Page 186 - Class-11-Physics-Part-1_Neat

172 PHYSICS

p = m v ,
Note L = L z + L ⊥ (7.44c)
+
l = (OC × m v ) (CP × m ) v The rigid bodies which we have mainly
The magnitude of the linear velocity v of the considered in this chapter are symmetric about
particle at P is given by v = ωr where r is the the axis of rotation, i.e. the axis of rotation is
⊥ ⊥
length of CP or the perpendicular distance of P one of their symmetry axes. For such bodies,
from the axis of rotation. Further, v is tangential for a given OC , for every particle which has a
i
at P to the circle which the particle describes. velocity v , there is another particle of velocity
i
Using the right-hand rule one can check that –v located diametrically opposite on the circle
i
CP × v is parallel to the fixed axis. The unit with centre C described by the particle. Together
i
vector along the fixed axis (chosen as the z-axis) such pairs will contribute zero to L and as a
is . Hence ⊥
k
)

CP × m v = r ⊥ (mv k result for symmetric bodies L ⊥ is zero, and
hence

2
= mr ω k (since υ = ωr ) =
⊥
⊥ L = L Iωk (7.44d)
z
Similarly, we can check that OC × v is For bodies, which are not symmetric about
perpendicular to the fixed axis. Let us denote the axis of rotation, L is not equal to L and
z
the part of l along the fixed axis (i.e. the z-axis) hence L does not lie along the axis of rotation.
by l , then
z Referring to Table 7.1, can you tell in which

2
l = CP × m v = mr ω k cases L = L will not apply?
z ⊥ z
Let us differentiate Eq. (7.44b). Since is a
and = l + OC × m v k
l
z fixed (constant) vector, we get
We note that l is parallel to the fixed axis,
z
but l is not. In general, for a particle, the angular d L ( z )=  d  µ
k
momentum l is not along the axis of rotation, dt   dt I ( ω )  
i.e. for a particle, l and ωω ωω ω are not necessarily Now, Eq. (7.28b) states
parallel. Compare this with the corresponding
fact in translation. For a particle, p and v are dL = τ ττ τ
always parallel to each other. dt
For computing the total angular momentum As we have seen in the last section, only
of the whole rigid body, we add up the those components of the external torques which
contribution of each particle of the body. are along the axis of rotation, need to be taken
l =
l +
Thus L = ∑ i ∑ iz ∑ OC × m i v i into account, when we discuss rotation about a
i

fixed axis. This means we can take τ ττ τ = τk .
We denote by L and L the components
⊥
z
Since L = L + L and the direction of L (vector
of L respectively perpendicular to the z-axis z ⊥ z
) is fixed, it follows that for rotation about a
and along the z-axis; k
⊥ ∑
L = OC × m i v i (7.44a) fixed axis,
i
where m and v are respectively the mass and dL

i i z = τk
the velocity of the i particle and C is the centre (7.45a)
th
i dt
of the circle described by the particle;
dL ⊥ = 0


µ
z ∑
i i 
and L = l = ∑ m r 2  ω k and dt (7.45b)
iz

i
Thus, for rotation about a fixed axis, the

or L = Iω k (7.44b) component of angular momentum
z
perpendicular to the fixed axis is constant. As
The last step follows since the perpendicular

distance of the i th particle from the axis is r ; L = Iωk , we get from Eq. (7.45a),
i z
and by definition the moment of inertia of the
d
Iω
body about the axis of rotation is I = ∑ m r . dt ( ) τ= (7.45c)
2
i i
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