Drag models approach different terminal speeds

A metal ball is released from the surface of the liquid in a tall cylinder. The ball falls vertically through the liquid and the distance, xx cm, that the ball has fallen in time tt seconds is measured. The speed of the ball at time tt seconds is vv cm s⁻¹. The ball is released in a manner such that x=0x=0 and v=0v=0 when t=0t=0.

(a) The motion of the ball is modelled by the differential equation d2xdt2+12dxdt−10=0\frac{d^2x}{dt^2}+\frac12\frac{dx}{dt}-10=0. It is given that v=dxdtv=\frac{dx}{dt}.

(i) Show that the differential equation can be written as dvdt=10−12v\frac{dv}{dt}=10-\frac12v. mark

(ii) Using the differential equation in (a)(i), find vv in terms of tt. Hence find xx in terms of tt. marks

(b) The metal ball is now released in another tall cylinder filled with a different liquid. However, for this liquid, the motion of the ball is modelled by dvdt=10−k2v2\frac{dv}{dt}=10-k^2v^2, where kk is a positive constant. It is given that x=0x=0 and v=0v=0 when t=0t=0.

(i) Find vv in terms of tt and kk. marks

(ii) When the ball falls through this liquid, its speed will approach its “terminal speed” which is the speed it will attain after a long time. Find the ball’s terminal speed in terms of kk. You must show sufficient working to justify your answer. marks

[2024/TJC/Prelim/I/11]

Total marks: 14

Solution

(a)(i)

Reducing the order

Show that d2xdt2+12dxdt−10=0\dfrac{d^2x}{dt^2}+\dfrac12\dfrac{dx}{dt}-10=0 can be written dvdt=10−12v\dfrac{dv}{dt}=10-\dfrac12v.

Hint

  1. v=dxdtv=\frac{dx}{dt} so dvdt=d2xdt2\frac{dv}{dt}=\frac{d^2x}{dt^2}.

Answer · (a)(i)

v′=10−v/2v\prime=10-v/2.

(a)(ii)

Separation of variables

Find vv in terms of tt, and hence xx in terms of tt.

Explore this idea

Hint

  1. dvdt=20−v2\frac{dv}{dt}=\frac{20-v}2.

  2. Separate, integrate with the modulus, then apply v(0)=0v(0)=0.

  3. Integrate vv for xx with x(0)=0x(0)=0.

Answer · (a)(ii)

v=20(1−e−t/2)v=20(1-e^{-t/2}); x=20t+40e−t/2−40x=20t+40e^{-t/2}-40.

Explore the idea

Linear drag: v against t

Each dash shows the slope the differential equation gives at that point. Change the starting value and the solution follows the dashes.

Starting value of v at t = 00
Read the solution at t4
Try an example
v at t = 4
17.293
Rate dv/dt there
1.353
  • Solution that starts at the purple dot
  • Closed-form answer
Key idea

Whatever the start, the speed tends to the terminal speed 20 cm/s, where dvdt=0\frac{dv}{dt}=0.

Ideas to try
  1. Start the ball with a speed above 20 and see it slow down to 20.
  2. Read the speed at t=4t=4.
(b)(i)

Quadratic drag

For the second liquid, dvdt=10−k2v2\dfrac{dv}{dt}=10-k^2v^2 with v=0v=0 when t=0t=0. Find vv in terms of tt and kk.

Explore this idea

Hint

  1. Separate: dv10−k2v2=dt\frac{dv}{10-k^2v^2}=dt.

  2. Use the partial-fraction/standard form 12aln⁡∣a+ua−u∣\frac1{2a}\ln\left|\frac{a+u}{a-u}\right| with a=10a=\sqrt{10}, u=kvu=kv.

Answer · (b)(i)

v=(10/k)tanh⁡(k10 t)v=(\sqrt{10}/k)\tanh(k\sqrt{10}\,t).

Explore the idea

Quadratic drag (k = 1)

Each dash shows the slope the differential equation gives at that point. Change the starting value and the solution follows the dashes.

Starting value of v at t = 00
Read the solution at t1
Try an example
v at t = 1
3.151
Rate dv/dt there
0.071
  • Solution that starts at the purple dot
  • Closed-form answer
Key idea

The speed approaches 10≈3.16\sqrt{10}\approx3.16 much faster than the linear model approaches its limit.

Ideas to try
Raise the starting speed above 3.16: the ball slows down towards the same limit.
(b)(ii)

Terminal speed

Find the terminal speed in terms of kk, with justification.

Hint

  1. Terminal speed is the limit of vv as t→∞t\to\infty.

Answer · (b)(ii)

Terminal speed 10/k\sqrt{10}/k cm/s.

Positive equilibrium approached from rest; k>0k>0.