Page 171 - Class-11-Physics-Part-1_Neat

SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 157

th
where ττ τ τ τ is the torque acting on the i particle; Conservation of angular momentum
i
ττ = r × F If τ τ τ τ τ
i i i = 0, Eq. (7.28b) reduces to
ext
th
The force F on the i particle is the vector
i dL =
sum of external forces F ext acting on the particle 0
i dt
or L = constant. (7.29a)
int
and the internal forces F exerted on it by the
i Thus, if the total external torque on a system
other particles of the system. We may therefore of particles is zero, then the total angular
separate the contribution of the external and momentum of the system is conserved, i.e.
the internal forces to the total torque remains constant. Eq. (7.29a) is equivalent to
ττ = ∑ ττ = ∑ i × F i as three scalar equations,
r
i
i i L = K , L = K and L = K 3 (7.29 b)
y
2
1
x
z
Here K , K and K are constants; L , L and
ττ = ττ + ττ , 1 2 3 x y
ext int L z are the components of the total angular
=
where ττ ext ∑ r i × F i ext momentum vector L along the x,y and z axes
i respectively. The statement that the total
angular momentum is conserved means that
=
int
r
and ττ int ∑ i × F i each of these three components is conserved.
i
Eq. (7.29a) is the rotational analogue of
We shall assume not only Newton’s third law
Eq. (7.18a), i.e. the conservation law of the total
of motion, i.e. the forces between any two particles
linear momentum for a system of particles.
of the system are equal and opposite, but also that
Like Eq. (7.18a), it has applications in many
these forces are directed along the line joining the
practical situations. We shall look at a few of
two particles. In this case the contribution of the the interesting applications later on in this
internal forces to the total torque on the system is chapter.
zero, since the torque resulting from each action-
reaction pair of forces is zero. We thus have, ττ ττ τ = u Example 7.5 Find the torque of a force
int
ˆ
ˆ
ˆ
0 and therefore ττ ττ τ = τ τ τ τ τ . 7i + 3j – 5k about the origin. The force
ext
Since ττ = ∑ ττ , it follows from Eq. (7.28a) ˆ acts on a particle whose position vector is
i
ˆ ˆ
that i – j + k.
dL ˆ ˆ ˆ
= ττ ext (7.28 b) Answer Here = − +r i j k
dt
ˆ
ˆ
ˆ
i
j
Thus, the time rate of the total angular and F = 7 + 3 − 5k .
momentum of a system of particles about a We shall use the determinant rule to find the
point (taken as the origin of our frame of torque ττ = r × F
reference) is equal to the sum of the external
torques (i.e. the torques due to external forces)
acting on the system taken about the same
point. Eq. (7.28 b) is the generalisation of the
single particle case of Eq. (7.23) to a system of
particles. Note that when we have only one
particle, there are no internal forces or torques. ˆ ˆ ˆ
or ττ = 2i +12j +10k t
Eq.(7.28 b) is the rotational analogue of
dP
= F (7.17)
dt ext u Example 7.6 Show that the angular
Note that like Eq.(7.17), Eq.(7.28b) holds momentum about any point of a single
good for any system of particles, whether it is a particle moving with constant velocity
rigid body or its individual particles have all remains constant throughout the motion.
kinds of internal motion.
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