Resistive Forces
1. A force that depends on the motion it is opposing
Every force so far has been fixed by something outside the motion: how heavy the object is, how rough the surface is, how hard somebody pushes. A resistive force is different. It depends on how fast the object is already going, and it points opposite to the velocity:
Fr = −k v
The minus sign is what makes it always oppose the motion, whichever way that motion happens to be. The constant k is measured in kilograms per second, and it describes the object and the fluid together, not the object alone.
This topic exists only in AP Physics C. AP Physics 1 never meets a force that changes as the motion changes, which is why it can do the whole of Unit 2 with algebra.
2. The second law becomes a differential equation
Apply the second law to an object falling under gravity with this force on it, and the unknown appears on both sides:
m dv/dt = mg − kv
That is not an equation you can solve for v by rearranging, because v is a function and its own derivative is in the way. It is a differential equation, and recognizing it as one, and saying so, is worth points on its own.
Notice immediately what it tells you without being solved. At the start v = 0, so the acceleration is g. As the object speeds up, the kv term grows and the acceleration shrinks. The motion is therefore not uniformly accelerated at any point, and none of the constant-acceleration equations from Unit 1 may be used here, not even approximately.
3. Terminal velocity, without solving anything
The easiest question in the topic can be answered in one line. As the speed grows, the resistive force grows with it, and eventually the two forces on the object are equal and the net force is zero. That is the terminal condition, and from it comes the terminal velocity:
mg = k vtermso vterm = mg / k
It is an asymptote, not an arrival. The object gets arbitrarily close and never gets there, and a question that asks when the object reaches terminal velocity is asking about a limit rather than about a moment.
An object dropped from rest. The three gold dots mark one, two and three time constants, and they always sit at the same fractions of the terminal velocity no matter what you do to the mass or the drag. Notice that the strobe images stop spreading out once the curve flattens.
4. Separating the variables
To get the velocity as a function of time, put everything containing v on one side and everything containing t on the other, then integrate both sides over limits that match the physical situation:
∫0v dv / (mg − kv) = ∫0t dt / m
Definite limits are cleaner than a constant of integration and they are what the rubric asks for. The lower limits say the object started from rest at time zero; change the physical situation and those limits change with it. Write them before you integrate, because most of the points lost in this topic are lost by not writing them at all.
Doing the integral gives the standard result:
v(t) = (mg/k)(1 − e−kt/m)with τ = m/k
The combination m/k has units of time and is called the time constant. After one of them the speed has reached 63 per cent of its terminal value, after two 86 per cent, after three 95 per cent, and those fractions do not depend on the mass or the drag constant at all.
5. Everything about the motion is exponential
Once the velocity function is known, the acceleration follows by differentiating and the position by integrating, using the initial conditions to fix the constants. All three are exponentials with asymptotes:
The velocity rises toward mg/k. The acceleration falls from g toward zero. The position becomes a straight line, because once the object is moving at very nearly a constant speed, it covers very nearly equal distances in equal times.
The case with no driving force is worth knowing too. An object coasting with only drag on it has v = v0e−t/τ, which never quite reaches zero, and yet the total distance it covers is finite and equal to v0τ. Never stopping and going a finite distance are not in conflict.
No gravity, no push, only drag. The strobe images crowd toward a line the object never crosses, and the speed curve approaches zero without touching it. Both facts come from the same exponential.
Drag the gold dot up and down the left edge to choose the speed the object starts with. The faint curves are the same object released at four other speeds. Every one of them ends up on the same dashed line, from below and from above, and none of them ever quite reaches it. Terminal velocity is not where the object is heading because something pushes it there; it is simply the one speed at which the drag has grown enough to cancel the weight.
m dv/dt = mg − bv, so v(t) = vT + (v0 − vT)e−t/τ, with vT = mg/b and τ = m/b
The exponential never gets to zero, so the object never actually reaches terminal velocity. What it does is close the same fraction of the remaining gap in every time constant: about 63 percent in the first, another 63 percent of what is left in the second, and so on.
6. Reading it wrong
Using a kinematics equation. The acceleration is not constant for a single instant. If a solution contains v = v0 + at anywhere, it is wrong before the arithmetic starts.
Integrating without limits. An indefinite integral leaves a constant nobody has fixed, and the initial condition is the only thing that can fix it. Put the limits on first.
Saying the object reaches terminal velocity at some time. It approaches it. Give the time to reach 95 per cent of it if a number is wanted, and name the fraction you used.
Treating k as a property of the object. It describes the object and the fluid together. The same ball has a different k in air and in water.
Check yourself
1. A 0.20 kg ball falls through a fluid with Fr = −kv and k = 0.40 kg/s. Find the terminal velocity and the time constant, and state the acceleration at the instant it is released.
vterm = mg/k = (0.20)(9.8)/0.40 = 4.9 m/s and τ = m/k = 0.50 s. At release the speed is zero, so the resistive force is zero and the acceleration is the full 9.8 m/s². The largest acceleration in the whole motion is the one at the very beginning.
2. Write the differential equation for that ball, separate the variables, and state the limits of integration for a ball released from rest.
m dv/dt = mg − kv. Separating: dv/(mg − kv) = dt/m. The limits are from 0 to v on the left and from 0 to t on the right, because the ball was at rest at time zero. Those two zeros are the initial condition, and they are the whole reason no constant of integration appears in the answer.
3. A student writes that the ball in question 1 reaches 4.9 m/s after 0.50 s. Correct the statement and give a defensible number instead.
After one time constant the speed is 4.9(1 − e⁻¹) = 3.1 m/s, which is 63 per cent of the terminal value rather than all of it. The ball never reaches 4.9 m/s exactly. A defensible answer names a fraction: it is within 5 per cent of terminal after three time constants, that is 1.5 s.
4. A puck of mass m slides on a frictionless surface through a fluid with Fr = −kv and no other horizontal force. Describe its speed as a function of time and find how far it travels in total.
With gravity absent from the horizontal direction the equation is m dv/dt = −kv, giving v = v0e−kt/m: an exponential decay with time constant m/k. Integrating the speed from zero to infinity gives a total distance of mv0/k, which is finite. The puck never formally stops and still covers only a limited distance, because the tail of an exponential has a finite area.
5. Two objects fall through the same fluid, one twice the mass of the other but otherwise identical, so they share the same k. Compare their terminal velocities and their time constants, and say which one is still accelerating noticeably after a given time.
Both scale with m: the heavier object has twice the terminal velocity and twice the time constant. So the heavier one is still accelerating noticeably when the lighter one has already settled, and it ends up going twice as fast. Note that this is the opposite of the vacuum result, where mass makes no difference at all, and the reason is that mass appears in the weight but not in the drag.
Topic 2.10, Circular Motion. Every force in this unit is now available. The last topic takes an object that is going at a steady speed and still accelerating, and asks which of these forces is doing it.