Two masses are arranged
as shown. Mass m_{1} 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 m_{2}
= 6.00 kg, is sitting on the horizontal surface and is being pulled to the right
by a force F, so that m_{2} is moving to the right at a constant speed.
The coefficient of sliding friction between m_{1} and m_{2}
is μ_{1} = 0.380, while the coefficient of sliding friction between m_{2}
and the horizontal surface is μ_{2} = 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 m_{2}
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 = μ_{1}m_{1}g = 14.9 N.

(b) F = μ_{1}m_{1}g + μ_{2}(m_{1}+m_{2})g = 64.9 N.

(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 2^{nd}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".

[(**∇∙F**_{g}/m) = -4πGρ, for a spherical mass distribution.

4πr^{2}(F_{g}/m) = 4πGM_{inside}, (F_{g}/m) = GM_{inside}/r^{2}.

Here**F**_{g}/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/(4πR^{3}). 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**= -k**r**, 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(R^{3}/(GM))^{½}.

For a satellite orbiting just above the surface we have

GMm_{s}/R^{2}= m_{s}v^{2}/R = m_{s}Rω^{2}, ω^{2}= GM/R^{3}, T = 2π/ω = 2π(R^{3}/(GM))^{½}.

The object in the tunnel and the satellite have the same period.

(b) a = Gρ(4/3)πR^{3}/(x + R)^{2}- Gρ(4/3)π(R/4)^{3}/(x + 5R/4)^{2}.

a = Gρ(4/3)πR^{3}[(x + R)^{-2}- (8x + 10R)^{-2}].

The direction of a is towards the center of the sphere.

A particle of mass m is given an initial velocity v_{0}**i**.
Assume that the particle is subject to a drag force **F** = -bv^{½}**i**.

(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 = -bv^{½}, dv/dt = -(b/m)v^{½}, dv/v^{½}= -(b/m )t,

∫_{v0}^{v(t)}dv/v^{½}= -(b/m) ∫_{0}^{t}dt = -bt/m,

2(v(t)^{½}- v_{0}^{½}) = -bt/m,

v(t) = (v_{0}^{½}- bt/(2m))^{2}.^{ }(b) v(t) = 0 --> t_{f}= 2mv_{0}^{½}/b.

After a time interval t_{f}the particle stops. In the time interval t_{f}the particle travels a distance Δx.

Δx = ∫_{0}^{tf}v(t) dt = ∫_{0}^{tf}(v_{0}^{½}- bt/(2m))^{2}dt = [(v_{0}^{½}- bt_{f}/(2m))^{3}- v_{0}^{3/2}]/(-3b/(2m)).

Δx = 2mv_{0}^{3/2}/(3b).

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/s^{2}
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).

L^{2}= 2Hd - H^{2}= (2*80*60 - 80^{2}) m^{2}= 3200 m^{2}. 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}= ½mv^{2}.

mgd - ½mg(d - L) = ½mv^{2}. v^{2}= 2gd - gd + gL = g(d + L) = 1165.7 m^{2}/s^{2}.

v_{max}= 34.14 m/s.

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 = 45^{o} give a numerical answer.

M_{Earth} = 5.97*10^{24} kg, R_{Earth} = 6378 km.

Solution:

- Concepts:

Motion in an accelerating frame - Reasoning:

Assume an observer at mid latitudes λ = 90^{o}- θ.

The equations of motion in a rotating coordinate system contain fictitious forces.

md**v**/dt =**F**_{inertial}- m**Ω**× (**Ω**×**r**) - 2m**Ω**×**v**

= -m**g**+**N**+**F**_{ext}- m**Ω**× (**Ω**×**r**) - 2m**Ω**×**v.**-m

**Ω**× (**Ω**×**r**) = centrifugal force.

(**Ω**×**r**) = ΩRsinθ**j**.**Ω**× (**Ω**×**r**) = -Ω^{2}Rsin^{2}θ**k**- Ω^{2}Rsinθcosθ**i**.

-2m**Ω**×**v**= Coriolis force.

**Ω**×**v**= -Ωvcosθ**j.** - Details of the calculation:

We want d**v**_{tangential}/dt = 0. We need 0 =**F**_{ext}+ mΩ^{2}Rsinθcosθ**i**+ 2mΩvcosθ**j**.

**F**_{ext}= -mΩ^{2}Rsinθcosθ**i**- 2mΩvcosθ**j**.

Plug in numbers.

Ω = 2π/(24*3600 s),**F**_{ext}= -1.7*10^{-2}N**i**- 5.1*10^{-2}N**j**.