Page 170 - Class-11-Physics-Part-1_Neat

156 PHYSICS

The physical quantities, moment of a force
and angular momentum, have an important An experiment with the bicycle rim
relation between them. It is the rotational
Take a
analogue of the relation between force and linear
bicycle rim
momentum. For deriving the relation in the
and extend
context of a single particle, we differentiate its axle on
l = r × p with respect to time, both sides.
Tie two
dl d
= ( × p ) s t r i n g s
r
dt dt at both ends
Applying the product rule for differentiation A and B,
to the right hand side, as shown
in the
d dr dp a d j o i n i n g
( × p ) = × p + r ×
r
dt dt dt figure. Hold
Now, the velocity of the particle is v = dr/dt both the
and p = m v Initially After s t r i n g s
together in
dr one hand such that the rim is vertical. If you
Because of this × p = v × m v = 0,
dt leave one string, the rim will tilt. Now keeping
the rim in vertical position with both the strings
as the vector product of two parallel vectors
in one hand, put the wheel in fast rotation
vanishes. Further, since dp / dt = F,
around the axle with the other hand. Then leave
d p one string, say B, from your hand, and observe
r × = r × F = ττ
dt what happens.
The rim keeps rotating in a vertical plane
d and the plane of rotation turns around the
Hence (r × ) p = ττ string A which you are holding. We say that the
dt
axis of rotation of the rim or equivalently
dl its angular momentum precesses about the
or = ττ (7.27) string A.
dt
The rotating rim gives rise to an angular
Thus, the time rate of change of the angular
momentum. Determine the direction of this
momentum of a particle is equal to the torque angular momentum. When you are holding the
acting on it. This is the rotational analogue of rotating rim with string A, a torque is generated.
the equation F = dp/dt, which expresses (We leave it to you to find out how the torque is
Newton’s second law for the translational motion generated and what its direction is.) The effect
of a single particle. of the torque on the angular momentum is to
make it precess around an axis perpendicular
to both the angular momentum and the torque.
Torque and angular momentum for a system
Verify all these statements.
of particles
To get the total angular momentum of a system
of particles about a given point we need to add particle has mass m and velocity v ) We may
i
i
vectorially the angular momenta of individual write the total angular momentum of a system
particles. Thus, for a system of n particles, of particles as
n L = l = r × p
L = l + l + ... + l = ∑ i ∑ i ∑ i i (7.25b)
l
i
n
2
1
i =1
This is a generalisation of the definition of
The angular momentum of the i particle angular momentum (Eq. 7.25a) for a single
th
is given by particle to a system of particles.
l = r × p
i i i Using Eqs. (7.23) and (7.25b), we get
where r is the position vector of the i particle
th
i
d
with respect to a given origin and p = (m v ) is dL = (∑ dl i
i i l ) = ∑ = ∑ ττ i (7.28a)
i
the linear momentum of the particle. (The dt dt dt
i i
2018-19