 m. The spring is held by the other end, and the ball is whirled around in a horizontal circle at a speed of 3.18 m/s. The spring remains nearly parallel to the ground during the motion and is observed to stretch by 0.009 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 in m
To solve this problem, we need to consider the forces acting on the ball in both scenarios: when it is whirled around in a horizontal circle and when it hangs motionless.
In the first scenario, when the ball is whirled around in a horizontal circle, the centripetal force required to keep the ball in circular motion is provided by the tension in the spring. The stretching of the spring can be attributed to the centripetal force.
We can start by finding the spring constant (k) of the spring using Hooke's law. Hooke's law states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position.
F = -kx
Where:
- F is the force exerted by the spring
- k is the spring constant
- x is the displacement from the equilibrium position
In this case, the displacement (x) is given as 0.009 m. The force exerted by the spring (F) is equal to the centripetal force required to keep the ball in circular motion.
F = mv^2 / r
Where:
- m is the mass of the ball
- v is the velocity of the ball
- r is the radius of the circular path (the length of the stretched spring)
Rearranging the equation for F, we have:
kx = mv^2 / r
We can rearrange this equation to solve for the spring constant (k):
k = mv^2 / (xr)
Now, let's move to the second scenario where the ball hangs motionless from the ceiling. In this case, the gravitational force acting on the ball is equal to the force exerted by the spring.
F = mg
Where:
- m is the mass of the ball
- g is the acceleration due to gravity
Since the displacement (x) is zero in this scenario, we want to find the force exerted by the spring, which is equal to the stretching force.
Let's substitute the values we know into the equation for the force exerted by the spring:
F = kx
Here, x is 0 because there is no displacement in the second scenario. We can solve for k:
k = F/x
Now, we can substitute the force of gravity (mg) for = mg / x
Finally, we can solve for the stretching (displacement) in the second scenario using Hooke's law:
x = F / k
Substituting mg for F and the calculated value of k, we get:
x = mg / k
Let's put all the values into the equation to calculate the stretching (displacement) in the second scenario:
x = (m * 9.8) / ((m * v^2) / (0.009 * v))
Simplifying the equation gives:
x = 0.009 * 9.8 / v^2
Now, we can substitute the given value of v (3.18 m/s) into the equation to get the final answer:
x = 0.009 * 9.8 / (3.18^2)
Calculating this, the stretching (displacement) of the spring when attached to the ceiling and the ball hangs motionless is approximately 0.00882 meters.