Question

A particle is moving under gravity in a medium whose resistance =
$mk{v}^{4}$
Find the motion where V is velocity of particle. ​

Answer 1

Answer:

We are here probably considering a small sphere falling slowly through a viscous liquid, with laminar flow around the sphere, rather than a skydiver hurtling through the air. In the latter case, the airflow is likely to be highly turbulent and the resistance proportional to a higher power of the speed than the first.

We'll use the symbol yy for the distance fallen. That is to say, we measure yy downwards from the starting point. The equation of motion is

y¨=g−γv,(6.3.8)

(6.3.8)y¨=g−γv,

where gg is the gravitational acceleration.

The body reaches a constant speed when y¨y¨ becomes zero. This occurs at a speed v^=gγv^=gγ , which is called the terminal speed.

To obtain the first time integral, we write the equation of motion as

dvdt=γ(v^−v)(6.3.9)

(6.3.9)dvdt=γ(v^−v)

or

dvv^−v=γdt.(6.3.10)

(6.3.10)dvv^−v=γdt.

dvv−v^=−γdt.(6.3.11)

(6.3.11)dvv−v^=−γdt.

DON'T! In the middle of an exam, while covering this derivation that you know so well, you can suddenly find yourself in inextricable difficulties. The thing to note is this. If you look at the left hand side of the equation, you will anticipate that a logarithm will appear when you integrate it. Keep the denominator positive! Some mathematicians may know the meaning of the logarithm of a negative number, but most of us ordinary mortals do not - so keep the denominator positive!

With initial condition v=0v=0 when t=0t=0 , the first time integral becomes

v=v^(1−eγt).(6.3.12)

(6.3.12)v=v^(1−eγt).

This is illustrated in Figure VI.5.

Students will have seen equations similar to this before in other branches of physics - e.g. growth of charge in a capacitor or growth of current in an inductor. That is why learning physics becomes easier all the time, because you have seen it all before in quite different contexts. Perhaps you have already noticed that third-year physics is easier than second-year physics; just think how much easier fourth-year is going to be! At any rate, vv approaches the terminal speed asymptotically, never quite reaching it, but reaching half of the terminal speed in time ln2γ=.693γln⁡2γ=.693γ (you have seen that before while studying radioactive decay), and reaching ( 1−e−11−e−1 ) = 63% of the terminal speed in time 1γ1γ .

If the body is thrown downwards, so that its initial speed is not zero but is v=v0v=v0 when t=0t=0 , you will write the equation of motion either as Equation 6.3.106.3.10 or as Equation 6.3.116.3.11 , depending on whether the initial speed is slower than or faster than the terminal speed, thus ensuring that the denominator is kept firmly positive. In either case, the result is

v=v^+(v0−v^)e−γt(6.3.13)

(6.3.13)v=v^+(v0−v^)e−γt

Figure VI.6 shows vv as a function of tt for initial conditions v0=0,12v^,v^,2v^v0=0,12v^,v^,2v^ .

Returning to the initial condition v=0v=0 when t=0t=0 , we readily find the second time integral to be

y=v^t−v^γ(1−e−γt).(6.3.14)

(6.3.14)y=v^t−v^γ(1−e−γt).

You should check whether this equation is what is expected for when t=0t=0 and when tt approaches infinity. The second time integral is shown in Figure VI.7.

The space integral is found either by eliminating t between the first and second time integrals, or by writing y¨y¨ as vdvdyvdvdy in the equation of motion:

vdvdy=γ(v^−v),(6.3.15)

(6.3.15)vdvdy=γ(v^−v),

whence

y=v^γln(1−vv^)−vγ.(6.3.16)

(6.3.16)y=v^γln⁡(1−vv^)−vγ.

This is illustrated in Figure VI.8. Notice that the equation gives yy as a function of vv , but only

numerical calculation will give vv for a given yy .