PC1433 MECHANICS and WAVES Module Lecturers Prof Wang Chien Ming
Engineering Science Programme and Department of Civil Engineering, National University of Singapore E-mail:
[email protected] Office: E3A-04-17B
Asst Prof Andrew Bettiol
Engineering Science Programme and Department of Physics, National University of Singapore E-mail:
[email protected] Office: E3A-04-21 PC1433 Mechanics and Waves
1
8. KINEMATICS OF RIGID BODIES 8.1 Introduction – Recall that we have studied particle dynamics. Particles refer to bodies for which only the motion as an entire unit is considered and any rotation is neglected. However, when rotation is not negligible, the bodies cannot be treated as particles Î Rigid Body Dynamics 8.2 Kinematics of Plane Motion (Plane Kinematics) All particles of the body move in parallel planes. We shall first consider two special cases: (1) Translation: Any straight line inside the body remains parallel to its original direction during the motion (i.e. no rotation) (b) Curvilinear translation
(a) Rectilinear translation A2
A2
B2
A1 B1
A1
PC1433 Mechanics and Waves
B2
B1
2
During translation, any vector joining the two points in the body is a constant vector. rB/A B rB = rA + rB/A A Recall v = dr/dt; a = dv/dt rA rB Therefore vB = vA and aB = aA O i.e. all the points of the body have the same velocity and the same acceleration at any given instant. (2) Rotation about a fixed axis: All particles move in circular paths about the axis of rotation. t Angular motion of a rigid body in plane motion v = rω Angular displacement θ in radians at = rα ω = dθ/dt rad/sec Angular velocity A Angular acceleration α = dω/dt rad/sec2 r a = rω 2 α n = d2θ/dt2 rad/sec2 n ω = ω dω/dθ rad/sec2 O
dω i.e. α = ω& = θ&& = ω dθ
⎛ ⎜ c.f. ⎝
a = v& = &x& = v
dv ⎞ ⎟ dt ⎠
PC1433 Mechanics and Waves
3
Motion of particle in rigid body rotating about a fixed axis : Velocity :
ω
v = rω
Tangential Acceleration :
at = rα
Normal Acceleration :
v2 = rω 2 an = r
O
In vector form :
α = ω& ω
Let k be the unit vector normal to the plane of rotation. Angular Displacement :
θ =θk
Angular Velocity :
ω = ω k = θ&k
Angular Acceleration :
α = α k = ω& k = θ&&k
θ& r
v A
v =ω×r
O
an = ω × (ω × r )
A
at = α × r
Motion of particle in rigid body rotating about a fixed axis : v = r& = ω × r Velocity : a = v& = ω × r& + ω& × r Acceleration : = ω × (ω × r ) + α × r = −ω 2 r + α k × r = a n + at PC1433 Mechanics and Waves
4
General Plane Motion = Translation with respect to a reference point + Rotation about the reference point Example: Point A is taken as the reference point
B1 → B1′
B1 ↓ B2
B1′
+
=
B2
ω,α
A1 → A2
A1 → A2 Plane Motion
=
vB = v A + vB/ A
A2
Translation with A + Rotation about A vA vB
vB/ A
⇒ vB ↓ = v A → + ωrB / A aB = a A + aB/ A ⇒ a B ↓ = a A → + ω rB / A + αrB / A 2
PC1433 Mechanics and Waves
aA aB
(a B / A )n (a B / A )t 5
Example: Point B is taken as the reference point
B1
B1
↓ B2
↓ B2
=
B2
+ A1
A1 → A2 Plane Motion
=
ω,α ↓ A1′ A1′ Translation with B + Rotation about B
A2
vA
v A = v B + v A/ B
vB
⇒ v A → = vB ↓ + ωrA / B
v A/ B aA
a A = a B + a A/ B ⇒ a A →= aB ↓ + ω 2 rA / B
+ αrA / B
PC1433 Mechanics and Waves
aB
(a A / B )n (a A / B )t 6
Mechanism consisting of several moving parts: • Treat each part as a rigid body (each has its own ω and α) • Pin-connected joints: bodies have same absolute velocity and acceleration at joints • Gear tooth or contact surfaces without slip: same absolute velocity and tangential acceleration, but may have different normal accelerations
vD 2 = vD 3 and aD 2 = aD 3 D 3
vB 2 = vB1 and aB 2 = aB1
vB1 = vB 2 and aB1 = a B 2
2 B
1 1
2
E
B 3
vE 2 = vE 3
(aE 2 )t = (aE 3 )t (aE 2 )n ≠ (aE 3 )n
A
PC1433 Mechanics and Waves
7
r = 0.3m rA = 0.2m
Example 8.1 Wheel rolls without sliding to the right. Find v A and v C .
30o
O vO = 3 m / s C
Solution :
x0 = θr →
For rolling without sliding, the geometric centre O moves at a velocity vO = ωr Proof :
Find v A :
v A = vO + v A/ O
where
v A / O = ωrA
Therefore Find v C :
d : v0 = θ&r = ωr → dt d : a0 = ω& r = αr → dt
ω,α
θ x0
v A/O 600
A
300
ω=
vO 3 = r 0.3
O
v A/O
⎛ 3 ⎞ vA = 3 → +⎜ ⎟ × 0.2 600= 4 → + 1.732 ↑= 4.36 m / s ⎝ 0.3 ⎠ 23.40 vC = vO + vC / O = vO → + vO ← = 0
Alternatively, since C is in contact with the supporting surface which is fixed and since there is no slipping, v C = 0. PC1433 Mechanics and Waves
vA
vO
O v C / O = ωr = vO ← 8
rA = 120 mm
Example 8.2 Initially A and B are separated. ωA = 0, ω B = 500rpm B will coast to a rest in 60 s (without contact)
A
At t = 0, disk A is started with α A = 3 rad / s 2 . Find (a) time when disks may be brought together without slipping and (b) ωA and ωB as contact is made.
