Chapter 2: Problem 16

Consider the case of free fall with a damping force proportional to the velocity, \(f_{D}=\pm k v\) with \(k=0.1 \mathrm{~kg} / \mathrm{s}\). a. Using the correct sign, consider a \(50-\mathrm{kg}\) mass falling from rest at a height of \(100 \mathrm{~m}\). Find the velocity as a function of time. Does the mass reach terminal velocity? b. Let the mass be thrown upward from the ground with an initial speed of \(50 \mathrm{~m} / \mathrm{s}\). Find the velocity as a function of time as it travels upward and then falls to the ground. How high does the mass get? What is its speed when it returns to the ground?

Short Answer

Expert verified

In scenario (a), the velocity as a function of time is given by \(v(t) = 490(1- e^{-0.002t})\) m/s, and the terminal velocity is 490 m/s. Under scenario (b), the velocity as a function of time when the mass climbs is \(v(t) = 490 + (50-490)e^{-0.002t}\) m/s, and when it falls is the same as scenario (a). The terminal velocity when falling back is also 490 m/s. The exact height reached would need to be calculated by integrating the velocity curve from 0 to the time when velocity is zero, and substitute the related parameters.

Step by step solution

Determine the Mathematical Model

The motion of the body is governed by Newton's second law of motion, where the net force is equal to mass \(m\) times acceleration \(a\). In the case of downward motion of the body, the forces are gravity acting downwards and the drag force acting upwards. Hence, the mathematical model according to Newton's second law is \(m \cdot a = m \cdot g - k \cdot v\). This is a first order non-homogeneous differential equation which be expressed as \(\frac{dv}{dt} = g - \frac{k}{m} \cdot v \).

Solve the differential equation

This is a first order linear differential equation, and its solution is given by: \(v(t) = \frac{mg}{k} + (v_0 - \frac{mg}{k}) e^{-kt/m} \). Since the body starts from rest, the initial velocity \(v_0\) is zero. Substituting the specific values, \(v(t)= 490(1- e^{-0.002t})\) m/s. The terminal velocity is reached when the speed no longer changes with time, i.e., \(\frac{dv}{dt}=0\). This would imply the body has fallen for a very long time, or \(t \rightarrow \infty\), giving the terminal velocity as \(v=\frac{mg}{k}=490\) m/s.

Being Thrown Upward

When the body is thrown upwards, it will first climb to a certain height then fall back to the ground. There are two phases to this process, climbing and falling. In the case of climbing, The differential equation that describes the motion is the same as outlined in step 1. With an initial velocity \(v_0 = 50\) m/s, the velocity as a function of time is \(v(t) = \frac{mg}{k} + (v_0 - \frac{mg}{k}) e^{-kt/m} = 490 + (50-490)e^{-0.002t}\). This is valid until \(v(t)=0\). The height from the ground when \(v(t)=0\) can be obtained by solving the integral equation \(h(t) = \int v(t) dt\). When \(v(t) = 0\), the mass will then fall to the ground, and the equation of motion will be the same as in step 2.

Terminal velocity and maximum height

The mass will reach terminal velocity when descending, at the same terminal speed of 490 m/s as in step 2. The maximum height can be found by counting the time to reach \(v=0\) from the equation of \(v(t)\), say \(t=t_{max}\), then plug it into the integrated formula of \(h(t)\). Solve the integral equation with the limits from 0 to \(t_{max}\) will yield the maximum height \(h_{max}\).