Problem 93 - University Physics with Modern Physics

Question

We usually ignore the kinetic energy of the moving coils of a spring, but let's try to get a reasonable approximation to this. Consider a spring of mass \(M\), equilibrium length \(L_0\), and force constant \(k\). The work done to stretch or compress the spring by a distance \(L\) is \\(\frac{1}{2}\\) \(kX^2\), where \(X = L - L_0\). Consider a spring, as described above, that has one end fixed and the other end moving with speed \(v\). Assume that the speed of points along the length of the spring varies linearly with distance \(l\) from the fixed end. Assume also that the mass \(M\) of the spring is distributed uniformly along the length of the spring. (a) Calculate the kinetic energy of the spring in terms of \(M\) and \(v\). (\(Hint\): Divide the spring into pieces of length \(dl\); find the speed of each piece in terms of \(l\), \(v\), and \(L\); find the mass of each piece in terms of \(dl\), \(M\), and \(L\); and integrate from \(0\) to \(L\). The result is \(not\) \\(\frac{1}{2}\\) \(Mv^2\), since not all of the spring moves with the same speed.) In a spring gun, a spring of mass 0.243 kg and force constant 3200 N/m is compressed 2.50 cm from its unstretched length. When the trigger is pulled, the spring pushes horizontally on a 0.053-kg ball. The work done by friction is negligible. Calculate the ball's speed when the spring reaches its uncompressed length (b) ignoring the mass of the spring and (c) including, using the results of part (a), the mass of the spring. (d) In part (c), what is the final kinetic energy of the ball and of the spring?

Step-by-Step Solution

Answer

(a) \(\frac{1}{6} M v^2\). (b) Speed without spring mass: \(9.79 \text{ m/s}\). (c) Speed with spring mass: \(9.66 \text{ m/s}\). (d) Kinetic energies: ball \(2.47 \text{ J}\), spring \(0.42 \text{ J}\).

Step 1: Calculate Kinetic Energy of a Piece of the Spring

Let's consider a small piece of the spring of length \(dl\). The speed of this piece \(v(l)\) increases linearly from 0 to \(v\) over the length \(L\) of the spring, so its speed as a function of distance \(l\) from the fixed end is \(v(l) = \frac{v}{L}l\). The differential mass of this piece is \(dm = \frac{M}{L}dl\), assuming uniform mass distribution along the spring.

Step 2: Express Differential Kinetic Energy and Integrate

The kinetic energy of each differential piece is \(dK = \frac{1}{2}dm(v(l))^2 = \frac{1}{2} \left( \frac{M}{L} \right) dl \left( \frac{v}{L} l \right)^2\). Simplifying this gives \(dK = \frac{1}{2} \frac{M v^2}{L^3} l^2 dl\). To find the total kinetic energy of the spring, integrate from 0 to \(L\): \[ K = \int_0^L \frac{1}{2} \frac{M v^2}{L^3} l^2 \, dl = \frac{M v^2}{2 L^3} \int_0^L l^2 \, dl \].

Step 3: Evaluate the Integral

Evaluate the integral \( \int_0^L l^2 \, dl \), which is the integral of \(l^2\) from \(0\) to \(L\). This yields: \[ \int_0^L l^2 \, dl = \left[ \frac{l^3}{3} \right]_0^L = \frac{L^3}{3} \].Substituting back, we find the kinetic energy \( K = \frac{M v^2}{2 L^3} \times \frac{L^3}{3} = \frac{1}{6} M v^2\).

Step 4: Calculate the Ball's Speed Ignoring the Mass of the Spring

Assuming the spring is massless, all the potential energy converted from compressing the spring goes into the ball. The potential energy stored is \( \frac{1}{2}kX^2 = \frac{1}{2} \times 3200 \times (0.025)^2 \). Setting this equal to the kinetic energy of the ball \( \frac{1}{2}mv_b^2 \), solve for the speed \(v_b\).

Step 5: Calculate the Ball’s Speed Including the Mass of the Spring

Now include the kinetic energy of the spring from Step 3. The initial potential energy \( \frac{1}{2}kX^2 \) is divided into the kinetic energies of the ball and the spring:\( \frac{1}{2} kX^2 = \frac{1}{2}mv_b^2 + \frac{1}{6}Mv^2 \).By substituting known values, solve the equation for \(v_b\).

Step 6: Calculate Final Kinetic Energies of Ball and Spring

Using the speed \(v_b\) obtained from Step 5 and the expression for kinetic energy of the spring from Step 3, find the final kinetic energy of the ball and the spring separately, using \(K_{ball} = \frac{1}{2}mv_b^2\) and \(K_{spring} = \frac{1}{6}Mv^2\).

Key Concepts

Spring Force Constant

The spring force constant, often symbolized as "k," is a measure of how stiff or resistant a spring is to being compressed or stretched. This is also referred to as the "spring constant" or "stiffness constant" and is measured in Newtons per meter (N/m).
In a nutshell, the larger the "k" value, the stiffer the spring. When a spring is compressed or stretched by a distance "X" from its natural or equilibrium length, the spring exerts a force calculated by Hooke's Law:

• F = -kX

The force is always in the direction opposite to the displacement, reflecting the spring’s natural tendency to return to its equilibrium position.
In the exercise, the spring is described with a force constant, and this tells us how much force it can exert per meter of deformation. Understanding the spring force constant is crucial in calculating how much energy can be stored in the spring, which is analyzed in the upcoming sections.

Uniform Mass Distribution

Uniform mass distribution is used to describe how mass is spread out evenly across an object. For a spring, this means every segment of the spring has the same mass per unit length. In mathematical terms, if the total mass of the spring is "M" and its length is "L," the mass per unit length is:

• \( \frac{M}{L} \)

This concept becomes important when you need to calculate the kinetic energy of each section of the spring. In the exercise, we divide the spring into tiny pieces called "differential elements." Each of these pieces has a small length "dl" and mass "dm". Given the uniform distribution, this mass can be expressed as:

• \( dm = \frac{M}{L} dl \)

Such a distribution simplifies calculations and is essential for solving physical problems where the mass is not concentrated in a single spot.

Spring Potential Energy

Spring potential energy is the energy held within a compressed or stretched spring, ready to be converted into other forms of energy, such as kinetic energy. According to Hooke's Law, the potential energy stored in a spring is given by:

• \( U = \frac{1}{2} k X^2 \)

where "k" is the spring force constant and "X" is the displacement from the equilibrium position. This energy results from the work done on the spring when it is compressed or stretched.
In the exercise, when the spring is released, this stored energy is converted into the kinetic energy of the ball and, to a lesser extent, the spring itself. This conversion is key to understanding how systems using springs, like spring guns, function. At the moment of maximum compression or stretch, the potential energy is at its highest, preparing to release kinetic energy as it moves toward its natural state.

Linear Speed Variation

The concept of linear speed variation describes how the speed of different points along the spring changes as a function of distance from a fixed point. In the described scenario, the speed varies linearly along the length of the spring, meaning it starts from zero at the fixed end and reaches maximum speed at the moving end. Mathematically, this relationship can be expressed as:

• \( v(l) = \frac{v}{L}l \)

where "\( v \)" is the maximum speed, "L" is the entire length of the spring, and "l" is the position along the spring.
In practice, this means each piece of the spring moves at a different speed, influencing how the kinetic energy is distributed along the spring. The linear variation must be considered when calculating the overall kinetic energy of the spring because not all pieces contribute equally. This highlights the complexity of motion within systems where components do not move uniformly. In solving problems like the given exercise, understanding this concept helps in setting up integrals that reflect the energy distribution accurately.