How do you relate the angular acceleration of the object to the linear acceleration of a
particular point?
There are actually two perpendicular components to the linear acceleration.
One is due to the change in direction of a rotating object. It is inward directed along the
radius and is therefore called the radial component of the linear acceleration.
v2
ar 
r
This is simply the inward directed
centripetal acceleration of an object
moving in UCM
Remember v  r
ar 
ar 
r 2
r
 r
2 2
r
This is the component of the linear
acceleration that is due to changes
in direction.
v2
ar    2 r
r
The other component is due to any changes in speed of the rotating object. It acts
tangent to the circle and is therefore known as the tangential component of the linear
acceleration.
v
at 
t

But  
t
Remember v  r
r 
at 
t
at  r
This is the component of the linear
acceleration that is due to changes
in speed.

at 
r since r is constant
t
v2
ar    2 r
r
at  r
at
ar
ar
at
v2
ar    2 r
r
at  r
at
ar
ar
at
Rolling with Uniform Acceleration
Consider a disk rolling without slipping with a Uniform Acceleration.
While most points both rotate and move linearly, the center of mass is moving linearly
with a constant acceleration acm

acm
Rolling with Uniform Acceleration
Both a point on the outside of the
disk and the center of mass must
move the same distance with the
same linear velocity for the disk
to roll without slipping!
s  R
vcm   R
v R

t
t
v  
  R
t t
acm   R
R
s
s


acm
Rolling with Uniform Acceleration
Both a point on the outside of the
disk and the center of mass must
Rolling Condition – must hold for
move the same linear distance,
an object to roll without
with the same linear velocity and
slipping.
the same linear acceleration for
the disk to roll without slipping!
s  R
vcm   R
Radian measure
acm   R
R
s
s


acm