Download Lecture 15

Frictional Rolling Problems
This is a special class of problems involving wheels, cylinders or
similar bodies that roll on a rough plane surface.
GENERAL PLANE MOTION
If an inertial x – y coordinate system is chosen, the three
equations of motion may be written as:
∑F = ma
∑ F = ma
∑M = I α
x
Furthermore, it may not be known if the body rolls without slipping
or if it slides as it rolls.
Gx
y
Gy
G
Consider a homogeneous disk of mass m
and subjected to a known horizontal force P.
G
In some problems it is more convenient to sum moments about
a point P different from G (e.g. to eliminate unknown forces). In
this case the three equations of motion become:
∑F = ma
∑F = ma
∑ M = ∑ (M )
x
Gx
y
Gy
k
P
P
Gx
→ P − F = m aGx
y
Gy
→ N −W = 0
G
→ F ⋅ r = IG α
F ≤ μs N
aG = α r
Case2: Slip
F = μk N
EXAMPLES
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= I Gα ⇒ − 50 × 0.25 + FA × 0.4 = 9α
P
So independently of whether we have slip
or no-slip:
0.25
FA = 100 aGx
(1)
N A = 931
(2)
0.4 FA − 9α = 12.5 (3)
NA
2
1. The spool has a mass of 100 kg and a radius of gyration
kG = 0.3 m. If the coefficients of static and dynamic friction
at A are μ s = 0.2 and μk = 0.15, respectively determine the
angular acceleration of the spool if P = 50 N and is directed
upwards.
Data:
m = 100 kg
W = 981 N
kG = 0.3 m
μ s = 0.2
SOLUTION
μk = 0.15
If F > μ s N we must "re-work" the problem because there is
slip. In this case α and aG are not independent of each other.
The extra relation is
3) Assuming No-Slip:
x
G
Case 1: No Slip
No slip occurs if
and in this case:
G
∑F = ma
∑F = ma
∑M = I α
1
Here we have four unknowns, F , N , aGx , and α .
Therefore we need one more equation.
∑M
From the equations of motion we get:
aG
W
FA
1) First find the moment of inertia
I G = m kG2 = 100 × 0.32 = 9 kg ⋅ m 2
∑F
∑F
x
= m aGx ⇒ FA = 100 aGx
0.25
y
= 0 ⇒ 50 + N A − 981 = 0
NA
And solving the equations:
OK
FA
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Data:
W = 30 lb
m = 0.93168 slug
μ s = 0.2
μk = 0.15
θ max = ?
SOLUTION
1) Moment of Inertia:
4) Check if the no-slip assumption is correct:
( FA )max = μ s N A = 0.2 × 931 = 186.2 > FA
aG
W
2. The wheel has a weight of 30 lb and a radius of gyration
of kG = 0.6 ft. If the coefficients of static and dyamic friction
between the wheel and the plane are μ s =0.2 and μk = 0.15,
determine the maximum angle θ of the inclined plane so
that the wheel rolls without slipping
aGx = 0.4 α
(4)
α = 0.5 rad / s 2
aGx = 0.2 m / s 2
N A = 931 N
FA = 20 N
P
2) Equilibrium equations
5
I G = m kG2 = 0.93168 × 0.62 = 0.3354 slug ⋅ ft 2
6
1
From (1) 6cosθ − 30sin θ = −26.042cosθ ⇒ tan θ = 1.0681
2) Equilibrium Equations:
∑ Fx = m aGx ⇒ Fμ − W sin θ = m aGx
∑ Fy = 0 ⇒ − W cosθ + N = 0
W
∑ M G = IGα ⇒ Fμ × r = IGα
θ
•
Fμ
θ
x
N
3) For no-slip aGx = rα and maxθ occurs when Fμ = μ s N
The equilibrium equations become:
0.2 N − 30sin θ = 0.93168 × (1.25α ) (1)
N = 30cosθ
(2)
0.2 N × 1.25 = 0.3354α
(3)
1) Moments of Inertia:
1
1
2
2
I A = I B = I = m r 2 = × 8 × 0.092
I = 0.0324 kg ⋅ m 2
T
Initially we must have ω A = ωB = 0
A•
•C
B•
2) Roll A:
3) Roll B:
=
αA
A
•
•C
•
+
αA = ?
αB = ?
aB = ?
T =?
αB
B•
A
•
D
x •
N W
•
= I Bα B ⇒ T r = I Bα B ⇒ 1.5T = 0.0324α B
T = 0.36α B
(3)
4) We need one more equation, look at the kinematics:
a D = aC = −α A r j = −0.09α A j
A•
•C
Br•B / D •D
aB = 7.85 m / s 2
T = 1.57 N
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2) Moments about G:
∑M
G
= − I G α ⇒ − N x = − I G α (2)
x • α
N W aG
( m g − m x α ) x = I Gα → ( m x 2 + I G ) α = m g x
•
3) Substitute (1) into (2) and maximize.
•G
1
m A2
12
Differentiate α with respect to x:
dα
= mg
dx
mg − 2mxα
2mxα + (mx 2 + I G )
SOLUTION
FBD
B
9
A
I=
∑M
8
α = 4.36 rad / s 2
4. A slender bar of mass m and length A rests on an ledge as
shown. Find the length x that maximizes the angular
acceleration α when the bar is let go from rest in the
horizontal position and without slip.
