# A particle of mass m and radius a falls from rest under gravity in a ﬂuid whose resiWatch

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#1
A particle of mass m and radius a falls from rest under gravity in a ﬂuid whose resistance to motion,
given by Stokes’ drag law, is 6πηav. Here η is the ﬂuid viscosity and v the speed of the particle. Show that
the particle ultimately attains a speed of mg/6πηa (i.e. the terminal velocity), and ﬁnd the time taken to
reach half this speed.
Hint: The equation to be solved is dv/dt = g − kv where k = 6πηa/m and g is the constant acceleration due
to gravity.

---

I don't know how to approach this question, Do I literally start at

dv/dt = g - kv ?

What do I do with the given stokes drag law and how do I show the terminal velocity?

I'm really confused, any help much appreciated
0
5 years ago
#2
(Original post by lollage123)
A particle of mass m and radius a falls from rest under gravity in a ﬂuid whose resistance to motion,
given by Stokes’ drag law, is 6πηav. Here η is the ﬂuid viscosity and v the speed of the particle. Show that
the particle ultimately attains a speed of mg/6πηa (i.e. the terminal velocity), and ﬁnd the time taken to
reach half this speed.
Hint: The equation to be solved is dv/dt = g − kv where k = 6πηa/m and g is the constant acceleration due
to gravity.

---

I don't know how to approach this question, Do I literally start at

dv/dt = g - kv ?

What do I do with the given stokes drag law and how do I show the terminal velocity?

I'm really confused, any help much appreciated
Solve the differential equation to get v in terms of t. The terrminal velocity is the the limiting value of v as t tends to infinity.
0
5 years ago
#3
(Original post by lollage123)
Do I literally start at dv/dt = g - kv ?
Yes, it's a separable differential equation.

What do I do with the given stokes drag law
Nothing - it's all done for you; the deriving of the differential equation.

and how do I show the terminal velocity?
As the other poster said.
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