Newtonian Mechanics

Most problems have moved to:

bullet http://electron9.phys.utk.edu/phys513/Modules/module_1.htm
bullet http://electron9.phys.utk.edu/phys513/Modules/module_3.htm

Additional Problems:

Problem:

bullet (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.
bullet (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 r.  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:

bulletConcepts, principles, relations that apply to the problem:
Newton's law of gravity, Newton’s 2nd law, the principle of superposition, uniform circular motion
bulletWhy do they apply?
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”.  [(Ñ×Fg/m) = -4pGr, for a spherical mass distribution.  4pr2(Fg/m) = 4pGMinside, (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.
bulletHow do they apply?
(a)  For the object of mass m in the tunnel F = (4/3)Gmprr, with r = 3M/(4pR3).  M and R are the mass and radius of the planet, respectively.
 For an object moving in the tunnel we therefore have F = -kr, k =( 4/3)Gmpr.
The force on the object obeys Hooke's law.  The object will oscillate with angular frequency w = (k/m)1/2 = ((4/3)Gpr)1/2.
Its period is T = 2p/w = (3p/Gr)1/2 = 2p(R3/GM)1/2.
For a satellite orbiting just above the surface we have
GMms/R2 = msv2/R = msRw2,  w2 = GM/R3, T = 2p/w = 2p(R3/GM)1/2.
The object in the tunnel and the satellite have the same period.
bulletDetails of the calculation:
(b)  a = Gr(4/3)2pR3/(x + R)2 - Gr(4/3)2p(R/4)3/(x + 5R/4)2.
The direction of a is towards the center of the sphere.

Problem:

A hole is drilled straight through the earth, passing through its center.  The mass of the Earth is M  = 6*1024 kg and its radius is R = 6400 km; G = 6.67*10-11 Nm2/kg2.
(a)  Find the force on a particle of mass m as function of its distance r from the center. Assume that the density of the Earth is constant.
(b)  Write the differential equation for the motion of the particle.
(c)  If you drop the particle in the hole, what is the period of its motion?   Make a numerical estimate.
(d)  What is this type of motion called?

Solution:

bulletConcepts, principles, relations that apply to the problem:
Newton's law of gravity, Newton’s 2nd law
bulletWhy do they apply?
Assume a tunnel is bored through the center of the Earth.  The gravitational force on a test particle of mass m in the tunnel, a distance r from the center is in the –r direction and its magnitude is found using “Gauss’ law”.

4pGMinside

bulletHow do they apply?
(a)  F4pr2 = m4pG(4/3)pr3r,  F = (4/3)Gmprr = GmMr/R3.
(b)  For a particle moving in the tunnel we therefore have F = -kr, k = GmM/R3.  The force on the particle obeys Hooke's law. 
The differential equation for the motion is d2z/dt2 = -(k/m)z, where z is the axis of a coordinate system centered at the center of the earth and pointing along the drilled tunnel.
bulletDetails of the calculation:
(c)  The particle will oscillate with angular frequency w = (k/m)1/2 = (GM/R3)1/2. 
Its period is T = 2p/w = 5085s = 85min.
(d)  This type of motion is called simple harmonic motion.

Problem:

A uniform circular disk of radius a and mass nm rotates without friction about a fixed axis through its center.  Initially, an insect of mass m is at the lowest point of the disk and the system is at rest.  The insect begins to crawl along the circumference of the disk with a velocity V relative to the disk and at any time is at an angle q relative to the vertical line through the axis of the disk.

(a)  Find the initial value of dq/dt.
(b)  The insect continues to crawl at a constant speed relative to the disk.  Show that, if the insect is to reach the top of the disk, that V2 > 8(n + 2)ga/n2.

Image522.gif (2338 bytes)

Solution:

bulletConcepts, principles, relations that apply to the problem:
t = dL/dt
bulletWhy do they apply?
The disk can rotate freely about an horizontal axis.  In a gravity-free environment the total angular momentum of the disk-insect system would be conserved.  With gravity, the weight of the insect exerts a torque about the axis of rotation for q ¹ 0.
bulletHow do they apply?
(a)  Let the x-axis point out of the page.  t = dL/dt = (-i)mgasinq.
L = ma2(dq/dt)i + I0(dq/dt - V/a)i

At t = 0, t = 0 and L = 0.  (dq/dt) = -(I0/ma2)(dq/dt - V/a).
(dq/dt)(1 + n/2) = Vn/(2a), (dq/dt) = Vn/[(n+2)a] = w0.
bulletDetails of the calculation:
(b)  t = dL/dt
-mgasinq = ma2(d2q/dt2) + (nma2/2)(d2q/dt2)
(d2q/dt2) =  -2gsinq/[(n+2)a]
(dw/dt) =  -2gsinq/[(n+2)a]
(dw/dt) = (dw/dq)(dq/dt),  dw/dt = w(dw/dq)
w dw = (dw/dt)dq = -2gsinqdq/[(n+2)a]

For the insect to reach the top we need

i.e. wf must be a real number.
This requires 4g/[(n+2)a] < 2V2n2/[(n+2)2a2], V2 > 8(n+2)ga/n2.