Rotational Equilibrium
and Rotational Dynamics

Physics 2053
Lecture Notes
Rotational Motion 1 of 25
Rotational Equilibrium and Dynamics
Topics
9-01 Torque
8-02 Torque and the Conditions for Equilibrium
8-03 Center of Gravity
8-04 Examples of Objects in Equilibrium
Dynamics: Newton’s Laws of Motion
Torque
Torque
The force used to open a door produces a torque
r
τ  rF sinθ
τ  rF
F
Static Equilibrium
Rotational Equilibrium
Two equal forces are applied to a door. The first force
is applied at the midpoint of the door; the second force
is applied at the doorknob. Both forces are applied
perpendicular to the door. Which force exerts the
greater torque?
A) the first at the midpoint
B) both exert equal non-zero torques
C) both exert zero torques
D) the second at the doorknob
Rotational Equilibrium
Torque
Torque
r
q
F
R
R  r sin θ 
τ  rF sinθ
τ  RF
Static Equilibrium
Rotational Equilibrium
Two equal forces are applied to a door at the doorknob.
The first force is applied perpendicular to the door; the
second force is applied at 30° to the plane of the door.
Which force exerts the greater torque?
A) the first applied perpendicular to the door
B) the second applied at an angle
C) both exert equal non-zero torques
D) both exert zero torques
Rotational Equilibrium
Torque and the Conditions for Equilibrium
Conditions for Equilibrium
SF = 0
St = 0
Static Equilibrium
Torque and the Conditions for Equilibrium
N=?
A massless
meter stick
x3 =
20 cm
What is the
Normal
DistanceForce
x
x2 = ?
2
o
1st Condition
of Equilibrium
3 kg
m 3g
F  0
N  m 3g  m 2g  0
N  m 3  m 2 g
N  3 kg  2 kg 9.8 m/s 2
N  49 N
2 kg
m 2g
2nd Condition
of Equilibrium
τ  0
 τo  0
m 3gx 3  m 2gx 2  0
m 3x3
x2 
m2
3 kg 20 cm
x2 
2 kg
 30 cm
Static Equilibrium
Center of Gravity
xcm
y
m2
m1
x
cg
x1
x2
 mi xi
xcg  i
 mi
m1x1  m 2 x 2
xcg 
m1  m 2
i
Linear Momentum
Problem
Find the center of gravity of the three-mass system shown in
the diagram. Specify relative to the left-hand 1.00 kg mass.
y
1.0 kg
1.5 kg
1.1 kg
x
0.5 m
0.25 m
Linear Momentum
Center of Gravity
L
N
L
2
F
xcg
W
L = 150 cm
W = 600 N
Wb = 50 N
F = 200 N
Wb
Find xcg
Linear Momentum
Center of Gravity
 to  0
N
L
FL  Wxcg  Wb    0
 2
L
L
2
F
x cg
W
L
FL  Wb  
2

xcg 
W
 150 cm 
200 N150 cm   50 N

 2 
xcg 
600 N
Wb
L = 150 cm
W = 600 N
Wb = 50 N
F = 200 N
xcg  44 cm
Linear Momentum
Examples of Objects in Equilibrium
Truck on a Bridge
N1
N2
t
WB
WT
1st Condition
F  0
N1  N 2  WB  WT  0
N1  N 2  WB  WT
2st Condition
t  0
L
L
WB  WT  N1L  0
2
2
W  WT
N1  B
2
Static Equilibrium
Examples of Objects in Equilibrium
Truck on a Bridge
N1
N2
t
WB
1st Condition
F  0
WT
N1  N 2  WB  WT  0
N1  N 2  WB  WT
2nd Condition
L
L
WB  WT  N1L  0
t  0
2
4
WB 3WT
WB WT
N2 

N1 

2
4
2
4
Static Equilibrium
Examples of Objects in Equilibrium
Hanging a Sign
T
F

q
N
WB
Physics
WS
Static Equilibrium
Examples of Objects in Equilibrium
Hanging a Sign
Find Tension in cable
1st Condition
 Fx  0
N  T cosq   0
F
A
d = L sin(q)
T
d
q
N
L
WB
 Fy  0
WS
F  T sinq   WB  WS  0
2nd Condition
 tA  0
L
TL sinq   WB    WS L  0
 2
T
WS 
WB
sin q 
2
Static Equilibrium
Examples of Objects in Equilibrium
How far (d) can a student
walk on beam before
cable breaks (Tmax)
T
F

q
N
d
WS
WB
L
Static Equilibrium
Examples of Objects in Equilibrium
L
Walking Student
d
Tmax
F
When T reaches Tmax
o
 to  0
q
N
L


 to  TmaxL sin q  WSd  WB  0
WS WB
2
2TmaxL sin q   WB L
d
2WS
Static Equilibrium