Classical Mechanics - Homework Assigment 1 Alejandro G´ omez omez Espinosa
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September 19, 2012
Goldstein, Ch.1, 12 The escape velocity of a particle on Earth is the minimum velocity
required at Earth’s surface in order that the particle can escape from Earth’s gravitational field. Negle Neglecting cting the resistanc esistancee of the atmospher atmosphere, e, the system is conserva conservative. tive. From the conservation theorem for potential plus kinetic energy, show that the escape velocity for Earth, ignoring the presence of the Moon, is 11.2 km/s. From the conservation of energy: E before b efore = E after a fter . At the beggining, the particle has only potential potential energy energy.. Once this particle escapes from Earth, it cannot feel Earth’s gravitational field and its potential energy is zero, having only kinetic energy. Then, V b = T a mM 1 2 G = mv r 2 GM v2 = r 2 where, G is the gravitational constant, M is the mass of the Earth, m is the mass of the particle, r is the distance between the Earth and the particle, and v is the velocity of the particle. Solving for v and replacing with the values: v2 =
2GM 2(6. 2(6.67 × 10 11)(6 × 1024) = r 6 × 106 −
v = 1118 m/s From we can confirm that the escape velocity from Earth is 11. 11.2 km/s.
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Goldstein, Ch.1, 14 Two points of mass m are joined by a rigid weightless rod of lenght
l, the center of which is constrained to move on a circle of radius a. Expr Expres esss the kinetic energy in generalized coordinates. A graphical description of this problem is sketch in Figure 1. Here, we can split the kinetic kinetic energy energy in two: two: a circul circular ar motion of a point point in the plane plane x − y and the tridimensional motion of the rigid weightless rod with two masses in the plane x − y −z .
Figure 1: Sketch of the problem 14. Here, two points of mass are joined by a rigid rod and the center is constrained to move on a circle. For the first part of the kinetic energy, the total mass of the system is the sum of the two point masses. In additon, we define the angle between the position r and x as α: 1 1 2 T 1 = M vcm = (2m (2m)(a )(aα˙ )2 = ma2α˙ 2 2 2 Then, for the second part of the kinetic energy we used the other coordinates system where the rod is divided in the middle. We set the position of one of the point masses masses in spherical spherical coordinates: coordinates: L sin θ cos φ 2 L y = sin θ sin φ 2 L x = cos θ 2 where φ is the angle in the plane x − y and θ is the angle azimuthal. Here, when we define the position of one of the particles particles the other is fixed. Therefore, Therefore, to calculate 2 2 2 2 the velocity v = x˙ + y˙ + z˙ we find the temporal derivative of each variable: x =
L (cos φ cos θθ˙ − sin θ sin φφ˙ ) 2 L y˙ = (sin φ cos θθ˙ + sin θ cos φφ˙ ) 2 L z˙ = − sin θθ˙ 2
x˙ =
2
2
x˙
=
y˙ 2 = z˙ =
L2 (cos2 φ cos2 θθ˙ 2 − 2sin θ sin φ cos φ cos θθ˙ φ˙ + sin2 θ sin2 φφ˙ 2 ) 4 L2 (sin2 φ cos2 θθ˙ 2 + 2 sin sin θ cos φ sin φ cos θθ˙ φ˙ + sin2 θ cos2 φφ˙ 2 ) 4 L2 sin2 θθ˙ 2 4
Then, the total velocity is: v2 = = =
x2 + y˙ 2 + z˙ 2 L2 (cos2 φ cos2 θθ˙ 2 + sin2 θ sin2 φφ˙ 2 + sin2 φ cos2 θθ˙2 + sin2 θ cos2 φφ˙ 2 + sin2 θθ˙2 ) 4 L2 ˙ 2 (θ + sin2 θφ˙ 2 ) 4
