Review for Test #3
 Responsible for:
- Chapters 9 (except 9.8), 10, and 11 (except 11.9)
- The spring (6.2, 7.3, 8.2-8.3)
- Problems worked in class, class notes
- Homework assignments
 Test format:
- 13 problems, 11 simple, 1 intermediate, 1
advanced, 7.7 pts each
- Time: 50 minutes only
 Test materials:
- Pencil, paper, eraser, and calculator
- No formulae sheet
- Closed textbook and notes
Material Covered
 Chapter 9: Impulse and Momentum
- Impulse-momentum theorem
- Conservation of Linear momentum
- 1 and 2D collisions
- Center of mass
 Chapter 10: Rotational Kinematics and Energy
- , , , ar, and at
- Rotational kinematic equations
- Rolling motion (tire)
- Moment of inertia
- Rotational work and kinetic energy
- Conservation of Energy with rotation
 Chapter 11: Rotational Dynamics and Static Equilibrium
- , =I
- Applications of =0, F=0
- Center of gravity
- Angular momentum, conservation of ...
 The Spring
- Force due to a spring (Hooke’s Law)
- Work and potential energy of a spring
Grades returned by next Monday?
Example Problem
A 0.200-m bar with a mass of 0.750 kg is released
,
from rest in the vertical position. A spring is
attached, initially unstrained, and has a spring
constant of 25.0 N/m. Find the tangential speed
with which the free end strikes the horizontal
surface. (drawing to be provided)
Solution:
Bar rotating with axis at one end  rotational KE,
no translational KE
Bar falls from some height  gravitional PE (Ug)
A spring is attached to bar  spring PE (Us)
Bar  rigid body  need moment of inertia
 Use Conservation of Energy
E  KR U g Us
 i  0, xi  0
yih since this would
h
mean all mass of rod
Ei  mg , Ei  mgh is at y =h, but mass is
i
2
distributed. So, take
y f  0, U g , f  0
mass to be located at
2
2
1
1
center of gravity
E f  2 I f  2 kx f
2
1
I rod  3 mL , v f  v t  r f  L f


2
 vt  1 2
Ef 
mL    2 kx f
L
2
2
1
1
E f  6 mv t  2 kx f
1 1
2 3
2
xf  ?
From geometry of problem
x f  0.2  0.1  0.1  0.1236 m
2
2
Return to Conservation of Energy and solve for vt
Ef 
1
6
mv
vt 
vt 
vt 
2
t
h
L
mv  kx  Ei  mg  mg
2
2
L 1 2
 mg  2 kx f
2
k 2
3 gL  3 x f
m
25.0
2
3(9.80)(0.20)  3
(0.1236)
0.750
5.88  1.528  2.09 m/s
1
6
2
t
1
2
2
f