Newtonian Mechanics

Problem:
A rocket of initial mass m0 is shot vertically upward.  Assume the motion occurs under constant gravitational acceleration g, and that the initial velocity of the rocket on the surface of the earth is zero.  The rocket expels 1/100 of its initial mass per second for 50 seconds.  The exhaust velocity is 2000m/s relative to the rocket.  What is the maximum height reached by the rocket?  Neglect friction.

Solution:

bulletConcepts, principles, relations that apply to the problem:
Motion in one dimension, Newton's 2nd law, F = dp/dt, dp = p(t + dt) - p(t)
bulletWhy do they apply?
Choose a coordinate axis with the positive direction upward.  The system consists of the rocket body and the fuel.  The speed of small amount of fuel dm decreases by -v' in the time interval dt.  In the same time interval the speed of the rocket and the remaining fuel increases by dv.  The total momentum change of the system in the time interval dt therefore is
dp = p(t + dt) - p(t) = (m - dm)(v + dv) + dm(v - v') - mv = mdv + dmv' (only first order terms are retained). 
By Newton's 2nd law dp/dt is equal to the external force F = -mg.
bulletHow do they apply?
 -mg = m(dv/dt) + (dm/dt)v'
Here dm/dt = - km0, k = 1/100s.  Therefore dv/dt = -g + km0v'/m, with v' = 2000m/s.
For 0 < t < 50s we have m = m0 - km0t.  Therefore
dv/dt = -g + km0v'/(m0 - km0t) = -g + kv'/(1 - kt).
At time t = 50s the fuel is used up and the rocket becomes a projectile
bulletDetails of the calculation:
For t < 50s, dv/dt = -g + kv'/(1 - kt).
Let t' = 1 - kt.  Then dv/dt' = g/k - v'/t'.  We can integrate to find v(t').


Let t1' = 1 - kt1, t0' = 1.  v(t1) = -gt1 - v'ln(1 - kt1).

At time t1 the height of the rocket is
.
h(t1) = 2*105m [(1 - t1/100)ln(1 - t1/100) + t1/100] - (1/2)gt12 (t1 in seconds).
v(t1) = -gt1 - 2000m/s ln(1 - t1/100).
At t1 = 50s we have h(t1) = 18422m, v(t1) = 896m/s.
The rocket gains an additional height Dh, where mgDh = (1/2)mv2(50s).
Dh = 40918m.  The maximum height reached is 59340m.

Problem:
Moisture condenses at the constant rate l units of mass per unit time on a falling raindrop.  If the drop falls from rest and has initial mass M, find the distance it has fallen in time t.  Neglect air resistance.

Solution:

bulletConcepts, principles, relations that apply to the problem:
Newton's 2nd law:  F = dp/dt, dp = mdv + vdm
bulletWhy do they apply?
The system consists of the moisture, with zero momentum. and the rain drop with momentum p.  In a time interval dt a small amount dm of the moisture receives an impulse vdm.  In the same time interval the velocity of the drop changes by dv.  The total momentum change of the system is dp = mdv + vdm.
bulletHow do they apply?
Choose a coordinate system such that the z-axis points downward and F = mgk.  The problem then is a one-dimensional problem.
p = mv,  dp/dt = mdv/dt + vdm/dt = mg.  dv/dt = g - (v/m)l = g - (v/(m0 + lt))l
bulletDetails of the calculation:
Let t' = m0 - lt.  Then dv/dt' = g/l - v/t'.
[Look at the equation dx/dt + nx/t = B.  If B = 0, the solution is x(t) = At-n.  It can be found by simply integrating
dx/x = -ndt/t.  With B not equal to zero we are looking for a solution dx/dt = constant, x/t = constant.  We need x = ct with c = B/(n+1).
The differential equation dx/dt + nx/t = B therefore has the general solution
x = At-n + Bt/(n+1).]
For our problem we therefore have
v(t') = At'-1 + (gt'/(2l)).
v(t) = A/(m0 + lt) + g(m0 + lt)/(2l).
v(0) = 0 -->  A = -gm02/(2l).  Therefore v(t) = g(m0 + lt)/(2l) - gm02/[(2l)(m0 + lt)].
Distance fallen in time t:  d(t) = òm0m0+lt[At'-1 + (gt'/(2l))]dt' = Aln(t')|m0m0+lt + (gt'2/(4l))|m0m0+lt.

