Direct analogies between (linear) translational and rotational motion:
Quantity or Principle
Linear
Rotation
Displacement
x

Velocity
v

Acceleration
a

Inertia (resistance to
acceleration)
mass (m)
moment of
inertia (I)
Momentum
P = mv
L = I
dP/dt = Fnet
dL/dt = net
F = ma
 = I
Work
F•Ds
•D
Kinetic energy
(1/2)mv2
(1/2)I2
Momentum rate of change
Stated as Newton’s 2nd Law:
1/22/16
Oregon State University PH 212, Class 9
1
A simple application of rotational energy considerations: A
professional pitcher and catcher are testing a new design for a
baseball. The mass and radius of the new ball (B) are the same as the
current ball (A), but A is a solid sphere, and B has a hollow center.
Q: How could this skilled pitcher and catcher duo tell the two
baseballs apart?
A:
1/22/16
Oregon State University PH 212, Class 9
2
When KT and KR may both be present
We have seen that the total kinetic energy of an object that is
rotating around any fixed axis is “pure rotational energy:”
Ktotal = KR.fixed-axis = (1/2)Ifixed-axis2
Now note:
Ifixed-axis = Icm + Md2
So:
Ktotal = (1/2)Icm2 + (1/2)Md22
But:
d is the speed, vc.m., of the center of
mass as it rotates around the fixed axis.
So:
Ktotal = (1/2)Icm2 + (1/2)Mvcm2 = KT.cm + KR.cm
(parallel axis theorem)
In general (fixed axis or free rotation): Ktotal = KT.cm + KR.cm
1/22/16
Oregon State University PH 212, Class 9
3
When KT and KR may both be present
Option 1:
Ktotal = KR.fixed-axis = (1/2)Ifixed-axis2
Option 2:
Ktotal = KR.cm + KT.cm = (1/2)Icm2 + (1/2)Mvcm2
Option 1 is valid only for an object rotating around a fixed
axis, but that includes an axis that is only momentarily fixed
(i.e. its v = 0 for just an instant).
Option 2 is valid for either an object rotating around a fixed axis
or a freely rotating object (i.e. rotating around its c.m.).
After class 9 notes will go through a couple of examples to
demonstrate each option.
Note: When an object is rotating around a moving axis that is not the center of mass,
Ktotal is not generally a constant value; it is changing in time, because the axis pin is
doing work on the object. (So, why doesn’t an unmoving axis pin do work on an object?)
1/22/16
Oregon State University PH 212, Class 9
4
Rotational kinetic energy of a rolling object
Rolling: A common example of translation and rotation at
the same time. ASSUMPTION: no slipping – the center of
mass moves one circumference forward—much like the
string on a pulley rim or the chain on a bike sprocket.
vcm = Rω
Notice that the point of contact on the ground is stationary!
1/22/16
Oregon State University PH 212, Class 9
5
Prep 3, item 6b: A bowling ball rolls without slipping, first along
a level track, then up a ramp onto another level section of the track,
gaining 0.340 m in altitude. If its translational speed along the
lower track level was 3.15 m/s, find its translational speed at the
upper track level. [Isolid.sphere.center = (2/5)MR2]
1/22/16
Oregon State University PH 212, Class 9
6