Problem 103 - University Physics with Modern Physics

Question

CALC A Spring with Mass. 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 spring constant \(k .\) The work done to stretch or compress the spring by a distance \(L\) is \(\frac{1}{2} k X^{2}\) , where \(X=L-L-L_{0}\) . 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 the \(M\) and \(v .\) (Hint: Divide the spring into pieces of length \(d l ;\) find the speed of each pivide in terms of \(l, v,\) and \(L ;\) find the mass of each piece in terms of \(d l, M,\) and \(L ;\) and integrate from 0 to \(L .\) The result is \(n o t \frac{1}{2} M v^{2},\) since not all of the spring moves with the same speed.) In a spring gun, a spring of mass 0.243 \(\mathrm{kg}\) and force constant 3200 \(\mathrm{N} / \mathrm{m}\) is compressed 2.50 \(\mathrm{cm}\) from its unstretched length. When the trigger is pulled, the spring pushes horizon- tally on a 0.053 -kg ball. The work done by friction is negligible. Calculate the ball's speed when the spring reaches its uncom- pressed length (b) ignoring the mass of the spring and (c) includ- ing, 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) K.E. of spring = \(\frac{1}{6} M v^2\). (b) Ball's speed ignoring spring mass: \(\sqrt{\frac{k X^2}{m}}\). (c) Ball's speed with spring mass evaluated using total energy conservation. (d) Final K.E. of ball and spring calculated based on derived ball speed.

Step 1: Define Variables and Energy Components

The problem involves calculating the kinetic energy of a spring. We have a spring with mass \(M\) and one end fixed. The speed varies linearly with distance from the fixed end. We will account for the variable speed of spring segments. We need to calculate the kinetic energy for a moving spring with mass distribution.

Step 2: Breakdown of Spring and Define Segment Mass

Consider a small segment of the spring \(dl\). The mass \(dm\) of this segment is given by \(dm = \frac{M}{L} \, dl\), assuming the mass \(M\) is distributed uniformly along the length \(L\) of the spring.

Step 3: Determine Velocity of Spring Segments

The velocity \(v_l\) of a segment at distance \(l\) from the fixed end is proportional to \(l\). The velocity is \(v_l = \frac{l}{L} v\), where \(v\) is the speed of the end of the spring.

Step 4: Integrate for Kinetic Energy

The kinetic energy \(dK\) of a small segment is given by \(dK = \frac{1}{2} dm \cdot v_l^2\). Substituting \(dm\) and \(v_l\), we get \(dK = \frac{1}{2} \left(\frac{M}{L} \, dl\right) \left(\frac{lv}{L}\right)^2\). When simplified and integrated from \(0\) to \(L\), we find total kinetic energy \(K = \frac{1}{6} M v^2\) for the spring.

Step 5: Solve Ball Speed Without Spring Mass

When ignoring the spring's mass, the ball's speed can be calculated using energy conservation. Initially, the potential energy is stored in the spring: \(\frac{1}{2} k X^2\). At the point of release, this energy converts to kinetic energy of the ball: \(\frac{1}{2} m v_{ball}^2\). Therefore, \(v_{ball} = \sqrt{\frac{k X^2}{m}}\), where \(m\) is ball's mass and \(X = 0.025 \, m\).

Step 6: Solve Ball Speed Including Spring Mass

Include the spring's kinetic energy: initial total energy is \(\frac{1}{2} k X^2 + \frac{1}{6} M v^2\). Equate this to the kinetic energy of ball and spring system: \(\frac{1}{2} m v_{ball}^2 + \frac{1}{6} M v^2\). Solve for \(v_{ball}\) using this total energy equation.

Step 7: Calculate Final Kinetic Energies in Part (c)

For the final kinetic energies: the ball's kinetic energy is \(\frac{1}{2} m v_{ball}^2\). The spring's final kinetic energy, calculated before, is \(\frac{1}{6} M v^2\). Both contribute to total kinetic energy at equilibrium length.

Key Concepts

Mass Distribution in a Spring

In understanding the kinetic energy of a spring, it's crucial to consider how its mass is distributed along its length. If a spring has a total mass, denoted by \( M \), imagine dividing the spring into tiny segments each of length \( dl \).

• The mass \( dm \) of each small segment can be described by the relation \( dm = \frac{M}{L} \, dl \), where \( L \) is the total length of the spring.


• This relationship assumes a uniform distribution of mass, meaning each part of the spring carries an equal share of the total mass.

Understanding this mass distribution is significant because it affects how each part of the spring contributes to the total kinetic energy when the spring is in motion. By considering this distribution, we can calculate how the mass of each segment plays a role in the spring's kinetic dynamics.

Variable Speed of Spring Segments

When a spring is stretched or compressed, different parts move at different speeds. This variation is generally linear, meaning the closer a segment is to the fixed end, the slower it moves.

• For a segment located at a distance \( l \) from the fixed end, its velocity \( v_l \) can be expressed as \( v_l = \frac{l}{L} v \), where \( v \) is the velocity of the moving end of the spring.


• This equation indicates that the speed of any segment is a fraction of the spring's end speed. The further from the fixed end, the larger the fraction.

These varying speeds affect each segment's kinetic energy. Together with the individual masses, these speeds cumulatively determine the kinetic behavior of the whole spring. Calculating the kinetic energy involves integrating these effects over the entire length of the spring.

Energy Conservation

Energy conservation is a principle that helps us equate the total energy in a system to the sum of its kinetic and potential energies. For a spring, when you ignore the effect of the spring's own mass, this is straightforward.

• Initially, the potential energy stored in the spring, given by \( \frac{1}{2} k X^2 \), is converted into kinetic energy as it releases.


• When including the mass of the spring, part of the initial potential energy should also convert into the kinetic energy of the moving spring itself \( \frac{1}{6} M v^2 \).


• The energy balance equation then becomes \( \frac{1}{2} k X^2 = \frac{1}{2} m v_{ball}^2 + \frac{1}{6} M v^2 \), considering both the ball's and the spring's kinetic energy.

Using energy conservation principles helps in solving the exact speeds and energy distributions after a spring releases its energy, providing a holistic view of the energy transformations.