1. OH NO!! I don’t know the mass???
PEi + KEi + Wc = PEf + KEf + Wn
a) BEFORE
AFTER
KEf = ½mv2
PEi = KEf
PEf = 0
mgh = ½mv2
OH NO!! I don’t know the mass???
KEi = 0
4.0 m
PEi = mgh
Wc = 0
Wn = 0
Substitute:
(9.8)(4) = ½v2
v = 8.85 m/s

2. How do we account for that?
This is a simplified scenario aside from friction, what else did we neglect?
The ball is rolling
How do we account for that?
Rotational KE KE = ½Iω2

3. Moment of Inertia Mass analog is moment of inertia, I
r defined relative to rotation axis
SI units are kg m2

4. More about Moment of Inertia
I depends on both the mass and its distribution.
If mass is distributed further from axis of rotation, moment of inertia will be larger.

5. Moment of Inertia of a Uniform Ring
Divide ring into segments
The radius of each segment is R

6. Other Moments of Inertia
bicycle rim
rolling bar
baton
swinging bar
basketball
bowling ball

7. Rotational Kinetic Energy
Each point of a rigid body rotates with angular velocity w.
Including the linear motion
KE due to rotation
KE of center-of-mass motion

8. Substitute in the I expressions
PEi + KEi + Wc = PEf + KEf + Wn
b) BEFORE
AFTER
KEf = ½mv2 + ½Iω2
PEf = 0
PEi = KEf
mgh = ½mv2 + ½Iω2
Now what???
KEi = 0
4.0 m
PEi = mgh
Wc = 0
Can’t cancel the mass (yet !)
Wn = 0
Substitute in the I expressions
ball
mgh = ½mv2 + ½( mr2)ω2
ring
Now what???
v = ω.r
mgh = ½mv2 + ½(mr2)ω2
so ……..
ω = v
disk r
mgh = ½mv2 + ½(½mr2)ω2

9. v ω = r Algebra !!! ball mgh = ½mv2 + ½( mr2)ω2 ring
disk
mgh = ½mv2 + ½(½mr2)ω2
ball
mgh = ½mv2 + ½( mr2)( )2
Algebra !!!
ring
mgh = ½mv2 + ½(mr2)( )2
disk
mgh = ½mv2 + ½(½mr2)( )2

10. Now cancel the m’s ball mgh = ½mv2 + ½(mr2)( ) mgh = ½mv2 + ½( mv2)
ring
mgh = ½mv2 + ½(mr2)
mgh = ½mv2 + ½(mv2)
disk
mgh = ½mv2 + ½(½mr2)
mgh = ½mv2 + ½(½mv2)
ball
mgh = mv2
mgh = ½mv2 + mv2
Now cancel the m’s
ring
mgh = mv2
mgh = ½mv2 + ½mv2
disk
mgh = ¾mv2
mgh = ½mv2 + ¼mv2

11. ball
(9.8)(4) = v2
vball = 7.48 m/s
gh = v2
ring
(9.8)(4) = v2
vring = 6.26 m/s
gh = ¼v2
disk
(9.8)(4) = ¾v2
vdisk = 7.22 m/s
gh = ¾v2