16.4 Scalar differentiation of a vector 455
where v
x, v
yand v
zare the components of velocity in the three cartesian directions.
Velocityis therefore the rate of change of position. The magnitude of the velocity, the
speed, is then
(16.26)
and the kinetic energy of the body is
(16.27)
The accelerationof the body is the rate of change of velocity,
(16.28)
The vector form of Newton’s second law of motion is thenF 1 = 1 ma, equivalent to
three scalar equations, one for each component,
Linear momentum and force
In mechanics the momentum pof a body of mass mmoving with velocity vis defined
asp 1 = 1 mv. The direction of plies along the direction of the line of motion, and pis
often referred to as the linear momentum. By Newton’s second law of motion, the
force acting on a body is given by the rate of change of momentum that it induces:
(16.29)
Equation (16.29) shows that when no external forces act on a system whose linear
momentum is pthendp 2 dt 1 = 10 , and pis a constant vector. This is Newton’s first law
of motion; the principle of conservation of linear momentum.
EXAMPLE 16.10A body of mass mmoves along the curver 1 = 14 t 1 i 1 + 1 cos 12 t 1 j. Find
(i) the velocity of the body, (ii) its acceleration, (iii) the force acting on the body.
Describe the motion of the body (iv) in the x-direction, (v) in the y-direction and
(vi) overall.
(i) , (ii) , (iii) F==−mmtaj42cos
aj==−
d
dt
t
v
42cos
v==−d
dt
t
r
422 ijsin
F
p
=
d
dt
Fm
dx
dt
Fm
dy
dt
Fm
dz
dt
xyz===
222222,,,
=
d
dt
d
dt
d
dt
xzv
v
ij
yv
=
ki j
dx
dt
dy
dt
dz
dt
222222
k
a
r
==
d
dt
d
dt
v
22Tm m
xyz== ++
( )
1
2
1
2
2222vvvv
vvvv== ++||v
xyz222