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Chap. 18
Planer Kinetics of a
Rigid Body:
Work and Energy
APPLICATIONS
The work of the torque (or moment) developed by the driving
gears on the two motors on the concrete mixer is transformed
into the rotational kinetic energy of the mixing drum.
If the motor gear characteristics are known, could the velocity
of the mixing drum be found?
18-2
18.1 Kinetic Energy
1
2
T = ∫m dmvi
2
vi = v P + v i / P
= (v P ) x i + (v P ) y j + ω k × ( xi + y j )
= [(vP ) x - ωy ]i + [(vP ) y + ωx] j
18-3
⇒ v i ⋅ v i = vi = [( v P ) x − ω y ]2 + [( v P ) y + ω x ]2
2
= (v P ) x − 2(v P ) x ω y + ω 2 y 2 + (v P ) y + 2(v P ) y ω x + ω 2 x 2
2
2
= v P − 2(v P ) x ω y + 2(v P ) y ω x + ω 2 r 2
2
1
1 2
2
∴ T = ( ∫ dm ) v P − (v P ) x ω ( ∫ ydm ) + ( v P ) y ω ( ∫ xdm ) + ω ( ∫ r 2 dm )
m
m
m
2 m
2
1
1
2
= m v P − ( v P ) x ω y m + ( v P ) y ω x m + I Pω 2
2
2
若 P 為質心 G , 則 x = y = 0
1
1
2
T = mv G + I G ω 2
2
2
(平移)
(旋轉)
18-4
Translation
Qω = 0
0
1
1
2
∴T = mvG + I Gω 2
2
2
1
2
T = mvG
2
18-5
Rotation about a Fixed Axis
1
1
2
T = mvG + I Gω 2
2
2
1
1
2
= m ( rGω ) + I Gω 2
2
2
1
2
2
= ( mrG + I G )ω
2
1
2
T = I Oω
2
18-6
General Plane Motion
1
1
2
T = mv G + I G ω 2
2
2
18-7
18.2 The Work of a Force Work of a Variable Force
U F = ∫ F cos θds
s
18-8
Work of a Constant Force
U Fc = ∫ FC cos θds =( FC cos θ ) s
18-9
Work of a Weight
若物體向上移動 , 即 Δy > 0
則 U W = −WΔy 為負
若物體向下移動 , 即 Δy < 0
則 U W > 0 為正
18-10
Work of a Spring Force
Fs = -ks
伸長時(往外),彈簧力向內
壓縮時(往內),彈簧力向外
即位移和力相反
1 2 1 2
Us = - ( ks2 - ks1 )
2
2
1 2 1 2
= ks1 - ks2
2
2
18-11
Forces That Do No Work
(1)外力作用於固定點
(2)外力與位移的方向垂直
18-12
18.3 The Work of a Couple
r
r
dU M = F ( dθ ) + F ( dθ )
2
2
= ( Fr)dθ
= Mdθ
18-13
θ2
U M = ∫ Mdθ
θ1
18-14
若M為constant
U M = M (θ 2 − θ1 )
18-15
18-16
18-17
18.4 Principle of Work and
Energy
T1 + ΣU 1− 2 = T2
Procedure for analysis:
1. Kinetic energy (Kinematic diagram)
translation, rotation
2. Work (Free-body diagram)
3. Principle of Work and Energy
18-18
18-19
18-20
18-21
18-22
18-23
18-24
p. 470, 18-2
The double pulley consists of two parts that are attached to
one another. It has a mass of 25 kg and a radius of gyration
about its center of kO = 0.24 m. If it rotates with an angular
velocity of 20 rad/s clockwise, determine the kinetic energy of
the system. Assume that neither cable slips on the pulley.
18-25
p. 471, 18-8
The drum has a mass of 50 kg and a radius of gyration about the pin at O
of kO = 0.23 m. If the 15-kg block is moving downward at 3 m/s, and a
force of P = 100 N is applied to the brake arm, determine how far the
block descends from the instant the brake is applied until it stops.