Bv
rB = 80 mm
Solution : For no slipping as disks touch, edge speeds must be equal , i.e. v A = v B at contact point. Therefore rAω A = rBω B
(ω A )0 = 0
Disk A : α A = 3
ωA
.....(1)
∴ ω A = (ω A )o + α At = 3t
A
Disk B : (ω B )0 = 500 rpm = 52.36 rad / s
At t = 60, (ω B )60 = 0 = (ω B )0 + α B t ⇒ α B = −0.873
ω B = 52.36 − 0.873t (a) Bring disks together : Eq. (1) ⇒ rAω A = rBω B At time t ,
B
rAω A rBω B
ωB
∴ 0.120 × 3t = 0.080 × (52.36 − 0.873t ) ⇒ t = 9.75 s
(b) At t = 9.75 s : ω A = 3t = 29.25 rad / s
ω B = 52.36 − 0.873t = 43.85 rad / s PC1433 Mechanics and Waves
9
Example 8.3 B
A
Given ωAB = 3 rad / s Find ωBD and ωDE
Solution:
ω AB
Bar AB.
6 cm
D
vB
8 cm
10 cm
A 10 cm B vB = ( AB)ωAB = 10× 3 = 30 ↑ 8 cm E Bar DE 6 cm vD = (DE)ωDE = 10ωDE ωDE 3 D
5
Bar BD vD = vB + vD/ B
4
vB
= 30 ↑ +15ωBD ←
4 ∴ωDE = 3.75rad / s Resolve: ↑ 10ωDE × = 30 5 3 ← 10ωDE × = 15ωBD ∴ωBD = 1.5 rad / s 5
B
B
vB
D
PC1433 Mechanics and Waves
vD
B (fixed)
ωBD
+
=
where vD/ B = (BD)ωBD = 15ωBD ← 10ωDE
9 cm
E
D
vB
vD/ B
15 cm D
10
Instantaneous Centre of Rotation For general plane motion of rigid body, as far as velocities are concerned, the body at an instant seems to rotate about a certain point called the instantaneous centre at which v = 0 at that instant. Once this center C is located, the velocity at any point A is given by
v A = ωrCA
(recall : v A = v C + v A / C )
C – instantaneous centre of rotation
ω
rCA
Special case: translation only (ω = 0), the centre C is at infinity
rCB vB = ωrCB
v A = ωrCA
Caution: In general, the acceleration at C may not be zero. The acceleration of any point cannot be determined as if the slab was rotating about C.
PC1433 Mechanics and Waves
11
Example 8.4
A
θ
By activation of hydraulic cylinder E,
E
plunger F has a a downward velocity
200mm
of 2 m/s. Find ωAD and v A at θ = 60o. A
vA
C
α
ω AD
D
of rotation because v A → and vB ↓ v D is perpendicular to CD Therefore v D = vD
B
α
10 0mm
C is the instantaneous centre
30o 0.2m v B
0.1m
B
2m/ s
α
Resolve ↓: vDy = 2 = vD cos α
0.1m D
vD α
F
(
)
From geometry : ΔACD ⇒ 30o + 90o + α + α = 180o ∴ α = 30 o 2 = 2.309 m / s Therefore v D = cos 30o But vD = ω AD (CD ) = ω AD (2 × 0.1 cos α ) ∴ ω AD = 13.33 rad / s
(
)
v A = ω AD (CA) = 13.33 × 0.2 cos 30o = 2.31 m / s → PC1433 Mechanics and Waves
12
For acceleration, we can use a A = aC + a A/ C where a A / C = −ω 2 r A / C + α k × r A / C an
at
Rolling without sliding on a fixed surface : The contact point is the instantaneous centre
B
vD D
rDC
A
vB vA
vA
C
No sliding at C ⇒ v C = 0 v A = ωr AC →, v B = ωrBC →, v D = ωrDC However, a C ≠ 0
→ (a C )t = 0,
↑ (a C )n ≠ 0
Proof : Consider a C = a A + a C / A = a A → +ω 2 rAC ↑ +αr AC ← But a A = αr AC , ∴ a C = ω 2 r AC ↑
PC1433 Mechanics and Waves
13
Example 8.5 A
Given ωAB = 3 rad / s
B
E
Find (a) I.C. R. for BD, (b) ωBD and ωDE (c) α BD and α DE
D
10 cm Solution (a)
A
vB 3x10=30 cm/s
ω BD
B
vD
C
E
9 cm
6 cm 8 cm
BC 8 = 15 6 ∴ BC = 20 cm Similarly, DC = 25 cm
D
30 = 1.5 rad / s 20 vD = ωBD × DC ∴vD = 1.5× 25 = 37.5 cm/ s
(b) BD: vB = ωBD × BC ∴ωBD =
DE: vD = ωDE × DE ∴ωDE =
37.5 = 3.75rad / s 10
PC1433 Mechanics and Waves
14
ωAB = 3 α AB = 0
(c) Accelerations AB. 2 ( AB ) = 90 cm / s 2 ← a B = ω AB
DE 2 (DE ) + α DE (DE ) a D = ω DE
= 140.625
+ 10α DE
A 10 cm
B
ωDE = 3.75 αDE = ? D
BD
B (fixed)
ωBD = 1.5 15 cm αBD = ?
