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Chapter 8
Rotational Equilibrium and Rotational dynamics
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Torque and Equilibrium
� First Condition of Equilibrium
� The net external force must be zero
This is a statement of translational
0
0 0x y
or
and
Σ =
Σ = Σ =
F
F F
r
r r
� This is a statement of translational equilibrium
•The Second Condition of Equilibrium states
• The net external torque must be zero
0τΣ =r
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A hobbit house
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•The magnitude of the force•The position of the application of the force•The angle at which the force is applied
Three Factors affect torque
τ = rF sin θ
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τ = rF sin θF sinθ
F cosθ
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Torque and Equilibrium
� First Condition of Equilibrium
� The net external force must be zero
This is a statement of translational
0
0 0x y
or
and
Σ =
Σ = Σ =
F
F F
r
r r
� This is a statement of translational equilibrium
•The Second Condition of Equilibrium states
• The net external torque must be zero
0τΣ =r
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Torque direction: Right hand rule again
Force turns it in the counterclockwise directioncounterclockwise direction
Force turns it in the clockwise direction
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Center of Gravity
� In finding the torque produced by the force of gravity, all of the weight of the object can be weight of the object can be considered to be concentrated at a single point
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Center of gravity
i i i icg cg
i i
mx myx and y
m m
Σ Σ= =Σ Σ
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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
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Example� Find the cg of a 4x8
uniform sheet of plywood with the upper right quadrant removed.
)2,2(),(;2 == yxMm
x
y
4
a
4 8)2,2(),(;2 211 == yxMm 4 8
)1,6(),(; 222 == yxMm
ftM
M
MM
MM
mm
xmxmxcg 3
10
6
10
2
622
21
2211 =⋅⋅=
+⋅+⋅=
++=
ftM
M
MM
MM
mm
ymymycg 3
5
3
5
2
122
21
2211 =⋅⋅=
+⋅+⋅=
++=
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Torque and Equilibrium
� First Condition of Equilibrium
� The net external force must be zero
This is a statement of translational
0
0 0x y
or
and
Σ =
Σ = Σ =
F
F F
r
r r
� This is a statement of translational equilibrium
•The Second Condition of Equilibrium states
• The net external torque must be zero
0τΣ =r
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Example of a Free Body Diagram--Ladder
� free body diagram shows normal force and force of static friction acting on the ladder at the ground
75N 75N
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In-class quiz 18-1Find the force P of the wall on the top of the10 meter ladder that weights 75 N
A. 50 N
B. 25N
C. 30 N
D. 21 N
E. 45 N
40° 40°75N 75N
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In-class quiz 18-1Find the force P of the wall on the top of the10 meter ladder that weights 50 N
A. 50 N
B. 25N
C. 30 N
D. 21 N
E. 45 N
40° 40°
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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
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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
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Example: a ladder against a wall
� What minimum angle can the ladder make with the floor before it slips?
0
0
=
=
∑
∑
y
x
F
F 0=− Pfs
0=− mgn
P
N
mg
y
x
0
0
=
=
∑
∑τ
yF 0=− mgn
0cos2
sin =− θθ LmgPL
mgnf sss µµ ==
0cos2
sin =− θθµ LmgmgLs µs sinθ − 1
2cosθ = 0
sµθ
2
1tan =
θfs
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Torque and Angular Acceleration
analogous to ∑F = ma
I = moment of inertia
Newton’s Second Law for a Rotating Object
Iτ αΣ =
2 2
i iI mr MR= Σ =For Uniform Ring i iI mr MR= Σ =For Uniform Ring
moment of inertia depends on quantity of matter and its distribution and location of axis of rotation
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Other Moments of Inertia