Page 177 - Class-11-Physics-Part-1_Neat

SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 163

Pythagoras theorem, BC = 2 2 m. The forces linear velocity is υ = r ω . The kinetic energy of
i i
on the ladder are its weight W acting at its centre motion of this particle is
of gravity D, reaction forces F and F of the wall
1 2 1 2 1 2 2
and the floor respectively. Force F is k = m υ = m r ω
i i
i
i i
1 2 2
perpendicular to the wall, since the wall is
where m is the mass of the particle. The total
frictionless. Force F is resolved into two i
2 kinetic energy K of the body is then given by
components, the normal reaction N and the
the sum of the kinetic energies of individual
force of friction F. Note that F prevents the ladder
particles,
from sliding away from the wall and is therefore
directed toward the wall. n 1 n
K = ∑ k = ∑ (m r ω 2 )
2
For translational equilibrium, taking the i i i
i= 1 2 i = 1
forces in the vertical direction,
Here n is the number of particles in the body.
N – W = 0 (i) Note ω is the same for all particles. Hence, taking
Taking the forces in the horizontal direction, ω out of the sum,
F – F = 0 (ii) 1 n
1 K = ω 2 ( ∑ m r 2 )
For rotational equilibrium, taking the 2 i = 1 i i
moments of the forces about A,
We define a new parameter characterising
2 2 F − (1/2) W = 0 (iii) the rigid body, called the moment of inertia I ,

1
given by
n
Now W = 20 g = 20 × 9.8 N = 196.0 N I = ∑ m r 2 (7.34)
From (i) N = 196.0 N i= 1 i i
With this definition,
From (iii) F = W 4 2 = 196.0/4 2 = 34.6 N
1 1
From (ii) F = F = 34.6 N K = Iω 2 (7.35)
1 2
Note that the parameter I is independent of
2
F = F + N 2 = 199.0 N
2 the magnitude of the angular velocity. It is a
The force F makes an angle α with the characteristic of the rigid body and the axis
2
horizontal, about which it rotates.
Compare Eq. (7.35) for the kinetic energy of
−
tanα = N F = 4 2 , α = tan (4 2) ≈ 80 t
1

a rotating body with the expression for the
kinetic energy of a body in linear (translational)
7.9 MOMENT OF INERTIA motion,
We have already mentioned that we are K = 1 m υ 2
developing the study of rotational motion 2
parallel to the study of translational motion with Here, m is the mass of the body and v is its
which we are familiar. We have yet to answer velocity. We have already noted the analogy
one major question in this connection. What is between angular velocity ω (in respect of rotational
the analogue of mass in rotational motion? motion about a fixed axis) and linear velocity v (in
respect of linear motion). It is then evident that
We shall attempt to answer this question in the
the parameter, moment of inertia I, is the desired
present section. To keep the discussion simple,
rotational analogue of mass in linear motion. In
we shall consider rotation about a fixed axis
rotation (about a fixed axis), the moment of inertia
only. Let us try to get an expression for the
plays a similar role as mass does in linear motion.
kinetic energy of a rotating body. We know We now apply the definition Eq. (7.34), to
that for a body rotating about a fixed axis, each calculate the moment of inertia in two simple
particle of the body moves in a circle with linear cases.
velocity given by Eq. (7.19). (Refer to Fig. 7.16). (a) Consider a thin ring of radius R and mass
For a particle at a distance from the axis, the M, rotating in its own plane around its centre

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