Classical Mechanics
Classical Mechanics
5th Edition
ISBN: 9781891389221
Author: John R. Taylor
Publisher: University Science Books
Question
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Chapter 4, Problem 4.39P

(a)

To determine

Show that τ=τ01π0Φdϕsin2(Φ/2)sin2(ϕ/2)=τ02π01du1u21A2u2.

(a)

Expert Solution
Check Mark

Answer to Problem 4.39P

It is proved that τ=τ01π0Φdϕsin2(Φ/2)sin2(ϕ/2)=τ02π01du1u21A2u2.

Explanation of Solution

Write the expression for the potential energy

    U(ϕ)=mgl(1cosϕ)        (I)

Here, U is the potential energy, m is the mass, g is the acceleration due to gravity, l is the length and ϕ is the angle.

Write the expression for the kinetic energy of the perpendicular

  K=12mv2        (II)

Here, K is the kinetic energy, v is the linear velocity.

Write the formula for angular velocity

    ω=vR        (III)

Here, ω is the angular velocity and R is the radius of the circle.

Where, ω=ϕ˙ and the length of the perpendicular is R=l. Equation (II) becomes,

K=12ml2ϕ˙2        (IV)

The total energy of the system is E=K+U.

Substitute equation (I) and (IV) in the above relation

E=12ml2ϕ˙2+mgl(1cosϕ)12ml2ϕ˙2=Emgl(1cosϕ)ϕ˙2=2Eml22gl(1cosϕ)ϕ˙=2Eml22gl(1cosϕ)

But t=dϕϕ˙

When ϕ becomes maximum, ϕ˙=0. And at ϕ˙=0 when E=U(ϕ). But ϕ˙=0 when ϕ=Φ.

Therefore, at ϕ=Φ

E=mgl(1cosΦ)

Using the above equation, ϕ˙ becomes

ϕ˙=2mglml2(1cosΦ)2gl(1cosϕ)=2gl(1cosΦ)2gl(1cosϕ)=gl[2(1cosΦ)2(1cosϕ)]=gl[2(2sin2Φ2)2(2sin2ϕ2)]

ϕ˙=2gl[sin2Φ2sin2ϕ2]

Substitute the above equation in t=dϕϕ˙

t=dϕ2gl[sin2Φ2sin2ϕ2]=12lgdϕsin2Φ2sin2ϕ2

Multiply and divide the above equation with 2π [Note: τ0=2πlg ]

=2π2×2πlgdϕsin2Φ2sin2ϕ2t=14πτ0lgdϕsin2Φ2sin2ϕ2

Since τ=4t,

τ=4[14πτ0lgdϕsin2Φ2sin2ϕ2]τ=τ01πdϕsin2Φ2sin2ϕ2        (V)

Let sin(ϕ2)=Au and sin(Φ2)=A.

Then,

 12cos(ϕ2)dϕ=Adudϕ=2Aducos(ϕ2)dϕ=2Adu1sin2(ϕ2)dϕ=2Adu1A2u2

And,

sin2(Φ2)sin2(ϕ2)=A2(Au)2=A2(1u2)

When ϕ=0, u=sin0=0

When ϕ=Φ, ϕ2=Φ2

sin(ϕ2)=sin(Φ2)Au=Au=1

Substituting the above relations in equation (V)

τ=τ01π012Adu1A2u21A1u2=τ02π01du1u21A2u2

Conclusion:

Hence it is proved that τ=τ01π0Φdϕsin2(Φ/2)sin2(ϕ/2)=τ02π01du1u21A2u2.

(b)

To determine

Show that for very small-amplitude period, the equation from part (a) gives familiar result τ=τ0=2πlg.

(b)

Expert Solution
Check Mark

Answer to Problem 4.39P

The assertion is proved that τ=2πlg for very small-amplitude period.

Explanation of Solution

Write the equation from part (a),

τ=τ01π0Φdϕsin2(Φ/2)sin2(ϕ/2)=τ02π01du1u21A2u2

When the amplitude, Φ is very small. Then A is also very small. Therefore, A2u2 can be neglected as it is very small compared with 1.

The equation becomes,

τ=τ02π01du1u2=τ02π[sin1(u)]01=τ02π[sin1(1)sin1(0)]=τ02π[π20]

=τ0τ=2πlg

Conclusion:

Hence, the assertion is proved that τ=2πlg for very small-amplitude period.

(c)

To determine

Show that τ=τ0[1+14sin2(Φ2)] for small amplitude.

(c)

Expert Solution
Check Mark

Answer to Problem 4.39P

The assertion is true that for amplitude that are small but not very small τ=τ0(1+14sin2(Φ2)).

Explanation of Solution

Given that 11A2u2 can be approximated to 1+12A2u2 and A is small but not very small.

Write the equation from part (a),

τ=τ02π01du1u21A2u2

The equation becomes,

τ=τ02π01du1u2(1+12A2u2)=τ02π[01du1u2+A2201(u21u2)du]=τ02π[01du1u2+A2201(u211u2+11u2)du]=τ02π[(1+A22)01du1u2+A22011u2du]

=τ02π[(1+A22)(π2)+A22[u21u2+12sin1(u)]01]=τ02π[(1+A22)(π2)+A22[0+12sin1(1)012sin1(0)]]=τ02π[(1+A22)(π2)+A22[12π2]]=τ02π[(1+A22A24)π2]

=τ0(1+A24)τ=τ0(1+14sin2(Φ2))

Conclusion:

Hence, the assertion is true that for amplitude that are small but not very small τ=τ0(1+14sin2(Φ2)).

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