aD = aB + aD/ B 4 3 → 140.625 × − (10α DE )× = −90 − 15α BD 5 5 3 4 ↑ 140.625 × + (10α DE )× = 33.75 5 5 By solving the above two equations, one obtains αBD = −16.03 rad / s 2
E
vD/ B
D
α DE = −6.328 rad / s 2
PC1433 Mechanics and Waves
15
Example 8.6
a B = 2.4m / s 2 ↓
Inextensible cord
A
Double pulley : v B = 0.6m / s ↓ y
B
Find aD 80mm
ω
Solution Velocity: B
C
0.08m
vB
D
200mm
Instantaneous centre of rotation is at C. ∴ vC = 0 vB = ω (BC ) ⇒ ω =
0.6 = 7.5 rad / s 0.08
Acceleration Since centre B moves in a straight line, a B ↓
Since cord is inextensible, (a C )y = 0 and ∴ a C →
PC1433 Mechanics and Waves
16
ω
ω a C B 0 .2 m C
=
α
0.08m a = 2.4m / s 2 B
C
B
ω = 7.5rad / s
α +
C
B
aC / B
aB
aB
fixed
0.08m
aC = a B + aC / B aC →= 2.4 ↓ +0.08α ↑ +0.08 × 7.52 → Resolve :
ω = 7.5rad / s
α = 30
→ aC = 4.5 m / s 2 ↓ 0 = 2.4 − 0.08α ⇒ α = 30 rad / s 2 Point D aD = aB + aD/ B
B fixed
D
aD/ B
= 2.4 ↓ +0.2α ↓ +0.2ω 2 ← = 14.04 m / s 2 36.70 PC1433 Mechanics and Waves
17
Summary :
Plane Kinematics of Rigid Bodies
Translation : all points have same v and a at any given instant. Rotation about a fixed axis :
α = ω& = θ&& = ω
dω dθ
ω α
ω 2r
v = r& = ω × r
αr
a = v& = − ω r + α k × r an at 2
General plane motion = Translation with A + Rotation about A vB = v A + vB A
Velocity
where v B A = ω × r B A Acceleration
aB = a A + aB A
where a B A = −ω 2 r B A + α k × r B A an
at
Instantaneous Centre : v c = 0 at a given instant but a C may not be 0. v A = ωrCA PC1433 Mechanics and Waves
18
9. PLANE MOTION OF RIGID BODIES: NEWTON’S 2ND LAW 9.1 Introduction We have studied particle kinetics, i.e. relation between forces and motion for particle. There are three approaches : (1) Newton' s 2nd Law of Motion :
∑ F = ma
(2) Work and Energy Method :
T1 + U1→2 = T2
(3) Impulse and Momentum Method : mv1 + Imp1→2 = mv 2
For rigid body kinetics, we shall consider the linear motion ( x, v, a ) of the body
as a whole as well as the rotation (θ , ω , α ) of the body about its mass centre.
A rigid body can be considered to be made of a large number of particles (m1 , m2 ,..., mn ). The total mass is thus m = ∑ mi = ∫ dm
y
(9.1)
i
G
Let r = x i + y j be the position vector of the mass centre G. Thus
mi
r
mx = ∑ mi xi ; my = ∑ mi yi i
(x, y )
ri
i
⇒ mr = ∑ mi r i = ∫ r dm (9.2) i
PC1433 Mechanics and Waves
(xi , yi ) x 19
If the origin of position vectors is located at G, then r = 0.
∑m r
Eq. (9.2) ⇒
i i/G
=0
i
Recall from rigid body kinematics that we can write ai = aG + ai / G = a + α × ri/G − ω 2 ri/G
F2
F3 Free = Body Diagram
G F1
F4
G
miai
Effective Force Diagram
Apply Newton' s 2nd Law to all particles as a single system : FBD ⇒ L.H .S . = ∑ F
EFD ⇒ R.H .S . = ∑ (mi a i ) i
= ∑ mi ai + α × ∑ mi r i / G − ω 2 ∑ mi r i / G i
i
Note : m(α × r ) = α × (mr )
i
= ma Therefore
∑ F = ma
(9.3) PC1433 Mechanics and Waves
20
Now take moment about mass centre G : (recall M = r × F ). FBD ⇒ EFD ⇒
∑ M = ∑ (r × F ) ∑ r × (m a ) = ∑ m r G
i/G
i
i
i
i
i
=0
=0 i/G
× ai + ∑ mi r i / G × (α × r i / G ) − ω 2 ∑ mi (r i / G × r i / G ) i
i
⎡ ⎤ = ⎢∑ mi ri 2/G ⎥α = I α ⎣ i ⎦ Therefore ∑ M G = I α (9.4) where I = ∑ mi ri 2/G = ∫ r 2 dm is the mass moment of inertia about i
mass centre G and r is measured from G. In summary, the effective force system of a rigid body in plane motion can be represented by (1) a linear vector m a attached at G in the plane, and (2) a couple vector I α normal to the plane.
