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1. E n g in eerin g Mec h an ic s
Engineering Mechanics
Dynamics
First year

2. E n g in eerin g Mec h an ic s
Dynamics
Kinematics of
a Rigid body
Rotational
motion
General plane
motion
Kinetics of a
Rigid body
Force, mass
and
Acceleration
Work and
Energy
Impulse and
Momentum

3. E n g in eerin g Mec h an ic s
Chapter Four
Plane Motion of Rigid Bodies
Sheet seven: Energy Method

4. E n g in eerin g Mec h an ic s
Work and Energy
• Method of work and energy will be used to analyze the plane motion of rigid
bodies and systems of rigid bodies.
• Principle of work and energy is well suited to the solution of problems involving
displacements and velocities.

5. E n g in eerin g Mec h an ic s
+
+ if the displacement is downward
if the displacement is upward

6. E n g in eerin g Mec h an ic s
Work of Friction Force
UFr = - Fr . s Rolling
motion
Rolling without slipping : U Fr = 0
Rolling with slipping :
U Fr = - Fr . S = -( μk N ) . S

7. E n g in eerin g Mec h an ic s

8. E n g in eerin g Mec h an ic s
Conservation of Energy

9. E n g in eerin g Mec h an ic s
The spool has a mass of 100 kg and a radius of gyration of 400 mm about its
center of mass O. If it is released from rest, determine its angular velocity after
its center O has moved down the plane a distance of 2 m. The contact surface
between the spool and the inclined plane is smooth
example 1
h
Vo
r O/IC
IC

10. E n g in eerin g Mec h an ic s
example 2
The system consists of a 20-lb disk A, 4-lb slender rod BC, and a 1-lb smooth collar C. If
the disk rolls without slipping, determine the velocity of the collar at the instant θ=30o.
The system is released from rest when θ=45o
30°
45°
Conservation of Energy (only weight)
At θ=45 : Σ T1 = 0 ( the system released from rest )
At θ=30 : Σ T2 = ( T of disk A ‘GPM’ ) + ( T of rod BC ‘GPM’ ) + ( T of collar ‘Translation’)
Σ T2 = ( 1
2
mA 𝑉2B +
1
2
IB 𝜔2A ) + ( 1
2
mBC 𝑉2G +
1
2
IG 𝜔2BC ) + ( 1
2
mcollar 𝑉2
collar)
At θ=45 : ΣV1 = Vdisk + VBC + Vcollar = 0 + mghG1 + mghcollar1
At θ=30 : ΣV2 = Vdisk + VBC + Vcollar = 0 + mghG2 + mghcollar2

11. E n g in eerin g Mec h an ic s
1
2
45°
30°
IC of dick A
r B/IC
20 Ib
4 Ib
1 Ib
hG1
h
collar1
At θ=45
Σ T1 = 0 ( the system released from rest )
ΣV1 = Vdisk + VBC + Vcollar = 0 + mghG1 + mghcollar1 =
At θ=30
ΣV2 = Vdisk + VBC + Vcollar = 0 + mghG2 + mghcollar2 =
20 Ib
4 Ib
1 Ib
hG2
hcollar2 from disk ,
from Rod BC
B
C
IC
30°
60°
1.5 ft
2.598 ft
VB
VC
G

12. E n g in eerin g Mec h an ic s
At θ=30
Σ T2 = ( 1
2
mA 𝑉2B +
1
2
IB 𝜔2A ) + ( 1
2
mBC 𝑉2G +
1
2
IG 𝜔2BC ) + ( 1
2
mcollar 𝑉2
collar) =
30°
45°

13. E n g in eerin g Mec h an ic s
example 3
The 12-Ib lever OA with 10-in. radius of gyration about O is initially at rest in the vertical
position (θ=90o), where the attached spring of stiffness k= 3Ib/in. is unstreched. Determine
the constant moment M applied to the lever through its shaft at O which will give the lever
an angular velocity ω= 4 rad/s as the lever reaches the horizontal position θ=0.
15 15
15
h= 8
1
2
ΣU=
ΣU=
ΣU= (12 x 8) + -( 1
2
3 (30 − 21.213)2 −0) + (M
π
2
)
Σ T1 = 0 ( the system released from rest )
Σ T2 = ( T of lever OA) =
1
2
Io 𝜔2 =
1
2
(
12
32.2x12
102) (42)
( -19.81 + M
π
2
) = 24.8 M= 28.4 Ib.in

14. E n g in eerin g Mec h an ic s
E n g in eerin g Mec h an ic s
E n g in eerin g Mec h an ic s