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negligible mass; it has a mass m at its end. At time t = 0, the system is held in a horizontal
position in the constant gravitational field of the Earth.
At t = 0:
g Ó!
m L M
x = 0 at t = 0
At t = 0, the mass m is released and falls under gravity. At time t, we have
N E W T O N I A N D Y N A M I C S 89
g Ó!
vx
Æ
x x = 0
L
vx LsinÆ(dÆ/dt)
LcosÆ(dÆ/dt) L(dÆ/dt) = instantaneous velocity of m
about M
There are no external forces acting on the system in the x-direction and therefore the
horizontal momentum remains zero:
M(dx/dt) + m((dx/dt)  LsinÆ(dÆ/dt)) = 0. (4.56)
Integrating, we have
Mx + mx + mLcosÆ = constant. (4.57)
If x = 0 and Æ = 0 at t = 0, then
mL = constant, (4.58)
therefore
(M + m)x + mL(cosÆ  1) = 0,
so that
x = mL(1  cosÆ)/(M + m). (4.59)
We see that the instantaneous position x(t) is obtained by integrating the momentum
equation.
The equation of conservation of energy can now be used; it is
90 N E W T O N I A N D Y N A M I C S
(M/2)vx2 + (m/2)(vx  LsinÆ(dÆ/dt))2 + (m/2)(LcosÆ(dÆ/dt))2 = mgLsinÆ.
(The change in kinetic energy is equal to the change in the potential energy).
Rearranging, gives
(M + m)vx2  2mLsinÆvx(dÆ/dt) + (mL2(dÆ/dt)2  2mgLsinÆ) = 0. (4.40)
This is a quadratic in vx with a solution
(M + m)vx = mLsinÆ(dÆ/dt)[1 ± {1  [(M + m)(mL2(dÆ/dt)2 
2mLgsinÆ)]/[m2L2(dÆ/dt)2sin2Æ]}1/2].
The left-hand side of this equation is also given by the momentum equation:
(M + m)vx = mLsinÆ(dÆ/dt).
We therefore obtain, after substitution and rearrangement,
dÆ/dt = {[2(M + m)gsinÆ]/[L(M + mcos2Æ)]}1/2, (4.41)
the angular velocity of the rod of length L at time t.
PROBLEMS
4-1 A straight uniform rod of mass m and length 2 is held at an angle ¸0 to the vertical.
Its lower end rests on a perfectly smooth horizontal surface. The rod is released and
falls under gravity. At time t after the motion begins, we have
gÓ! ¸0
Initial position
¸
m Mass m, length 2
mg
N E W T O N I A N D Y N A M I C S 91
If the moment of inertia of the rod about an axis through its center of mass,
2
perpendicular to the plane of the motion, is m /3, prove that the angular velocity of
the rod when it makes an angle ¸ with the vertical, is
d¸/dt = {6g(cos¸0  cos¸)/ (1 + 3sin2¸)}1/2.
4-2 Show that the center of mass of a uniform solid hemisphere of radius R is 3R/8 above
the center of its plane surface.
4-3 Show that the moment of inertia of a uniform solid sphere of radius R and mass M
about a diameter is 2MR2/5.
4-4 A uniform solid sphere of radius r can roll, under gravity, on the inner surface of a
perfectly rough spherical surface of radius R. The motion is in a vertical plane.
At time t during the motion, we have:
g Ó!
É
¸
R
"
mg
rolling sphere, radius r
Show that
d2¸/dt2 + [5g/(7(R  r))]sin¸ = 0.
As a preliminary result, show that rÉ = (R  r)(d¸/dt) for rolling motion without
92 N E W T O N I A N D Y N A M I C S
slipping.
4-5 A particle of mass m hangs on an inextensible string of length and negligible
mass. The string is attached to a fixed point O. The mass oscillates in a vertical plane
under gravity. At time t, we have
O
¸ É = d¸/dt
Tension, T
m
mg
Show that
1) d2¸/dt2 + (g/ )sin¸ = 0.
2) É2 = (2g/ )[cos¸  cos¸0], where ¸0 is the initial angle of the string with respect
to the vertical, so that É = 0 when ¸ = ¸0. This equation gives the angular velocity
in any position.
4-6 Let be the natural length of an elastic string fixed at the point O. The string has a
negligible mass. Let a mass m be attached to the string, and let it stretch the spring
until the equilibrium position is reached. The tension in the string is given by Hooke s
law:
Tension, T = »(extension)/original length, where » is a constant for a given material.
The mass is displaced vertically from its equilibrium position, and oscillates under
N E W T O N I A N D Y N A M I C S 93
gravity. We have
O Equilibrium General position
g Ó!
yE
y(t)
TE
mg
T
mg
Show that the mass oscillates about the equilibrium position with simple harmonic
motion, and that
y(t) = + (mg /»){1  cos[t "»/m ]}.
0 0 0
5
INVARIANCE PRINCIPLES AND CONSERVATION LAWS
5.1 Invariance of the potential under translations and the conservation of
linear momentum
The equation of motion of a Newtonian particle of mass m moving along the x-axis
under the influence of a force Fx is
md2x/dt2 = Fx . (5.1)
If Fx can be represented by a potential V(x) then
md2x/dt2 =  dV(x)/dx . (5.2)
In the special case in which the potential is not a function of x, the equation of motion
becomes
md2x/dt2 = 0,
or
md(vx)/dt = 0. (5.3)
Integrating this equation gives
mvx = constant. (5.4)
We see that the linear momentum of the particle is constant if the potential is
independent of the position of the particle.
5.2 Invariance of the potential under rotations and the conservation of angular
momentum
I N V A R I A N C E P R I N C I P L E S A N D C O N S E R V A T I O N L A W S 95
Let a Newtonian particle of mass m move in the plane about a fixed origin, O,
under the influence of a force F. The equations of motion, in the x-and y-directions, are
md2x/dt2 = Fx and md2y/dt2 = Fy. (5.5 a,b)
If the force can be represented by a potential V(x, y) then we can write
md2x/dt2 =  "V/"x and md2y/dt2 =  "V/"y . (5.6 a,b)
The total differential of the potential is
dV = ("V/"x)dx + ("V/"y)dy.
Let a transformation from Cartesian to polar coordinates be made using the standard linear
equations
x = rcosÆ and y = rsinÆ .
The partial derivatives are
"x/"Æ =  rsinÆ =  y, "x/"r = cosÆ, "y/"Æ = rcosÆ= x, and "y/"r = sinÆ .
We therefore have
"V/"Æ = ("V/"x)("x/"Æ) + ("V/"y)("y/"Æ) (5.7)
= ("V/"x)( y) + ("V/"y)(x)
= yFx + x( Fy)
= m(yax  xay) (ax and ay are the components of acceleration)
= m(d/dt)(yvx  xvy) (vx and vy are the components of velocity).
If the potential is independent of the angle Æ then
"V/"Æ = 0, (5.8)
in which case
96 I N V A R I A N C E P R I N C I P L E S A N D C O N S E R V A T I O N L A W S
m(d/dt)(yvx  xvy) = 0
and therefore
m(yvx  xvy) = a constant. (5.9)
The quantity on the left-hand side of this equation is the angular momentum (ypx  xpy) of
the mass about the fixed origin. We therefore see that if the potential is invariant under
rotations obout the origin (independent of the angle Æ), the angular momentum of the
mass about the origin is conserved. [ Pobierz całość w formacie PDF ]