F2 G F1
F4
F3 Free = Body Diagram
ma G
Iα
PC1433 Mechanics and Waves
Effective Force Diagram 21
Example 9.1
D
System is initially at rest. A couple M = 180 Nm is applied at the end of link AB. Mass of plate = 30 kg Neglect link masses. Find force at pin C for θ = 60
o
600 mm 900 mm
C M
A
600 mm
600mm
θ
B
Plate
Solution C
For links, m = 0
θ = 60o
∑ F = ma = 0 ⇒ same as static equilibrium equation
0.6 m
C M = 180 An
Link CD is a two - force member ⇒ force along CD Link AB :
∑ M A = 0 : 180 − 0.6 Bt = 0, ∴ Bt = 300 N
Bt
At
0.6 m
Bn
vC
Plate Kinematics : Instantaneous centre at infinity ∴ω = 0 ⇒ α = 0 ⇒ curvilinear translation ⇒ a = aC = aB
X
G
vB PC1433 Mechanics and Waves
22
Solution (cont’d) θ = 30
C
0
30 0
Bt
30
mg
0
0.6 m
Bn
=
ma n
mat
Iα
Iα = 0
an = ω 2 r = 0
0.9 m
Resolve 30 o
Bt − mg cos 30 o = mat
t
∴ at = 1.504 m / s 2
∑M
B
=∑M B : eff
C cos 30o × 0.6 − mg × 0.45 = mat cos 30o × 0.45 − mat sin 30o × 0.3 ∴ C = 276 N
PC1433 Mechanics and Waves
23
9.2 Mass Moment of Inertia I Consider plane motion only. The general definition for I with respect to the perpendicular axis through point O (not necessarily the mass centre) is I O = ∫ r 2 dm
G
(r measured from O)
O r
=k m 2 O
dm
IO = is called the radius of gyration about O. m If the reference point is the mass centre G, we use the notation I , i.e. the mass where kO =
motion of inertia about the centroidal axis is I = I G = ∫ r 2 dm
(r measured from G)
= k 2m Parallel - Axis Theorem : I O = I + md 2
(9.5)
O
where d is the distance between O and G. Example (Thin Rod) 1 I = I G = mL2 , 12
2
L 1 ⎛L⎞ 1 d = , ∴ I O = mL2 + m⎜ ⎟ = mL2 2 12 ⎝2⎠ 3 PC1433 Mechanics and Waves
L/2
G L/2
24
Mass moments of inertia of common geometric shapes
Slender rod
Iy = Iz =
1 mL2 12
(
)
1 m b2 + c2 12 1 I y = mc 2 12 1 I z = mb 2 12 Ix =
Thin rectangular plate
(
)
(
)
(
)
1 m b2 + c2 12 1 I y = m c2 + a2 12 1 I z = m a 2 + b2 12
Ix =
Rectangular prism
PC1433 Mechanics and Waves
25
Mass moments of inertia of common geometric shapes
Thin disk
1 mr 2 12 1 I y = I z = mr 2 4 Ix =
(For ring or hoop, take twice)
1 ma 2 2 1 I y = I z = m 3a 2 + L2 12 3 I x = ma 2 10 3 ⎛1 ⎞ I y = I z = m⎜ a 2 + h 2 ⎟ 5 ⎝4 ⎠ Ix =
Circular cylinder
(
Circular cone
Ix = I y = Iz =
Sphere
PC1433 Mechanics and Waves
)
2 2 ma 5
26
9.3 Constrained Plane Motion Often rigid bodies move under constraints. A kinematic analysis is first required to establish kinematic relations, e.g. between a and α , between interconnected bodies, etc. Special case (1) : Translation only
α =0
∴ ∑ M G = 0 and
∑ F = ma
Special case (2) : Rotation about a fixed point O (a) If point O is at point G ⇒ centroidal rotation a =0
∴ ∑ F = 0 and
∑M
G
= Iα
PC1433 Mechanics and Waves
27
(b) If point O is not at point G ⇒ non - centroidal rotation F2
Rx
F1
G O
=
r
Iα
mat
O
G man
F3
Ry
0
Recall a = a G = a O + a G/O = rα
+ rω 2
Take moment about O :
∑M
O
(
)
= I α + (mr α )r = I + mr 2 α = I Oα
Caution : This equation does not mean that the effective force system consists of only a couple vector I Oα . It simply means that the sum of moments of external forces about O is equal to I Oα . PC1433 Mechanics and Waves
28
Example 9.2 Pulley : m = 6 kg , k = 135 mm Each collar : m = 1.2 kg One collar is removed from cord A at t = 0 Neglect friction. Find (a) angular acceleration of pulley and (b) v A at t = 2.5 s
120 mm
B I = mk 2 = 0.10935 kgm 2
a A = rAα ↑ a B = rBα ↓ (translation) Pulley : a = 0 (rotation)
A 2.4g
∑M
(a)
C
= ∑ (M C )eff
(2.4 g )rB − (2.4 g )rA = I α + (2.4a A )rA + (2.4aB )rB
rA = 0.12 mm rB = 0.18mm
C
C
A
Kinematics:
Solution
180mm
∴α = 6.37 rad / s 2 (b) Collars at A :
C Iα
= B
A
2.4g
2.4aA
B 2.4aB
a A = rAα = 0.7648m / s 2 ↑ Constant acceleration : v A = v Ao + a At
PC1433 Mechanics and Waves
= 0 + 0.7648 × 2.5 = 1.912m / s ↑ 29
Example 9.3
B I=
Uniform slender bar of mass m is released from rest in the vertical plane. Neglect friction. Find the angular acceleration
1 mL2 12
θ
L
A Kinematics: ← x A = L sin θ ⇒ v A = x& A = L cos θ θ& ⇒ a A = v& A = L(− sin θ θ& 2 + cos θ θ&& ) Recall ω = θ&, α = θ&&
(
)
∴ a A = L − sin θ ω 2 + cos θ α = Lα cos θ ← L a = a G = a A + a G / A = Lα cos θ ← + α 2
Bx
C θ
Kinetics:
G θ
∑ M C = ∑ (M C )eff
L2 ⎛L ⎞ ⎛L ⎞ mg ⎜ sin θ ⎟ = m α + mLα cos θ ⎜ cos θ ⎟ 12 ⎝2 ⎠ ⎝2 ⎠ L L ⎛L ⎞ ⎛L ⎞ + m α sin θ ⎜ sin θ ⎟ − m α cos θ ⎜ cos θ ⎟ 2 2 ⎝2 ⎠ ⎝2 ⎠ 3g ⇒α = sin θ 2L
