Ph 425 (Classical Mechanics II) : Mid-Term Test (open-book exam)

2:00 – 3:30 pm, 26 April 2000 Dr. R. Egerton

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G = 6.67 x 10^-11 N m²/kg² e ₀ = 8.85 x 10^-12 F/m e = 1.60 x 10^-19 C

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1. A hole is drilled at the south pole, towards the center of the earth (mass M) to a depth of one half the earth’s radius R_e. A rock (mass m) is dropped into the hole from rest. Derive expressions for:

1. the force F(r) when the rock is a distance r from the center of the earth
2. the weight of the rock at the bottom of the hole,
3. the increase in gravitational potential f when the rock is at bottom of the hole, compared to its value at the surface
4. the kinetic energy of the rock when it reaches the bottom of the hole;
5. the impulse transferred to the earth when it hits the bottom and stays there;
6. the net velocity acquired by the earth (along its polar axis) as a result of dropping the stone.

2. (a) A planet is held in a circular orbit (radius a) by a central force:

F(r) = – k/r² – c/r⁴ . Find the minimum value of a for which the orbit is stable.

(b) Give one reason why the laws of mechanics were developed on a planet whose orbit has a low eccentricity.

3. (a) A mass m makes a direct collision (coefficient of restitution e ) with a stationary mass M and a fraction a of its kinetic energy is liberated as heat. What is the ratio m/M ?

(b) For what kind of direct collision could m/M not be calculated, knowing of a and e ?

4. Two uniform solid spheres, each of radius r and mass m, are joined by a thin rod of length L and mass M which is aligned along the line joining the centers of the spheres. Specify:

(a) the center of mass of the object, measured relative to the center of the rod;

2. the moment of inertia of the object, when rotated about an axis perpendicular to the rod and passing through its center.
3. For what axis of rotation will the moment of inertia of the above object have the smallest possible value.
4. A square lamina (side a) is rotated about a central axis perpendicular to its plane. What is its radius of gyration ?

5. 100keV alpha-particles are elastically scattered through 60° by gold nuclei (Z=79). Calculate: (a) the impact parameter of these 60° collisions;

1. (a) F(r) = G m(M.r³/r_e³)/r² = GmMr/r_e³

(b) Since F(r) is proportional to r, weight = mg (r/r_e) = mg/2

(c) Gain in gravitational PE = work done on rock = ò F(r)dr

= - ò GmMr/r_e³dr = - GmM[r_e²- r_e²/4] / r_e³ = - (3/4) GmM/r_e

Increase in gravitational potential = PE/m = - (3/4) GM/r_e

(d) K = loss of gravitational PE = (3/4) GmM/r_e

(e) Impulse P = change in momentum of rock = mv - 0 = (2mK)^1/2

= m(1.5GM/r_e)^1/2

(f) Momentum of system before stone is released = M.0 + m.0 = 0

= momentum of system after = M.v' + m v' , therefore v’ = 0.

2. (a) The condition for stability of a circular orbit is (Eq.6.13.7, p.245):

f(a) + (a/3) f ’(a) < 0 , giving:

- k/a² – c/a⁴ + (a/3) [ 2k/a³ + 4c/a⁵ ] < 0

- (1/3) k/a² + (1/3) c/a⁴ < 0 , giving a > (c/k)^1/2

(b) Intelligent life would be impossible if the orbit were highly elliptical, due to temperature variations as the earth-sun distance varied.

3. (a) From Eq.(7.5.7), Q = (1/2)[mM/(m+M)] v²(1-e ²)

= a (mv²/2) where v is the speed of mass m before the collision,

giving a = [M/(m+M)] (1-e ²) , m/M = [(1 - e ²)/a ] – 1

2. If a = 0, we have e = 1 and m/M = [(0)/0] – 1which is indeterminate.

This corresponds to an elastic collision.

1. (a) Take the x-axis as the axis of the rod, with x=0 at its center.

x_cm = S x_i m_i / S m_i = 0 from symmetry, so C of M is at the center of the rod.

(b) I_z = S m_i x_i² = I_rod + 2 I _sphere

I _sphere = M of I about an axis displaced (L/2 + r) from the center of the sphere

= (2/5)L² + m(L/2+r)² from the parallel-axis theorem.

Therefore total moment of inertia: I_z= ML²/12 + (2/5)r² + m(L/2 + r)²

3. Moment of inertia will be minimum when the mass is as close to the rotation axis as possible. This occurs when the rotation axis passes through the center of each sphere and through the C of M , i.e. coincides with the axis of the rod.
4. If x and y are axes parallel to the sides of the square, I_x = I_y = ma²/12.

From perpendicular-axis theorem, I_z = I_x + I_y = ma²/6, giving k = a/Ö 6

5. (a) E = 10⁵ eV = 10⁵ e Joule. From Eq.(6.12.9), p. 242,

impact parameter b = [Qq/(2E. 4p e ₀.)] cot(60° /2) in SI units

= [(2e)(79e) / (2 x 10⁵ . e . 4p e ₀.)] cot(60° /2)

[(158)(1.60x10^-19) / (2x10⁵)(1.11 x 10^-10)] (1.732)

= 1.97 x 10^-12 m

(b) In SI units, ds /dW = [(Ze)(2e)/(4p e ₀.4E)]² / [sin⁴(60deg/2)]

ds /dW = [2Ze/(4p e ₀ . 4x10⁵)]² / [sin⁴(30deg)]

= [(158)(1.6 x 10^-19)/(4p e ₀.4x10⁵)]² / [sin⁴(30deg)]

= [(158)(1.6 x 10^-19)/(1.11 x 10^-10)(4x10⁵)]² [16]

= [5.70 x 10^-13]² [16] = 5.20 x 10^-24 m²/steradian