(a) Write down the equation of motion for the point particle of mass m moving in the Kepler potential U(x) = -A/x+B/x² where x is the particle displacement in m. The potential, (b) Calculation of the dimension of the parameters (a) U(x) == x (c) The force is, Then, (c) =- mx. doc A 2B F=-33 Now, dx x ma= d²x B == V= + Introduce a dissipative term in the equation of motion assuming that the dissipative force acting on a particle is proportional to the partical velocity with the coefficient of proportionality v. Write down the modified equation of motion. (b) The dissipate force acts opposite to the acceleration, that is, it will try to reduce the acceleration. Then, dx dt B A 2B 2+2 2B 7+7 d²x A 1 2B 1 === -x-+-x X- m x m 2B ma=-=+ -Vv XX m A 1 2B 1 --x-+- m X- Determine the dimensions of parameters A, B, vin SI system of units. v dx m dt A 2B =-4+3 -vv ma=- [ma] = [Dimension of force] A 2B [+23-vv ] = [Dimension of force] -V [Dimension of force] = [1] [4] [M² L² T-^² ] = [~ 7² ] ⇒[4] = [M² L²¹ T~²][M°1²T°] [4]=[M¹ £ªT~²³] =kgm³/s² [Dimension of force] = 2B ⇒[x³]=[M² L¹ T~²][M°ËT°] =[M² L* T~²] [B]=kgm/s² [Dimension of force] =[Vv] [V]=[M°LT¹] [V][v]=[M² L¹ T-²] [v] = [M²I¹T²] [M°LT¹] =[M² LºT-¹] = kg/s
(a) Write down the equation of motion for the point particle of mass m moving in the Kepler potential U(x) = -A/x+B/x² where x is the particle displacement in m. The potential, (b) Calculation of the dimension of the parameters (a) U(x) == x (c) The force is, Then, (c) =- mx. doc A 2B F=-33 Now, dx x ma= d²x B == V= + Introduce a dissipative term in the equation of motion assuming that the dissipative force acting on a particle is proportional to the partical velocity with the coefficient of proportionality v. Write down the modified equation of motion. (b) The dissipate force acts opposite to the acceleration, that is, it will try to reduce the acceleration. Then, dx dt B A 2B 2+2 2B 7+7 d²x A 1 2B 1 === -x-+-x X- m x m 2B ma=-=+ -Vv XX m A 1 2B 1 --x-+- m X- Determine the dimensions of parameters A, B, vin SI system of units. v dx m dt A 2B =-4+3 -vv ma=- [ma] = [Dimension of force] A 2B [+23-vv ] = [Dimension of force] -V [Dimension of force] = [1] [4] [M² L² T-^² ] = [~ 7² ] ⇒[4] = [M² L²¹ T~²][M°1²T°] [4]=[M¹ £ªT~²³] =kgm³/s² [Dimension of force] = 2B ⇒[x³]=[M² L¹ T~²][M°ËT°] =[M² L* T~²] [B]=kgm/s² [Dimension of force] =[Vv] [V]=[M°LT¹] [V][v]=[M² L¹ T-²] [v] = [M²I¹T²] [M°LT¹] =[M² LºT-¹] = kg/s
Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter9: Multivariable Calculus
Section9.CR: Chapter 9 Review
Problem 54CR
Related questions
Question
![(a)
Write down the equation of motion for the point particle of mass m moving in
the Kepler potential U(x) = -A/x+B/x² where x is the particle displacement in m.
The potential,
(b)
Calculation of the dimension of the parameters
(a)
(c)
(c)
(d)
U(x) =
-A
==+3/1/2
x
(e)
The force is,
F=
Then,
(f)
mx
Now,
ܐ
Introduce a dissipative term in the equation of motion assuming that the
dissipative force acting on a particle is proportional to the partical velocity with the coefficient of
proportionality v. Write down the modified equation of motion.
ma=
dU
do
d
dx
A 2B
+3
(b)
The dissipate force acts opposite to the acceleration, that is, it will try to reduce the acceleration.
Then,
V_dx
dt
d²x
dt²
dt²=
B
2.B
2B
A 1
m x²
A 1 2B
m
m
-Vv
Determine the dimensions of parameters A, B, vin SI system of units.
2B 1 v dx
2
m
m dt
-Vv
[ma] =[Dimension of force]
2B
[- +23³ _ vv]=[Dimension of force]
-[+]
[]
ma=--
[Dimension of force] =
[M² L² T^²] =
⇒[4] =[M³ L¹ T~²][M°1²T°]
[4] = [M¹ Ľ³ T~²] =kgm³/s²
[Dimension of force] = []
⇒[x³]=[M² L¹ T~²][M°ËT°]
= [M² L²T²¹²]
[B]=kgm/s²
[Dimension of force]=[Vv]
[V]=[M°L¹T¹]
[V][v]=[M² L² T²³]
[M² L¹ T-²]
[M°LT¹]
=[M² L°T-¹]
= kg/s
Mimimise the number of independent parameters in the modified equation of
motion with dissipation by introducing dimensionless varibles and r for x and 1.
[v] =!
Present the equation derived in the form of an energy balance equation and
find the expressions for the effective kinetic and potential energies (7 and V), and dissipative
function F.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F88ea2a71-9ac4-4e85-921f-8e0d04f36029%2Fe70788ac-98dc-4895-b797-a44ff10af83f%2Fnz4ci2_processed.png&w=3840&q=75)
Transcribed Image Text:(a)
Write down the equation of motion for the point particle of mass m moving in
the Kepler potential U(x) = -A/x+B/x² where x is the particle displacement in m.
The potential,
(b)
Calculation of the dimension of the parameters
(a)
(c)
(c)
(d)
U(x) =
-A
==+3/1/2
x
(e)
The force is,
F=
Then,
(f)
mx
Now,
ܐ
Introduce a dissipative term in the equation of motion assuming that the
dissipative force acting on a particle is proportional to the partical velocity with the coefficient of
proportionality v. Write down the modified equation of motion.
ma=
dU
do
d
dx
A 2B
+3
(b)
The dissipate force acts opposite to the acceleration, that is, it will try to reduce the acceleration.
Then,
V_dx
dt
d²x
dt²
dt²=
B
2.B
2B
A 1
m x²
A 1 2B
m
m
-Vv
Determine the dimensions of parameters A, B, vin SI system of units.
2B 1 v dx
2
m
m dt
-Vv
[ma] =[Dimension of force]
2B
[- +23³ _ vv]=[Dimension of force]
-[+]
[]
ma=--
[Dimension of force] =
[M² L² T^²] =
⇒[4] =[M³ L¹ T~²][M°1²T°]
[4] = [M¹ Ľ³ T~²] =kgm³/s²
[Dimension of force] = []
⇒[x³]=[M² L¹ T~²][M°ËT°]
= [M² L²T²¹²]
[B]=kgm/s²
[Dimension of force]=[Vv]
[V]=[M°L¹T¹]
[V][v]=[M² L² T²³]
[M² L¹ T-²]
[M°LT¹]
=[M² L°T-¹]
= kg/s
Mimimise the number of independent parameters in the modified equation of
motion with dissipation by introducing dimensionless varibles and r for x and 1.
[v] =!
Present the equation derived in the form of an energy balance equation and
find the expressions for the effective kinetic and potential energies (7 and V), and dissipative
function F.
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