Problem:
A trainload of empty, non leaking coal cars is accelerating in an easterly direction due to a constant force F.  It starts to rain, and the rain has a velocity which has a horizontal component v0, in a westerly direction.  The rain fills the cars at a constant rate of l mass units per unit time.  Under these conditions the train will reach a maximum velocity.  Find the maximum velocity and express it in terms of F, v0, and l.

Solution:

bulletConcepts, principles, relations that apply to the problem:
Newton's 2nd law, F = dp/dt, dp = p(t+dt) - p(t)
bulletWhy do they apply?
Assume the train moves into the x-direction.  its mass is m(t) = m0 + lt.
In a small time interval Dt its momentum changes by Dp = Pafter - Pbefore = mDv + Dm(v + v0)
dp = mdv + dmv + dmv0.
bulletHow do they apply?
F = mdv/dt + (dm/dt)(v + v0) = (m0 + lt)dv/dt + l(v + v0).
dv/dt = F/(m0 + lt) - l(v + v0)/(m0 + lt).
Let t' = m0 + lt.  dv/dt = ldv/dt'.  dv/dt' = (F/l - (v + v0))/t'.
Let V = F/l - (v + v0), then dV = -dv.
dV/dt' = -V/t'.  dV/V = dt'/t'.   V(t') = B/t'.  B is a constant which depends on the initial conditions.
F/l - v(t) - v0) = B/(m0 + lt).  v(t) = F/l - v0 - B/(m0 + lt).
vmax = F/l - v0.
bulletDetails of the calculation:
None

Problem:

A uniform chain of length 4l and mass density r is held so half of it is coiled up in a heap at the edge of a smooth table while the other half is hanging as shown.  It is supported at a point A, which is at the same vertical height as the table.

The chain is released from rest.
(a)  With what speed does the last link leave the table?

Answer:

You may assume that all motion occurs in a single vertical line.

(b)  What force is exerted on the chain at point A, immediately prior to the instant when the chain first becomes vertical?

Solution:

bulletConcepts, principles, relations that apply to the problem:
Newton’s 2nd law
bulletWhy do they apply?
The system consists of the part of the chain that is coiled up, the part that is falling, and the part that is hanging at rest.  In a small time interval dt the momentum of the system changes because a small amount dm1 of the coiled up chain receives an impulse dm1v and starts falling, a small amount of the falling chain dm2 receives an impulse -dm2v and stops falling and the section m that is falling speeds up (mdv).  The forces acting on the uncoiled parts of the chain are gravity and the force from the support.
bulletHow do they apply?
(a) Let us picture each link of the chain as a box.

At time t:  p = ryv.  Here ry = m.  dp/dt = ry(dv/dt) + ry(dv/dt).  dy/dt = v/2.
The falling links travel a distance equal to the length of two links while the length of the chain y only increases by the length of one link.  The force on the part of the chain that is not coiled up is
F = 2ryg - Fs (gravity + force from the support).
Let Fs = F1 + ryg, then F = ryg - F1.  F1 provides the impulse that stops dm2. 
F1 = ry(dv/dt) =rv2/2.
F = dp/dt.  ryg - F1 = rv2/2 + ry(dv/dt).  ryg  = rv2 + ry(dv/dt).   ydv/dt +v2 = yg.
bulletDetails of the calculation:
We want to solve this differential equation for v(y).  We want to find v(2l).  We have to eliminate t from the equation.
dy/dt = v/2.  d/dt = (v/2)d/dy.  (vy/2)dv/dy +v2 = yg.  (y/4)dv2/dy +v2 = gy.
dv2/dy = -4v2/y + 4g.
Try v2 = ay + b/y4.  Then dv2/dy = a - 4b/y5 = -4v2/y + 5a.  Therefore a = 4g/5.
v2(l) = 0 --> al + b/l4 = 0, b = -al5.
v2(2l) = a2l -al5/(16l4) = =al(2 - 1/16) = al*31/16 = gl*31/20.

(b) 

p = r(y - y1)v,  dp/dt = rvdy/dt - vrdy1/dt  +r(y - y1)dv/dt.
dy/dt = v/2  dy1/dt = y.
F = r(y - y1)g - F1 = r(y - y1)g - rv2/2.
F = dp/dt --> r(y - y1)g - rv2/2 = -rv2/2  + r(y - y1)dv/dt.
dv/dt = g, dy/dt = v/2, (v/2)dv/dy = g, dv2/dy = 4g.
v2(4l) - v2(2l) = 4g*2l = 8gl.
v2(4l) = gl*31/20 + 8gl = 191/20.
F1(4l) = rv2(2l)/2 = rgl*191/40.
Fs = r4gl + rgl*191/40 = rgl*351/40 = total force exerted by the support.