·
And the second part of the kinetic energy, for the two masses: T 2 = 2
1 2
mv
2
L2 ˙2 = m (θ + sin2 θφ˙ 2 ) 4
Finally, the total kinetic energy in generalized coordinates is: T = T 1 + T 2 = ma2 α˙ 2 +
3
mL2 ˙2 (θ + sin2 θφ˙ 2 ) 4
are connecte onnected d by a string string Goldstein, Ch.1, 21 Two mass points of mass m1 and m2 are passing through a hole in a smooth table so that m that m1 rests on the table surface and m2 hangs suspended. Assuming m Assuming m2 moves only in a vertical line, what are the generalized coordin ordinate atess for the system system?. ?. Write Write the Lagrang agrangee equatio quations ns for the system system and, if possible, discuss the physical significance any of them might have. Reduce the problem to a single second-order differential equation and obtain a first integral of the equation. What is its physica physicall significanc significance? e? (Consider (Consider the motion only until m1 reaches the hole.) The mass m1 on the the table table surf surfac acee can can mo move ve only in that that plan plane. e. Ther Theref efor ore, e, we can describe the position of this mass with the distance r and the angle θ, since: x = r cos θ and y = r sin θ. The kinetic energy is: 1 T 1 = m1 (r˙ 2 + r2 θ˙2 ) 2 For the mass m2 , the motion is constrained in the vertical line. Its position is z = L−r where L is the total length of the string. Then, the kinetic and potential energy are: 1 1 T 2 = m2 z˙ 2 = m2 r˙ 2 2 2
V 2 = m2 g (−z ) = −m2 g (L − r)
Hence, the generalized coordinates for this system are r and θ, and the Lagrangian is: 1 L = T − V = [(m [(m1 + m2)r˙ 2 + m1 r2 θ˙2 ] + m2 g (L − r) 2 Then, we have to solve the Euler-Lagrange equation: ∂L ∂x
−
∂ ∂t
∂L ∂ ˙ ∂ x˙
=0
for each generalized coordinate. First for θ: ∂L ∂θ ∂L ∂ θ˙
∂ ˙ = m r θ ∂t ∂ ˙ m r θ =0 m r θ˙ = k (constant) ∂ ∂t
⇒
= 0 2
1
2
⇒
1
∂t For the other generalized coordinate:
∂ ∂t
∂L ∂r ∂L ∂ ˙ ∂ r˙
2
1
= m1 rθ˙2 − m2 g =
∂ ((m ((m1 + m2)r˙ ) = (m1 + m2 )¨ r ∂t
m1 r θ˙2 − m2 g − (m1 + m2 )¨ r=0 if we replace the value of θ˙ from the previous equation: θ˙ = k/( k/(m1 r2 ), then: ⇒
m1 r
k m1 r2
2
−
m2 g − (m1 + m2 )¨ r=0
(m1 + m2 )¨ r−
4
k2 + m2 g = 0 m1 r 3
Goldstein, Ch.1, 23 Obtain the equation of motion for a particle falling vertical under
the influence of gravity when frictional forces obtainable from a dissipation function 1 kv 2 are are present present.. Integr Integrate ate the equatio equation n to obtain the velocity velocity as a function function of time 2 and show that the maximum possible velocity for a fall from rest is v = mg/k. mg/k . If we consider the motion in the z -axis, our Lagrangian has the usual form: 1 L = mz˙ 2 − mgz 2 According to the problem the frictional force is F = 12 kv 2 = 12 kz˙ 2 . Then Then the EulerEulerLagrange equation, including dissipative forces is: ∂ ∂t
∂L ∂ ˙ ∂ z˙
−
∂L ∂F + =0 ∂z ∂ ˙ ∂ z˙
Solving Solving this equation: mz¨ − mg + k z˙ = 0
⇒
mz¨ = mg − k z˙
From this equation, we can conclude that when the total force goes to zero: mg = k z˙ . It means that the maximum possible velocity is reach when the total force goes to zero, zero, and has has the form: form: z˙ = mg . z
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