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Homework 1 : Classical Mechanic I

Harirak Intarak

5405282

Finding the force by knowing the potential energy

Our staring point we consider particle acted by conservative force F(r) , with corresponding potential energy U(r) ,

and examine the word done by F(r) in a small displacement from r to dr , From the definition of work ;

W (r → r + dr) = F (r) · dr

= F xdx + F ydy + F zdz (1)

for any small displacement dr with component (dx,dy,dz) On the other hand, From the definition of potential

energy is

U r+dr − U r = −

∫ r+drr

F (r′) · dr′

When dr is the small displacement that F is not vary appreciably over the path, the integral on the right side is

approximately to −F · drdU = −F · dr (2)

When we put (2) to (1) we get ;

W (r → r + dr) = −dU = −[U (r + dr) − U (r)]

= −[U (x + dx, y + dy, z + dz )− U (x,y ,z )] (3)

For functions of one variable, a difference like that in (3) can be expressed in terms of the derivative :

df = f (x + dx)− f (x) = df

dx dx

We have functions of three variable in (3) ,the corresponding result is

dU = U (x + dx, y + dy, z + dz ) − U (x,y ,z )

= ∂U

∂x dx +

∂U

∂y dy +

∂U

∂z dz (4)

Where the three derivatives are the partial derivatives with respect to the three in dependent variable (x,y,z)

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Substituting (4) into (3), we can find that the work done in the small displacement from r to r+dr is

W (r → r + dr) = −

∂U

∂x dx +

∂U

∂y dy +

∂U

∂z dz

(5)

We can choose dr to point in the x direction, in which case d y =d z = 0. We can see that F x = −∂U /∂x . Bychoosing dr to point in the y or z direction, we get corresponding results for F y and F z , and we conclude that

F x = −∂U /∂x, F y = −− ∂U /∂y, F z = −∂U /∂z (6)

This is F , is the vector whose three components are minus the partial derivatives of U with respect to x, y, and z. We

can write this result with this :

F = −x̂∂U

∂x − ŷ

∂U

∂y − ẑ

∂U

∂z (7)

The vector whose three components are the partial derivatives of f (r) is called the gradient of f , denoted∇f :

∇f = x̂∂f

∂x

+ ŷ∂f

∂y

+ ẑ∂f

∂z

(8)

From ( 7 ) we can write that

F = −

x̂

∂U

∂x + ŷ

∂U

∂y + ẑ

∂U

∂z

(9)

From (9) we can write x̂∂U ∂x

+ ŷ ∂U ∂y

+ ẑ ∂U ∂z

into (8) form :

F = −∇U (10)

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Problem 2 : Taylor 4.16 If a particle's potential energy is U (r) = k(x2 + y2 + z 2) ,where k is a constant, what isthe force on the particle?

Solution

From ⃗ F c = −⃗ ∇U

= −⃗ ∇{k{x2 + y2 + z 2}}

= −

x̂

∂

∂x + ŷ

∂

∂y + ẑ

∂

∂z

k(x2 + y2 + z 2)

= −2kxx̂− 2kyŷ − 2kz ̂z

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Problem 3: Kleppner 5.1

Find the forces for the following potential energies.

a. U = Ax2 + By2 + Cz 2

b. U = A ln(x2 + y2 + z 2) (ln = loge)

c. U = A cos θ/r2 ( plane polar coordinates)

Solution

a.

From ⃗ F = −⃗ ∇U

= −⃗ ∇

Ax2 + By2 + Cz 2

= −

x̂

∂

∂x + ŷ

∂

∂y + ẑ

∂

∂z

Ax2 + By2 + Cz 2

= −2Axx̂− 2Byŷ − 2Cz ̂z

b.

From ⃗ F = −⃗ ∇U

= −⃗ ∇

A ln

x2 + y2 + z 2

= −

x̂

∂

∂x + ŷ

∂

∂y + ẑ

∂

∂z

A ln

x2 + y2 + z 2

= −x̂A

1x2 + y2 + z 2

∂ ∂x (x

2

+ y2

+ z 2

)

+ ŷA

1

x2 + y2 + z 2

∂

∂y(x2 + y2 + z 2)

+ ẑA

1

x2 + y2 + z 2

∂

∂z (x2 + y2 + z 2)

= − 2Ax

x2 + y2 + z 2 x̂−

2Ay

x2 + y2 + z 2 ŷ −

2Az

x2 + y2 + z 2ẑ

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Problem 3: Kleppner 5.1

Solution

c.

From ⃗ F = −⃗ ∇U ( plan polar coordinates)

Because it is in plan polar coordinates, the ⃗ ∇ = r̂ ∂ ∂r

+ θ̂ 1r∂ ∂θ

= −⃗ ∇

A cos θ/r2

= −

r̂

∂

∂r + θ̂

1

r

∂

∂θ

A cos θ/r2

= −

r̂A cos θ

∂

∂r

1

r2

+ θ̂A

1

r3

∂

∂θ(cos θ)

= −r̂A cos θ−2r3 + θ̂A 1

r3 (− sin θ)

= 2A

cos θ

r3

r̂ + A

sin θ

r3

θ̂

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Problem 4 : Taylor 5.1 A massless spring has unstretched length l0 and force constant k . One end is now attached to

the ceiling and a mass m is hung from the other. The equilibrium length of the spring is now l1.

