Video Rolling Motion Video Lecture - Physics for JEE Main & Advanced

various motions possessed by a rigid
body a rigid body possesses different
types of motion for example
translational motion rotational motion
combination of translational and
rotational motion translational motion
the motion of the block sliding down an
inclined plane is translational motion
all the particles of the body moved
together that is they have the same
velocity at any instant of time
rotational motion if we fix the rigid
body shown in the above figure along a
straight line then the body will undergo
rotational motion the examples of
rotational motion are motion of a
ceiling fan a potter's wheel a Mary
ground etc in rotation of a rigid body
about a fixed axis every particle of the
body moves in a circle which lies in a
plane perpendicular to the axis and has
its center on the axis rolling motion
rolling motion is a combination of
translation and rotation motion let us
consider the rolling motion of a
circular disk on a leveled surface at an
instant the point of contact P knot of
the disk with the surface is at rest if
we see m bar is the velocity of center
of mass then the translational velocity
of disc is equal to Z cm bar which is
parallel to the level surface velocity V
2 bar a very point p2 of disc is vector
sum of V cm bar and VR bar which is the
linear velocity on account of rotation
judith via bar is equal to R Omega where
Omega is angular velocity of rotation of
the disc at Point P not in the figure
the linear velocity VR bar due to
rotation directed opposite to the
translational velocity V cm bar at P
naught is R Omega where R is the radius
of the disc as P naught is
instantaneously at rest therefore VC m
bar is equal to R Omega hence for the
disc to roll without slipping the
essential condition is zc m bar is equal
to R Omega kinetic energy of rolling
motion total kinetic energy K is equal
to ke of translational motion of center
of mass KT plus ke of rotational motion
about center of mass K R that is K is
equal to KT plus K R 1 if M is mass of
the body and VC M is velocity of center
of mass of the body then 3 cm square
minus V cm square KT is equal to 1 by 2
MV cm square and K R is equal to 1 by 2
pi Omega square where I is moment of
inertia of the body about this mystery
axis of the rolling body substituting in
equation 1 k is equal to 1 by 2 and 3 cm
square plus 1 by 2 I Omega square 2 VC M
is
equal to R Omega therefore Omega is
equal to V cm by r and i is equal to MK
square where k is corresponding radius
of gyration of the body substituting in
2 we obtain regime by R whole square k
is equal to 1 by 2 MV cm square plus 1
by 2 MK square v cm by R whole square K
is equal to 1 by 2 MV cm square of 1
plus K square by r square this is a
general equation which applies to any
rolling body a disc a ring a cylinder or
a sphere