mg
=
mLα cos θ
Iα
θ L m α 2
Ay
PC1433 Mechanics and Waves
30
n
Example 9.4 Uniform slender bar of mass m is acted upon by a torsional spring at O which exerts a counter clockwise couple M . Find n- and t- components of bearing reaction. Kinematics:
ω L
Rotation about a fixed point O. at =
Kinetics:
t
O
L α→ 2
an = ω 2
L ↑ 2 2
1 1 2 ⎛L⎞ 2 : where M I M I I mL m mL = α = α = + = ⎜ ⎟ ∑ O O O O 12 2 3 ⎝ ⎠ Rn 3M ∴α = M mL2 Rt 3M L → ∑ Ft = mat : Rt = mat = m α = 2 2L = G L 2 mg Iα ↑ ∑ Fn = man : Rn − mg = man = m ω 2 L2 ∴ Rn = m ω + mg 2 PC1433 Mechanics and Waves
ma n mat
31
9.3 Constrained Plane Motion Special case (3) : Rolling motion Rolling of wheel/disk/cylinder/ring/hoop/sphere on plane surface. First consider a balanced disk of radius r, i.e. mass centre G = geometric centre O. (a) Rolling without sliding Kinematics : x = xO = rθ ⇒ v = vO = rω ⇒ a = aO = rα Iα
W
P
=
G
G
C N
F
ma
C
Friction : No sliding ⇒ No relative motion between disk and ground at C. ⇒ F < μs N
(similar to a block at rest on a surface) PC1433 Mechanics and Waves
32
(b) Rolling, Sliding Impending Kinematics : Friction :
a = rα F = μs N
(same as above)
(c) Rotating and Sliding Kinematcs :
a and α are independent.
Friction :
F = μk N
(d) Motion Unknown > First assume case (a), i.e. no sliding If yes, assumption OK > Check whether F ≤ μs N > If no, sliding occurs ⇒ case (c), use F = μk N
PC1433 Mechanics and Waves
33
Now consider an unbalanced disk : Point O ≠ Point G Rolling without sliding or sliding impending : [Cases (a) and (b)] r Kinematics : a0 = r α but a ≠ a 0 a = aG = a0 +
aG 0
(a )
= a0 + ∴
G0 t
+
a = rα → + OG α
(a )
(9.6a )
G0 n
(9.6b )
+ OG ω 2
P
Iα
O
W
=
G
N
F ≤ μs N
0
maG 0
a0
n
maG 0 t ma0 = mrα
C
Rotating + sliding : (Case c) but a 0 ≠ r α
Kinematics :
a 0 is horizontal
Friction :
Eq. (9.6a) still holds but not Eq. (9.6b). F = μk N PC1433 Mechanics and Waves
34
ω = 2 rad s
Example 9.5 Unbalanced wheel rolls without slip. m = 30 kg, k = 200 mm ω = 2 rad/s for the instant shown.
75mm O
x
G
C
Find friction and normal reaction at C.
Solution Kinematics :
a0
15 0
15 0
a0 = rα
α
2 OG ω G
a
≡ aG = aO + aG = rα
(
225mm
O fixed
OGα O
150 + ( OG )α
150
= 0 . 225 α − 0 . 075 × 2
PC1433 Mechanics and Waves
+ (OG )ω 2
2
)
+ 0 . 075 α
15 0
150
35
Kinetics : 0.075
5 0.22
∑F
G
may
mg F
∑M ∑F
=
xO
G
Iα
ma X
y
x
N
= Iα :
I = mk 2 = 1.2 kgm2
0 .075 × N + 0 .225 F = I α
x
= m ax :
− F + mg sin 15 0 = m a x
y
= m a y:
mg cos 15 0 − N = m a y
--- (1)
= m (0 .225 α − 0 .3 ) --- (2 )
= m (0 .075 α )
--- ( 3 )
Solving
α = 14 .02
rad s 2
F = − 9 .47 N N = 253 N
F ≤ μs N min μ s =
PC1433 Mechanics and Waves
F N 36
Example 9.6 A drum (r = 0.08 m) is attached to a disk ( R = 0.16 m). Total mass m = 5 kg, k = 0.12 m P = 18 N, μs = 0.20, μ k = 0.15 Find (a) whether or not the disk slides; (b) α and a
G r P
R
Solution : * Assume no slipping Kinematics :
a = Rα = 0.16α Assume α then a is →
PC1433 Mechanics and Waves
α G
a
37
Kinetics : Iα W
G
=
I = mk 2 = 0.072 kg.m 2 ma
18N
C N
∑M
C
=
C
F
∑ (M ) C
eff
: 18 × 0 .08
= m a × 0 .16 + I α
∴ α = 7 .20 rad s 2 ,
a = 0 .16 α = 1.152 m / s 2 →
→ ∑ Fx = ma x : 18 − F = ma ∴ F = +12.24 N ← ↑ ∑ Fy = ma y :
N −W = 0
∴ N = 5 × 9.81 = 49.05 N ↑ Check whether F ≤ μs N Fmax = μs N = 0.20 × 49.05 = 9.81 N ∴ F > Fmax ⇒ Disk slides PC1433 Mechanics and Waves
38
Iα W
=
G
I = mk 2 = 0.072 kg.m 2 ma
18N
C N
C
F
Rotating and Sliding Kinematics : a and α are not related (a ≠ Rα ) Kinetics : F = μk N ( Direction is as before) y↑ G
N = 49.05 ∴ F = 0.15 × 49.05 = 7.36 F × 0.16 − 18 × 0.08 = I α ∴ α = − 3.65 rad s 2 ∴
x ⎯ ⎯→
18 - F = ma ∴ a = 2.13 m s 2 → PC1433 Mechanics and Waves
39
Summary : Plane Kinetics of Rigid Bodies (Newton' s 2nd Law)
F2
F1
ma
=
G
F3
G
Iα
F4
Effective force diagram
Free body diagram Equations of motion :
Linear : Angular :
∑ F = ma ∑ M = Iα G
(9.3) (9.4)
Note : Vector m a is attached at G and I is with respect to G. Plane motion : Scalar equations are
∑ F = ma ∑ F = ma ∑M =I α x
x
y
y
G
PC1433 Mechanics and Waves
40
Translation only :
α =0
Centroidal rotation :
a =0
Non - centroidal rotation :
∑M
0
= I 0α
Eqs. (9.3), (9.4) still valid
(about fixed point O) Rolling motion : (balanced a = a 0 ; unbalanced a ≠ a 0 ) (a + b) No sliding/impending : a0 = rα
F ≤ μs N
a0 ≠ rα
F = μk N
(c)
Sliding :
(d)
Unknown : Assume a + b; check F ≤ μs N ? If no, case c.
