Problem 11 - Physics

Question

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

Step-by-Step Solution

Answer

The spring would stretch approximately 0.229 m vertically.

Step 1: Understand the Forces in the Horizontal Circle

When the ball is whirled around in a horizontal circle, the centrifugal force (which is actually the tension in the spring providing the centripetal force) is acting on the ball. Given that the spring stretches by 0.010 m, this stretch is caused by the centripetal force needed for circular motion. We'll use this information to understand the dynamics when the spring is vertical.

Step 2: Calculate the Spring Constant

Using Hooke's Law, the force in the spring when stretched is given by \( F = k \cdot x \), where \( k \) is the spring constant and \( x \) is the extension. For the ball in circular motion, the force is also equal to the centripetal force \( F = \frac{mv^2}{r} \), where \( m \) is the mass of the ball, \( v \) is the speed, and \( r \) is the radius of the circle (0.20 m + 0.01 m). However, we don't need \( m \) as it will cancel out in a later step. We start by equating \( k \cdot 0.010 = \frac{mv^2}{0.210} \).

Step 3: Derive an Expression for "m/g/x"

When the spring is vertical, and motionless, the same tension in the spring balances the weight of the ball. Let's express this relation: \( F = mg = k \cdot x' \), where \( x' \) is the stretch in the spring in the vertical position. Substitute \( k \) obtained from Step 2 to simplify \( mg = \frac{mv^2}{0.210} / 0.010 \, \cdot x' \). Rearrange to solve for \( x' \): \( x' = 0.210 \cdot \frac{g}{v^2} \).

Step 4: Calculate Vertical Stretch

Use \( g = 9.81 \, \text{m/s}^2 \) and \( v = 3.00 \, \text{m/s} \) to substitute in our expression \( x' = 0.210 \cdot \frac{9.81}{3^2} \). Calculate \( x' \) to find how much the spring would stretch if it were vertical.

Key Concepts

Hooke's Law

Hooke's Law is a fundamental principle in spring mechanics that describes how the force exerted by a spring is proportional to the distance it is stretched or compressed. The formula used to express this is:

• \( F = k \cdot x \)

where:

• \( F \) is the force applied on the spring,
• \( k \) is the spring constant, a measure of the stiffness of the spring,
• \( x \) is the displacement of the spring from its equilibrium position.

In our exercise, the spring stretches 0.010 meters when the ball is swung in a circle. This stretch is due to the centripetal force pulling the ball outward, balanced by the spring force as per Hooke's Law. Understanding how stretching relates to force helps us calculate the spring constant, which is essential in determining how much the spring will stretch in different configurations.

Centripetal Force

In circular motion, centripetal force is crucial as it keeps an object moving along a circular path. This force is always directed towards the center of the circle. For the ball attached to the spring, the centripetal force comes from the tension in the spring itself.
This force can be calculated using:

• \( F_c = \frac{m v^2}{r} \)

where:

• \( F_c \) is the centripetal force,
• \( m \) is the mass of the object,
• \( v \) is the velocity,
• \( r \) is the radius of the circle, which in this case is the unstrained length of the spring plus the stretch \( (0.200 + 0.010) \) m.

In the context of our problem, this force provides a measure that was matched against the spring's stretching derived using Hooke's Law. Understanding this helps set up a pathway to examine what happens under vertical equilibrium.

Spring Constant

The spring constant, denoted as \( k \), tells us how stiff or flexible a spring is. A higher spring constant means the spring is harder to stretch or compress. In the exercise, determining the spring constant is essential to make predictions about how the spring behaves under different conditions.
Using the relationship given by Hooke's Law,

• \( F = k \cdot x \)

we equate it with the centripetal force:

• \( k \cdot 0.010 = \frac{m v^2}{0.210} \).

From this equation, even without knowing the mass \( m \), we can solve for \( k \) with relative conditions between horizontal and vertical setup given. This inferred spring constant is critical in cross-checking predictions in different setups, like when vertically hanging.

Circular Motion

Circular motion involves an object moving along a circular path, constantly changing direction because of the centripetal force acting on it. The object, in this case, the ball, continues in its circular path due to the tension in the spring, acting as the underlying centripetal force.
Circular motion is characterized by:

• A constant radius, here the sum of the unstrained length and the stretched part of the spring.
• Constant speed, indicating uniform circular motion.

When considering the ball and spring system, understanding circular motion gives insight into how centripetal force arises and influences stretching. This knowledge forms a basis for comparing forces in circular and straight-line (vertical) motion.

Vertical Equilibrium

Vertical equilibrium is where forces acting on an object in a vertical line balance out, leaving the object stationary. In our problem, this is considered when the ball hangs motionless from the ceiling.
For this, the downward force due to gravity \( (mg) \) is balanced by the upward spring force, in accordance with Hooke's Law:

• \( mg = k \cdot x' \)

where \( x' \) is the displacement due to the ball's weight. By using the previously calculated spring constant and other parameters, such as gravitational acceleration \( g \) and velocity \( v \), the vertical stretch in this static scenario can be predicted. Therefore, vertical equilibrium helps to translate what was learned from horizontal motion to predict static behavior.