Page 188 - Class-11-Physics-Part-1_Neat

174 PHYSICS

which is in contact with the surface is at rest At P , the linear velocity, v , due to rotation
o r
on the surface. is directed exactly opposite to the translational
We have remarked earlier that rolling motion velocity v . Further the magnitude of v here is
cm
r
is a combination of rotation and translation. We Rω, where R is the radius of the disc. The
know that the translational motion of a system condition that P is instantaneously at rest
o
of particles is the motion of its centre of mass. requires v = Rω. Thus for the disc the condition
cm
for rolling without slipping is
υ = R ω (7.47)
cm
Incidentally, this means that the velocity of
point P at the top of the disc (v ) has a
1 1
magnitude v + Rω or 2 v and is directed
cm cm
parallel to the level surface. The condition (7.47)
applies to all rolling bodies.
7.14.1 Kinetic Energy of Rolling Motion
Our next task will be to obtain an expression
for the kinetic energy of a rolling body. The
kinetic energy of a rolling body can be separated
into kinetic energy of translation and kinetic
energy of rotation. This is a special case of a
general result for a system of particles,
Fig. 7.37 The rolling motion (without slipping) of a according to which the kinetic energy of a
disc on a level surface. Note at any instant, system of particles (K) can be separated into
the point of contact P of the disc with the
0 the kinetic energy of translational motion of the
surface is at rest; the centre of mass of centre of mass (MV /2) and kinetic energy of
2
the disc moves with velocity, v . The disc
cm rotational motion about the centre of mass of
rotates with angular velocity ω about its the system of particles (K′). Thus,
axis which passes through C; v =Rω,
cm (7.48)
where R is the radius of the disc.
We assume this general result (see Exercise
Let v be the velocity of the centre of mass 7.31), and apply it to the case of rolling motion.
cm
and therefore the translational velocity of the In our notation, the kinetic energy of the centre
disc. Since the centre of mass of the rolling disc of mass, i.e., the kinetic energy of translation,
is at its geometric centre C (Fig. 7. 37), v is of the rolling body is mv 2 /2, where m is the
cm cm
the velocity of C. It is parallel to the level mass of the body and v is the centre of the
cm
surface. The rotational motion of the disc is mass velocity. Since the motion of the rolling
about its symmetry axis, which passes through body about the centre of mass is rotation, K′
C. Thus, the velocity of any point of the disc, represents the kinetic energy of rotation of the
like P , P or P , consists of two parts, one is the body; , where I is the moment of
0 1 2
translational velocity v and the other is the inertia about the appropriate axis, which is the
cm
linear velocity v on account of rotation. The symmetry axis of the rolling body. The kinetic
r
magnitude of v is v = rω, where ω is the angular energy of a rolling body, therefore, is given by
r r
velocity of the rotation of the disc about the axis
(7.49a)
and r is the distance of the point from the axis 2
(i.e. from C). The velocity v is directed Substituting I = mk where k = the
r
perpendicular to the radius vector of the given corresponding radius of gyration of the body
and v = R ω, we get
point with respect to C. In Fig. 7.37, the velocity cm
of the point P (v ) and its components v and
2 2 r
v are shown; v here is perpendicular to CP .
cm r 2
It is easy to show that v is perpendicular to the
z
line P P . Therefore the line passing through P
O 2 O or (7.49b)
and parallel to ωω ωω ω is called the instantaneous axis
of rotation.

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