PC1433 Mechanics and Waves
41
10. PLANE MOTION OF RIGID BODIES: ENERGY AND MOMENTUM METHODS Introduction Similar to particle kinetics, there are 3 methods for analysing plane motion of rigid bodies : Variables involved (a) Newton' s 2nd law of motion :
F , m, a
(b) Work and Energy method :
F , m, v, s
(c) Impulse and Momentum method :
F , m, v, t
WORK & ENERGY METHOD
Work of Forces Acting on a Rigid Body The work of a force F from point A to point A' is A2
s2
A1
s1
U1→2 = ∫ F • d r = ∫
(F
cos α ) ds
PC1433 Mechanics and Waves
(10.1)
42
Special case : Work of a couple Two equal and opposite forces F and − F forming a couple of moment M act on a rigid body. During translation, the sum of works by F and - F is zero. The couple does work only during rotation. θ2
U1→2 = ∫ M dθ θ1
(10.2) A1
dθ
A
−F Forces that do no work include :
BII
BI B
F
(1) Internal forces (2) Forces at fixed points (e.g. reactions at fixed supports) (3) Forces perpendicular to displacement (4) Weight moving horizontally (5) Rolling without sliding : friction at contact point PC1433 Mechanics and Waves
43
y
Kinetic Energy
(Plane Motion)
ω
Consider a rigid body that is made of a large number of particles ( mi ).
v
G
ri G
Let v = velocity at mass centre G
ω = angular ve locity of rigid body o
vi
mi
x
Kinematics : Velocity of particles :
vi = v + vi G = v + ω × r i G
v i • v i = v • v + 2v • (ω × r i G ) + (ω × r i G )• (ω × r i G ) = v 2 + 2v • (ω × r i G ) + ω 2 ri 2G
K.E. of particle : K.E. of rigid body :
1 mi vi2 2 1 T = ∑ mi vi2 i 2
Ti =
PC1433 Mechanics and Waves
44
1 ∑i 2mi v i • v i 1 = ∑ mi v 2 + 2v • (ω × r i G ) + ω 2 ri 2G i 2
T=
[
]
1 ⎛ ⎞ 1 2⎛ 2 2 ⎞ = ∑ mi v + v • ω × ⎜ ∑ mi r i G ⎟ + ω ⎜ ∑ mi ri G ⎟ 2 i ⎝ i ⎠ 2 ⎝ i ⎠ Recall
∑m
i
i
T=
=m ;
∑m r i
iG
i
1 1 m v 2 + I ω2 2 2
=0 ;
∑m r
2 i iG
=I
i
(10.2)
PC1433 Mechanics and Waves
45
v
Special case (1) : Translation only
ω =0
G
1 ∴ T = mv 2 2
ω
Special case (2) : Rotation about a Fixed Point O (a) If point O = point G ⇒ Centroidal rotation 1 v = 0 ∴ T = Iω 2 2 (b) If point O ≠ point G ⇒ Non - centroidal rotation
(
G
)
Substituti ng v = ω d d = OG in Eq. (10.2) : 1 1 T= md 2 + I ω 2 = I 0ω 2 2 2
(
)
Principle of Work and Energy for Rigid Bodies T1 + U1→2 = T2
(10.3) PC1433 Mechanics and Waves
46
Example 10.1 Platform mass = 15 kg. Mass of each leg = 3 kg Each spring k = 700 N/m, unstretched at θ = 0. A constant torque M = 18 Nm is applied at one leg, starting from rest with θ = 0. Find ω of legs at θ = 450. 450 mm
A 450 mm
450 mm
θ 450 mm
M
PC1433 Mechanics and Waves
47
1
Solution : 2
0.831m
θ
x 2 = 0.831 − 0.45 2 = 0.195m
0.45 cos 45 0
45
0
0.45m
1 1 Kinematics Each Leg : Rotation about fixed point I 0 = mL2 = × 3 × 0.452 = 0.2025 kgm 2 3 3 Platform : Curvilinear translation with v = 0.45ω PWE T1 + U1→2 = T2
( ) 1 ⎛ 45 ⎞ + k (x − x )× 2 + 18 × ⎜ ×π ⎟ 2 180
(
0 + 15 × 9.81× 0.45 − 0.45 cos 45o + 4 × 3 × 9.81× 0.225 − 0.225 cos 45o 2 1
∴
2 2
⎝
)
⎠
1 1 2 = ×15 × (0.45ω ) + 4 × × 0.2025 × ω 2 2 2 ω = 2.77 rad / s
PC1433 Mechanics and Waves
48
Example 10.2 Pulley A : m A = 5 kg , k = 0.15 m Collar B : mB = 1.8 kg
Frictionless Pulley
0.18m
A
Cylinder C : mC = 3 kg. Released from rest. After 0.3 m, B is removed. Find v C as cylinder strikes D. Solution : Pulley I = m A k 2 = 0.1125 kgm 2 1
PWE
At rest
Just before B is removed
0 + (mB g + mc g )× 0.3 =
1 1 2 2 Iω2 + (mB + mc )v2 2 2 Kinematics: v2 = ω2 r = 0.18ω2 = 1.848 m s ↓ 21 Just after B is removed
C
2
T1 + U1→2 = T2:
PWE
B
0.10m 0.20m
E
0.25m D
3 Just before C strikes D
T2 ' + U 2 '→3 = T3 : 1 1 1 1 2 2 2 2 mc v2 + I ω2 + mc g × 0.15 = mc v3 + I ω3 2 2 2 2 Kinematics: v3 = 0.18ω3 = 2.19 m/s ↓ PC1433 Mechanics and Waves
49
Note :
For several rigid bodies, each body may be considered separately . We can also apply PWE to several bodies as a system, in which case internal forces do no work.
