## Assignment 1, solutions

#### Problem 1:

Two masses are arranged as shown.  Mass m1 has a mass of 4.00 kg and is attached to the vertical surface on the left with a string in which the tension is T.  Mass m2 = 6.00 kg, is sitting on the horizontal surface and is being pulled to the right by a force F, so that m2 is moving to the right at a constant speed.  The coefficient of sliding friction between m1 and m2 is μ1 = 0.380, while the coefficient of sliding friction between m2 and the horizontal surface is μ2 = 0.510.
(a)  What will be the tension T in the string?
(b)  What will be the magnitude of the force F required to pull m2 to the right at a constant speed?

Solution:

• Concepts:
Newton's second law
• Reasoning:
The net force on each mass is zero.
• Details of the calculation:
(a)  T = μ1m1g = 14.9 N.
(b)  F = μ1m1g + μ2(m1 +m2)g = 64.9 N.

#### Problem 2:

(a)  Imagine that the earth were of uniform density and that a tunnel was drilled along a diameter.  If an object were dropped into the tunnel, show that it would oscillate with a period equal to the period of a satellite orbiting the earth just at the surface.
(b)  Find the gravitational acceleration at a point P, a distance x from the surface of a spherical object of radius R.  The object has density ρ.  Inside the object is a spherical cavity of radius R/4.  The center of this cavity is situated a distance R/4 beyond the center of the large sphere C, on the line from P to C.

Solution:

• Concepts:
Newton's law of gravity, Newton's 2nd law, the principle of superposition, uniform circular motion
• Reasoning:
In part (a) the acceleration of the of the objects is due to the gravitational force.  The gravitational force on a object in the tunnel, a distance r from the center is in the –r direction and its magnitude is found using "Gauss' law".
[(∇∙Fg/m) = -4πGρ, for a spherical mass distribution.
4πr2(Fg/m) = 4πGMinside, (Fg/m) = GMinside/r2
Here Fg/m is the gravitational force per unit mass.]
For part (b) we can use the principle of superposition. We compute the acceleration due to a sphere with radius R without a hole and subtract the acceleration due to a sphere with radius R/4 at the location of the hole.
• Details of the calculation:
(a)  For the object of mass m in the tunnel a distance r from the center of the earth, the magnitude of the gravitational force is F = (4/3)Gmπρr, with ρ = 3M/(4πR3).  M and R are the mass and radius of the earth, respectively.  The force points towards the center.
For an object moving in the tunnel we therefore have F = -kr, k = (4/3)Gmπρ.
The force on the object obeys Hooke's law.  The object will oscillate with angular frequency ω = (k/m)½ = (4Gπρ/3)½.
Its period is T = 2π/ω = (3π/(Gρ))½ = 2π(R3/(GM))½.
For a satellite orbiting just above the surface we have
GMms/R2 = msv2/R = ms2,  ω2 = GM/R3, T = 2π/ω = 2π(R3/(GM))½.
The object in the tunnel and the satellite have the same period.

(b)  a = Gρ(4/3)πR3/(x + R)2 - Gρ(4/3)π(R/4)3/(x + 5R/4)2.
a = Gρ(4/3)πR3[(x + R)-2 - (8x + 10R)-2].
The direction of a is towards the center of the sphere.

#### Problem 3:

A particle of mass m is given an initial velocity v0i.  Assume that the particle is subject to a drag force F = -bv½i.
(a)  Find v as a function of time.
(b)  How far does the particle travel before coming to rest?

Solution:

• Concepts:
Newton's second law
• Reasoning:
Given the acceleration and v(0) we can find v(t) and Δx(t) as a function of time.
• Details of the calculation:
(a)  mdv/dt = -bv½,  dv/dt = -(b/m)v½,  dv/v½  = -(b/m )t,
v0v(t) dv/v½  = -(b/m) ∫0tdt = -bt/m,
2(v(t)½ - v0½) = -bt/m,
v(t) = (v0½ - bt/(2m))2.

(b)  v(t) = 0 --> tf  = 2mv0½/b.
After a time interval tf the particle stops.  In the time interval tf the particle travels a distance Δx.
Δx = ∫0tfv(t) dt = ∫0tf(v0½ - bt/(2m))2dt = [(v0½ - btf/(2m))3- v03/2]/(-3b/(2m)).
Δx = 2mv03/2/(3b).

#### Problem 4:

A thrill-seeking student jumps off a bridge from the height H = 80.0 m above the water level.  The student is attached to the bridge by an elastic cord so that she reaches zero velocity just as she touches the water.  After some up-and-down bouncing, the student eventually comes to rest h = 20.0 m above the water level.  What is the maximum speed reached by the student during the fall?   Neglect air resistance and use 10.0 m/s2 for the acceleration due to gravity.

Solution:

• Concepts:
Gravitational and elastic potential energy
• Reasoning:
Let us assume that dissipative forces are negligible during the first descent and that mechanical energy is conserved during the first descent.  Let us also assume that the bungee cord obeys Hooke's law when stretched, with a spring constant k.
There are three distances we need to consider, the unstretched length of the cord, L, the equilibrium length of the cord with the student attached, d, and the maximum stretched length of the cord H.
• Details of the calculation:
Hooke's law gives for the equilibrium position
mg = k(d - L), mg/k = d - L.
At the end of the first descent v = 0, all the gravitational potential energy has been converted into elastic potential energy.
mgH = ½k(H - L)2, mg/k = d - L = (H - L)2/(2H).
L2 = 2Hd - H2 = (2*80*60 - 802) m2 = 3200 m2.  L = 56.57 m.

The jumper speed is a maximum when the acceleration is zero, which is at the equilibrium position.  Then, from energy conservation, mgd - ½k(d - L)2 = ½mv2.
mgd - ½mg(d - L) = ½mv2.  v2 = 2gd - gd + gL = g(d + L) = 1165.7 m2/s2.
vmax = 34.14 m/s.

#### Problem 5:

Assume a perfectly spherical Earth of radius R with a frictionless surface.  On the surface of this Earth an object with mass m is moving with constant speed v towards the north pole.  When the object is at latitude λ, find the external force required to keep it moving on that trajectory.
For m = 1 kg,  v = 500 m/s, λ = 45o give a numerical answer.

MEarth = 5.97*1024 kg, REarth = 6378 km.

Solution:

• Concepts:
Motion in an accelerating frame
• Reasoning:
Assume an observer at mid latitudes λ = 90o - θ.
The equations of motion in a rotating coordinate system contain fictitious forces.
mdv/dt = Finertial - mΩ × (Ω × r) - 2mΩ × v
= -mg + N + Fext - mΩ × (Ω × r) - 2mΩ × v.
-mΩ × (Ω × r) = centrifugal force.
(Ω × r) = ΩRsinθ j.  Ω × (Ω × r) = -Ω2Rsin2θ  k - Ω2Rsinθcosθ  i.
-2mΩ × v = Coriolis force.
Ω × v = -Ωvcosθ j.
• Details of the calculation:
We want  dvtangential/dt = 0.  We need 0 = Fext + mΩ2Rsinθcosθ  i + 2mΩvcosθ j.
Fext = -mΩ2Rsinθcosθ  i - 2mΩvcosθ j.
Plug in numbers.
Ω = 2π/(24*3600 s), Fext = -1.7*10-2i - 5.1*10-2 N j.