# Test 2 Covers: Ch. 6, Ch. 7, Secs. 8.1-8.5 Room

```Test 2 Thursday, March 25 8:20 pm
Covers: Ch. 6, Ch. 7, Secs. 8.1-8.5
Bring: calculator, ID, formula sheet
(no phones! formula sheet: 2-sided, handwritten)
Last name
A-G
H-M
N-U
V-Z
Room
TUR L007
CLB C130
WEIM 1064
LIT 0113
Office hours canceled this week: Hirschfeld, Chan (and next Mon),
Acosta, Goswami, Sabri, Dashti, Timmerwilke, Dasbiswas
Discussion Sections meet as usual
Ch. 8 General Torque Formula
Component of F  r
OR
 r F sin 
• The lever arm, d, is the perpendicular distance
from the axis of rotation to a line drawn along the
direction of the force
d = r sin 
so
 = F d = F r sin 
Torque and Equilibrium
• First Condition of Equilibrium
• The net external force must be zero

 F  0 or


 Fx  0 and  Fy  0
•The Second Condition of Equilibrium states
The net external torque must be zero

  0
In-class quiz
A meter stick has a fulcrum at its midpoint.
A 100 gram weight is hanging 20 cm from the
fulcrum.
A second weight is balancing at a spot 50 cm from
the other side of the fulcrum. What is the mass of
the second weight?
A. 40 gm
B. 30 gm
C. 20 gm
D. 5 gm
E. 2 gm
Center of mass and of gravity
• If an object is “symmetrical’’ then the center of
mass is just where you think that it should be.
• If an object is “complicated” then


 r m  MRCoM
The torque from gravity is “as if” the force
of gravity acts only at the CoM
Balancing Objects
• If an object is hanging freely from an axis,
then the CoM must be directly beneath the
axis.
• If a solid object is standing up, then the
CoM must be over the base of the object.
• A larger base is more stable.
Example of a Free Body Diagram--Ladder
L

free body diagram shows normal force and force of
static friction acting on the ladder at the ground
The last diagram shows the lever arms for the forces
from axis of rotation at ground
Pick axis and sum torques
Torque and Angular Acceleration
Newton’s Second Law for a Rotating Object
  I
analogous to ∑F
= ma
I = moment of inertia
2
2
I


m
r

MR
For Uniform Ring
i i
moment of inertia depends on
quantity of matter
and its distribution
and location of axis of rotation
Other Moments of Inertia
Newton’s Second Law for Rotation-- Example
• Draw free body diagrams of
each object
• Only the cylinder is rotating,
so apply  = I 
• The bucket is falling, but not
rotating, so apply F = ma
• Remember a =  r and
solve the resulting equations
In-class quiz
What is the torque applied to
the reel by the cord?
A.
B.
C.
D.
mgR
0
TR
MgR
Torque and Angular Acceleration
  I
∑F = ma
T - mg = ma
 12 Ma  mg  ma
mg
a
M
m
2



2
1
RT  2 MR 

R  a
 R/ T  12 MR/ a
Look at a as m0 or  or M0 or 
```