Potential Energy
(Recall U1→2 = V1 − V2 )
Force
P.E.
Sign
Weight
Vg = W y
y↑
Spring force
Ve =
Datum arbitrary
1 2 kx 2
x = 0 at undeformed state
Conservation of Energy for rigid bodies involving only conservative forces : ⇒
T1 + V1 = T2 + V2
(10.4)
If energy loss U L is involved, then T1 + V1 − U L = T2 + V2
(10.5)
PC1433 Mechanics and Waves
50
No work done by friction (no sliding) 2 r /π
Example 10.3
A half section of pipe of mass m and radius r is released from rest. Assume rolling without sliding. Find (a) ω after it rolled through 900. (b) reaction at horizontal surface.
Solution : For half pipe, I 0 = mr
m
I 0 = mr 2
r
Full pipe
2
Applying the parallel axis theorem : I 0 = I + m(OG )
⎛ 2r ⎞ ∴ I = mr 2 − m⎜ ⎟ = 0.5947 mr 2 ⎝π ⎠
2m O
2 2
o
G
r
I 0 = 2mr 2
PC1433 Mechanics and Waves
51
(a)
1
ω1 = 0 G
2
O
O
v2
r datum
ω2 2r
G C
π
v =0
COE : T1 + V1 = T2 + V2 0 + mgr =
1 1 2r ⎞ ⎛ 2 2 m v2 + I ω 2 + mg ⎜ r − ⎟ 2 2 π ⎠ ⎝
2r ⎞ ⎛ Kinematics : v2 = ⎜ r − ⎟ω 2 π ⎠ ⎝ ∴ ω 2 = 1.324
g r
PC1433 Mechanics and Waves
52
(b) Apply Newton' s 2nd Law of Motion :
∑ F = ma 0
0
=
G mg
ma x G C
C N
F
Iα
may
Kinematics : a 0 = r α
a = (CG ) α
←
a = aO + aG O = rα ←
∑M ∑F
x
C
+ rOG α → + rOG ω 2 ↑
2⎞ 2r 2 ⎛ ω ↑ ∴ a x = r α ⎜1 − ⎟ ← and a y = π π ⎝ ⎠ = ∑ (M C )eff : 0 = (CG )m a x + I α ∴ α = 0 and a x = 0
= ma x :
∑ F y = ma y :
− F = ma x = 0
⇒ F =0
N − mg = m a y = m
2r
π
ω 2 where ω = 1.324
g r
∴ N = 2.12 mg ↑ PC1433 Mechanics and Waves
53
IMPULSE AND MOMENTUM METHOD
Recall the impulse vector of F from t1 to t 2 is defined as =
I m p1→2
∫
t2
t1
F dt
(10.6)
Apply PIM to rigid body (treated as a system of many particles) : y
mi vi1
y
∫
(t = t1 )
Syst Momenta 1
mi vi 2
=
+ O
y
F dt
x
O
x
+ Syst Ext I mp1→ 2
where Syst Momenta = ∑ mi v i
O
(t = t2 )
x
= Syst Momenta 2
i
which can be reduced to an equivalent system comprising a linear momentum vector L and an angular momemtum couple H G
PC1433 Mechanics and Waves
L
mi vi
=
G
HG
54
Linearly, L = ∑ mi v i i
=
∑ m (v + ω × r ) iG
i
0
i
⎞ ⎛ + × m v m r ω ⎜ ∑i i ∑ i iG⎟ ⎝ i ⎠ ∴ L = mv =
(10.7)
Take momentum about G, HG =
∑ [r
iG
× mi v i 0
i
=
∑ [r
]
iG
]
× mi (v + ω × r i G )
i
=
∑m r i
i
∴H G = Iω
iG
⎛ 2 ⎞ × v + ⎜ ∑ mi r i G ⎟ω ⎠ ⎝ i
I
PC1433 Mechanics and Waves
(10.8) 55
For plane motion (ω = ω k ), Eq. (10.8) reduces to a scalar equation : H G = Iω Hence, PIM applied to a rigid body is
Linear :
mv 1 + ∑ I m p1→2 = mv 2
(10.9a)
Angular about G :
Iω1 + ∑ ∫ M G dt = Iω2
(10.9b)
y
+
I ω1 O
(t = t1)
1
x
∫F
y
mv1 G
2
dt
G
O
mv 2
y
G
= x
PC1433 Mechanics and Waves
I ω2 O
(t = t2 )
x
56
No sliding
Example 10.4 Each disk : m = 4 kg, r = 75 mm Mass of carriage = 6 kg. At rest initially.