(a) Write down the condition that determines l1. Suppose now the spring is stretched a further distance x beyond its new

equilibrium length. Show that the net force (spring plus gravity) on the mass is F = −kx. That is, the net force obeysHooke's law, when x is the distance from the equilibrium position - a very useful result, which lets us treat a mass on a vertical

spring just as if it were horizontal.

(b) Prove the same result by showing that the net potential energy (spring plus gravity) has the form

U (x) = const + 12kx2.

Solution a. Consider a mass m while it is in the equilibrium after hung with massless spring∑F = 0

kx = mg

k(l1 − l0) = mg (11)

Consider the mass m while it is in stretched a further distance x beyond its new equilibrium length.

F = −k(l1 − l0 + x) + mg= −k(l1 − l0) − kx + mg

Use (11) ;

F = −mg − kx + mg

= −kx

so that the net force on the mass is

F = −kx

Solution b. From the gravity potential energy U g(x) = mgx + C , when x is the vertical displacement fromarbitrary point that have potential energy is equal C and spring potential energy U sp(x) =

12

kx2 , when x is the distance

of the spring that has unstreched

Consider a mass m while is stretched a further distance x from the equilibrium position. Define that the reference point

of potential energy is the equilibrium point of spring.

U (x) = U g(x) + U sb(x)

= −mgx + C + 1

2k(l1 − l0 + x)

2

= −mgx + C + 12

k((l1 − l0)2 + 2(l1 − l0)(x) + x

2)

Use (11) mg = k(l1 − l0) ;

U (x) = −k(l1 − l0)x + C + 1

2k((l1 − l0)

2 + 2(l1 − l0)(x) + x2)

= C + 1

2k(l1 − l0)

2 + 1

2kx2

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Problem 4 : Taylor 5.1

Due to term of 12

k(l1 − l0) that is the constant. So we can write ;

U (x) = const + 1

2kx2

From ⃗ F = −⃗ ∇U and it is only in x-axis, so we can find net force on the mass ;

F = − d

dx

const +

1

2kx2

F = −kx

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Problem 4: Taylor 5.2 The potential energy of two atoms in a molecule can sometimes be approximated by the Morse

function, U (r) = A[(e(R−r)/s− 1)2− 1] where r is the distance between the two atoms and A, R, and S are positiveconstants with S ≫ R. Sketch this function for 0

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Problem 4: Taylor 5.2

To make sure that R = r0 is true, we can check from (13) by substitute r0 = R . U ′′(r0) must be positive. ;

U ′′(r0) = U ′′(R) =

2A

S 2

e(R−R)/S

2+

2A

S 2

e(R−R)/S − 1

e(R−R)/S

U ′′(R) = 2AS 2

(14)

U ′′(r0) is positive, so the system has a stable equilibrium position R = r0

Consider U (r0 + x) whereas x is small , when expanded in Taylor series. ;

U (r0 + x) = U (R + x)

= U (R) + U ′(R)x + 1

2U ′′(R)x2 +

1

3!U ′′′(R)x3 + ...

As long as x remains small, the first of three term in this series should be a good approximation. And U ′

(R) = 0 ;

U (r0 + x) = A

e(R−R)/S − 1

2− 1

+ 0 +

1

2U ′′(R)x2

= −A + 1

2U ′′(R)x2 ; From(14) U ′′(R) =

2A

S 2

= −A + 1

2

2A

S 2

x2

If the displacement x from stable equilibrium position is the lowest, the potential energy will be ;

U = const + 12

kx2

Whereas k = 2AS 2

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Problem 4: Taylor 5.3 Write down the potential energy U(0) of a simple pendulum (mass m, length 1) in terms of the

angle 0 between the pendulum and the vertical. (Choose the zero of U at the bottom.) Show that, for small angles, U has

the Hooke's law form U (0) = 12

kφ2,in terms of the coordinate φ.

What is k?

Solution From potential energy is U (h) = mgh ,when h is the different from the bottom to the height of massm can reach.

h = l − l cos φ

= l(1 − 1 cos φ)

F rom U (h) = mgh

U (φ) = mgl(1 − 1 cos φ)

When the angle φ is small then cos φ = 1− φ2

2

U (φ) = mgl(1 − (1 − φ2

2 ))

= mglφ2

2

= 1

2mglφ2

From the problem, U has the Hooke's law form U (φ) = 12

kφ2

We totally find k that

k = mgl

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Problem 4 : Magnitude VS Component

Force ⃗ W = m⃗g = mg(−ˆ j)

Magnitude of ⃗ W = mg

The component in ˆ j direction is 0 The component in −ˆ j direction is mg

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