### Chapter 8 Rotational Equilibrium and Rotational dynamics

```Chapter 8
Rotational Equilibrium and
Rotational dynamics
Torque and Equilibrium
First Condition of Equilibrium
The net external force must be zero
r
Σ F = 0 or
r
r
Σ Fx = 0 and Σ Fy = 0
This is a statement of translational
equilibrium
•The Second Condition of Equilibrium states
r
Στ = 0
• The net external torque must be zero
A hobbit house
Three Factors affect torque
•The magnitude of the force
•The position of the application of the force
•The angle at which the force is applied
τ = r F sin θ
F sinθ
τ = r F sin θ
F cosθ
Torque and Equilibrium
First Condition of Equilibrium
The net external force must be zero
r
Σ F = 0 or
r
r
Σ Fx = 0 and Σ Fy = 0
This is a statement of translational
equilibrium
•The Second Condition of Equilibrium states
r
Στ = 0
• The net external torque must be zero
Torque direction: Right hand rule again
Force turns it in the
counterclockwise direction
Force turns it in the
clockwise direction
Center of Gravity
In finding the torque produced by
the force of gravity, all of the
weight of the object can be
considered to be concentrated at a
single point
Center of gravity
xcg
Σmi xi
Σmi y i
=
and ycg =
Σmi
Σmi
A few pointers.
If a body has a
symmetry and it has
a uniform density
then the cg is on the
line of symmetry.
The center of
symmetry coincides
with the cg.
The cg might be
outside the object
Example
Find the cg of a 4x8
uniform sheet of
plywood with the
removed.
m1 = 2 M ; ( x1 , y2 ) = ( 2,2)
y
4
x
a
4
m2 = M ; ( x2 , y2 ) = (6,1)
xcg =
m1 x1 + m2 x2 2M ⋅ 2 + M ⋅ 6 10 ⋅ M 10
=
=
=
ft
m1 + m2
2M + M
6⋅ M
3
ycg =
m1 y1 + m2 y2 2M ⋅ 2 + M ⋅1 5 ⋅ M 5
=
=
= ft
2M + M
3⋅ M 3
m1 + m2
8
Torque and Equilibrium
First Condition of Equilibrium
The net external force must be zero
r
Σ F = 0 or
r
r
Σ Fx = 0 and Σ Fy = 0
This is a statement of translational
equilibrium
•The Second Condition of Equilibrium states
r
Στ = 0
• The net external torque must be zero
Example of a Free Body Diagram-Ladder
75N
75N
free body diagram shows normal force
and force of static friction acting on the
In-class quiz 18-1
Find the force P of the wall on the top of the
10 meter ladder that weights 75 N
40°
A.
B.
C.
D.
E.
50 N
25N
30 N
21 N
45 N
75N
40°
75N
In-class quiz 18-1
Find the force P of the wall on the top of the
10 meter ladder that weights 50 N
40°
A.
B.
C.
D.
E.
50 N
25N
30 N
21 N
45 N
40°
A 100-N uniform ladder, 8.0 m long, rests
against a smooth vertical wall. The
coefficient of static friction between
ladder and floor is 0.40. What minimum
angle can the ladder make with the floor
before it slips?
A. 42o
B. 22o
C. 18o
D. 51o
E. 39o
A 100-N uniform ladder, 8.0 m long, rests
against a smooth vertical wall. The
coefficient of static friction between
ladder and floor is 0.62. What minimum
angle can the ladder make with the floor
before it slips?
A. 42o
B. 22o
C. 18o
D. 51o
E. 39o
Example: a ladder against a wall
What minimum angle can the ladder make with the
floor before it slips?
P
∑F = 0
∑F = 0
∑τ = 0
x
y
fs − P = 0
y
x
n − mg = 0
L
PL sin θ − mg cos θ = 0
2
f s = µ s n = µ s mg
L
µ s mgL sin θ − mg cos θ = 0
2
N
mg
θ
fs
1
1
µs sinθ − cosθ = 0 tan θ = 2µ
s
2
Torque and Angular Acceleration
Newton’s Second Law for a Rotating Object
Στ = Iα
analogous to
∑F = ma
I = moment of inertia
2
2
I
=
Σ
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
```