Neglect the thickness of the handle. The coefficient of kinetic friction at
the brake pad is μK = 0.5.
18-26
18-27
p. 474, 18-20
The15-kg ladder is placed against the wall at an angle of θ = 45° as
shown. If it is released from rest, determine its angular velocity at the
instant just before θ = 0° . Neglect friction and assume the ladder is a
uniform slender rod.
18-28
1.2
18-29
18-30
p. 476, 18-32
The assembly consists of two 7.5-kg slender rods and a 10-kg
disks. If the spring is unstretched when θ = 45° and the assembly
is released from rest at this position, determine the angular velocity
of rod AB at the instant θ = 0° .The disk rolls without slipping.
18-31
18-32
Conservation of Energy (能量守
恆)
總位能 : V = Vg + Ve
保守力所作的功 : (ΣU1→ 2 ) cons = V1 − V2
又T1 + (ΣU1→ 2 ) cons + (ΣU1→ 2 ) noncons = T2
⇒ T1 + (V1 − V2 ) + (ΣU1→ 2 ) noncons = T2
⇒ T1 + V1 + (ΣU1→ 2 ) noncons = T2 + V2
若 (ΣU1→ 2 ) noncons = 0
⇒ T1 + V1 = T2 + V2
( 如摩擦力 )
稱為能量守恆
18-33
EXAMPLE 18.6
Given:The rod AB has a mass of
10 kg. Piston B is
attached to a spring of
constant k = 800 N/m.
The spring is un-stretched
when θ = 0°. Neglect the
mass of the pistons.
Find: The angular velocity of rod AB at θ = 0° if the rod is
released from rest when θ = 30°.
Plan: Use the energy conservation equation since all forces are
conservative and distance is a parameter (represented
here by θ). The potential energy and kinetic energy of
the rod at states 1 and 2 will have to be determined.
18-34
Solution:
Initial Position
EXAMPLE
(continued)
Final Position
Potential Energy:
Let’s put the datum in line with the rod when θ = 0°.
Then, the gravitational potential energy and the elastic potential
energy will be zero at position 2. => V2 = 0
Gravitational potential energy at 1: - (10)( 9.81) ½ (0.4 sin 30)
Elastic potential energy at 1: ½ (800) (0.4 sin 30)2
So V1 = - 9.81 J + 16.0 J = 6.19 J
18-35
Initial Position
EXAMPLE
(continued)
Final Position
Kinetic Energy:
The rod is released from rest from position 1
(so vG1 = 0, ω1 = 0). Therefore, T1 = 0.
At position 2, the angular velocity is ω2 and
the velocity at the center of mass is vG2 .
18-36
EXAMPLE
(continued)
Therefore,
T2 = ½ (10)(vG2)2 + ½ (1/12)(10)(0.42)(ω2)2
At position 2, point A is the instantaneous
center of rotation. Hence, vG2 = r ω = 0.2 ω2
.
Then, T2 = 0.2 ω22 + 0.067 ω22 = 0.267 ω22
Now apply the conservation of energy equation and solve for
the unknown angular velocity, ω2.
T1 + V1 = T2 + V2
0 + 6.19 = 0.267ω22 + 0
=> ω2 = 4.82 rad/s
18-37
p. 484, 18-40
At the instant shown, the 25-kg bar rotates clockwise at 2 rad/s.
The spring attached to its end always remains vertical due to the
roller guide at C. If the spring has an unstretched length of 0.6 m
and a stiffness of k = 100 N/m, determine the angular velocity of the
bar the instant it has rotated 30° clockwise.
18-38
18-39
p. 486, 18-52
The 25-kg square plate is pinned at corner A and attached to a
spring having a stiffness of k = 300 N/m. If the plate is released
from rest when θ = 0°, determine its angular velocity when θ = 90°.
The spring is unstretched when θ = 0°.
18-40
18-41
18-42
p. 489, 18-64
The 12.5-kg slender rod AB is attached to spring BC which has an
unstretched length of 1.2 m. If the rod is released from rest when θ
30°, determine its angular velocity at the instant θ = 90°.
18-43
18-44