P = 10 N
Solution:
vC
ω
B
A
Find velocity of carriage at 2.5 s later.
Kinematics
mC = 6 kg
v = v C = ωr
v
ω B
A
ω=
v vC = r r
Apply the Principle of Impulse and Momentum
Carriage
Pt
0
WC t
+
= Ht
Syst Momenta x ← ⎯⎯
1
mC(vC)2
Ht
+ Syst Ext Imp 1→ 2 = Syst Momenta
0 + Pt − 2 Ht = m C (vC )2
PC1433 Mechanics and Waves
2
(1) 57
One Disk
0
Wt
+ Ht
G C
Nt
C
G
=
r
Iω2
mv 2
Ft
0 + Ht r = I ω 2 + (m v 2 )r 1 2 (v C )2 mr 2 r
(2)
(v C )2
By solving (1) & (2), one obtains ( vC ) 2 = 1 .389 m / s
←
PC1433 Mechanics and Waves
58
Special case (1) : Translatio n only
ω = 0 ∴ H G = 0; L = mv Special case (2) : Rotation about a fixed point O (a) If point O = point G ⇒ Centroidal rotation v = 0 ∴ L = 0; H G = I ω (b) If point O ≠ point G ⇒ Non - centroidal rotation It can be shown by taking moment about O that I Oω1 + ∑ ∫ M O dt = I Oω 2 2
1
(10.10)
Caution : Eq. (10.9) is still valid. Advisable to use Eq. (10.9) which is more general Special case (3) : ∑ F = 0 ∴ ∑ I m p = 0 (10.9a) ⇒ mv 1 = mv 2 Conservation of Linear Momentum (10.9b) ⇒ I ω1 = I ω2
Conservation of Angular Momentum about G PC1433 Mechanics and Waves
59
F2
∑F =0:
F3
F2
∑F ≠ 0
G O
F1
Special case (4) :
F4
F1
∑ F ≠ 0 but resultant pass through point O
(10.10) ⇒ I 0ω1 = I 0ω 2 Conservation of angular momentum about point O Alternatively, use Eq. (10.9) but ∑ ∫ M G dt may not be 0. 2
1
PC1433 Mechanics and Waves
60
Eq. (10.9) can be generalise d to a system consisting of several rigid bodies. Internal forces would not enter the equation.
∑m v
Linear :
k
k1
K
+
∑ I mp
=
1→ 2
∑m v k
k2
k
Angular about O (arbitrary point) :
∑ r k × mk v k 1 + ∑ I k ω k 1 + ∑ ∫ k
2
1
k
m1 v 11
M O dt = ∑ r k × mk v k 2 + ∑ I k ω k 2 k
F3
F2
Iω21 I 1ω 11
Ox
m1v12
= I1ω12
+ m2 v 21 Ox
F
1
PC1433 Mechanics and Waves
Ox
I2ω22 m 2 v 22
61
Eccentric Impact mass centre(s) not on the line of impact As opp. to central impact
n
A n
B A
vB vA
B
A
uB
n
(a)
n
uA
(b)
n
B
v′B
n
v′A
(c)
Under impact, the body will deform and then restore its shape. Recall that the coefficient of restitutio n is the ratio of restoring force ( ∫ R dt) to deformation force ( ∫ P dt). Assume no friction at contact surfaces. It can be shown : Eq. (7.15) (for particles impact) still holds (whether motion is constrained or not), i.e. the e - equation : (v'B )n − (v' A )n = e[(v A )n − (vB )n ] (10.11) where A and B are the two points of contact. PC1433 Mechanics and Waves
62
Example 10.6
Knob
Slender bar, of mass m, pivoted at A is released from rest in a vertical position. The rod rebounds to a horizontal position after striking the vertical surface. Find the coefficient of restitution e between the knob K and the surface.
L
B
r A
Vertical Surface
PC1433 Mechanics and Waves
63
Solution : Before Impact Conservation of Energy :
1
T1 + V1 = T2 + V2 1 L 1 L = mv22 + Iω22 − mg 2 2 2 2 1 L 2 Noting that I = mL and v2 = ω2 , we have 12 2 6g ω2 = L 0 + mg
Knob : vk 2
6g = rω2 = r L
datum
2
→
PC1433 Mechanics and Waves
64
After impact Applying COE
3
T2 ' + V2 ' = T3 + V3
datum
L 1 2 1 2 mv2′ + I ω 2 ' − mg = 0 + 0 2 2 2 3g ∴ ω 2' = L Knob :
vk 2 ' = r
2'
3g ← L
Coefficient of restitutio n :
vk 2 ' 1 e= = = 0.707 vk 1' 2
PC1433 Mechanics and Waves
65
Summary :
Energy and Momentum Methods F3
F3
F2
F2
v1
ω2
ds
ω1
G
G
v2 F1
F1
PWE : T1 + U1→2 = T2 where U1→2 =
∫
(2)
(1)
F cos α ds and for couple U1→2 =
∫
(2)
(1)
M dθ
1 1 mv 2 + I ω 2 2 2 COE : T1 + V1 = T2 + V2 T =
PC1433 Mechanics and Waves
66
mv 1 G
+
G
=
I ω1
Iω2
( 2 )
∫F
dt
mv 2
(1 )
t = t2
t = t1
PIM : Linear :
mv1 + ∑ I m p1→ 2 = mv 2
Angular about G :
I ω1 + ∑ ∫ M G dt 2
1
= Iω2
Impact : (v ' B ) n − (v ' A ) n = e [(v A ) n − (v B ) n ]
PC1433 Mechanics and Waves
67