O
SOLUTION
Replacing into (2) 0.36α − 78.48 = −8 × (0.18α )
T − 78.48 = −8 aBy (2)
x
mA = mB = 8 kg
WA = WB = 78.48 N
rA = rB = 0.09 m
From (1) and (3) α B = −α A = α
Substituting in (4) aB = 0.18α
= I Aα A ⇒ − T × 0.09 = −0.0324α A
T = 0.36α A (1)
∑ Fy = m aBy ⇒ T − W = m aBy
•
3. Along strip of paper is wrapped into two rolls, each having
a mass of 8 kg . Roll A is pin supported about its center
whereas roll B is not centrally supported. If B is brought
into contact with A and released from rest, determine the
initial tension in the paper between the rolls and the angular
acceleration of each roll. For the calculation, assume the rolls
to be approximated by cylinders.
a B = a D + α B × rB / D
−aB j = −0.09α A j + (α B k ) × (−0.09 i )
aB = 0.09 × (α A + α B ) (4)
D
∑M
( N = 20.5 lb, aGx = 19.1 ft / s 2 , α = 15.3 rad / s 2 )
Data:
0.2 × 1.25
N = 0.74538 × 30cosθ = 22.361cosθ
0.3354
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From (3) α =
θ max = 46.9D
α
aGy = x α
dα
=
=0 ⇒
dx
mx 2 + I G
1) Equilibrium in the y-direction:
∑ Fy = N − m g = −m aGy ⇒N = m g − m x α (1)
11
x=
g
2α
N=
mg
2
(3)
4) Solve the equations
From (1): N = mg − m
g
α ⇒
2α
12
2
From (2):
mg g
1
12 g 2
3g
⋅
= m A 2α ⇒ α 2 =
⇒ α=
2 2α 12
4 A2
A
x=
Finally
A
WORK AND ENERGY
Consider a planar body as shown. First
we need to find the kinetic energy of the
body including ROTATIONAL kinetic
energy
x • α
•
N W aG
2 3
If instead of taking moments about G we do it about O:
For a particle i of mass dm in the body, the kinetic energy
1
is T = vi2 d m and the kinetic energy for the entire body is
2
1
T = ∫ vi2 d m.
2 m
∑ M O = −mgx = −mxaG − IGα ⇒ − mgx = −mx 2α − IGα
(mx 2 + I G )α = mgx, the same as above.
Hence
Check that x gives a max. Using α = Nx / I G we get
Expressing vi in terms of the velocity of point P we have:
v i = v P + v i / P = vPx i + vPy j + (ωk ) × ( x i + y j)
d 2α −4mNx(mx 2 + I G ) − 2mx( mgI G − 2mNx 2 )
=
dx 2
I G (mx 2 + I G )2
d 2α −4mNxI G − 2m 2 g xI G
=
<0
dx 2
I G (mx 2 + I G )2
or v i = ( vPx − ω y ) i + (vPy + ω x) j
and taking the square of the magnitude
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v i ⋅ v i = vi2 = (vPx − ω y ) 2 + (vPy + ω x) 2
Kinetic energy of a planar rigid body:
= v − 2vPxω y + ω y + v + 2vPyω x + ω x
2
Px
2
2
2
Py
2
2
2 2
Substitute into the expression for the total kinetic energy to get:
(
)
(
)
(
)
(
2) Rotation about a fixed axis
)
1
1
T = ∫ dm vP2 − vPxω ∫ y dm + vPyω ∫ x dm + ω 2 ∫ r 2 dm
m
m
m
m
2 2 m
ym
xm
IP
1
1
Therefore T = mvP2 − vPxω ym + vPyω xm + I Pω 2
2
2
1 2 1
and if P = G
T = mvG + I Gω 2
2
2
1) Translational kinetic energy:
1
T = mvG2
2
= v − 2vPxω y + 2vPyω x + ω r
2
P
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1
1
T = mvG2 + I Gω 2
2
2
1
Or from vG = rGω we have T = ( I G + mrG2 )ω 2
2
so using the parallel axis theorem
1
T = m I Oω 2
2
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3) Kinetic Energy in General Plane Motion
16
Work of a constant force:
1
1
T = mvG2 + I Gω 2
2
2
U FC = ( FC cosθ ) s
Note: Because the energy is a scalar, the total
kinetic energy for a system of connected rigid bodies is the SUM
of the kinetic energies of each of the moving parts
Work of a Weight:
UW = −W Δy
REVIEW: Work of a Force.
Work of a variable force F:
Work of a Spring Force:
U F = ∫ F ⋅ dr = ∫ F cosθ ds
1
⎛1
⎞
U s = − ⎜ k s22 − k s12 ⎟
2
⎝2
⎠
s
s
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The Work of a Couple
A couple of forces acting on a body produce
a moment, and a moment is always equivalent
to a couple, so we can write
The work done by a moment M when a plane body rotates
an angle from θ1 to θ 2 measured in radians is
θ2
U M = ∫ M dθ
θ1
M =Fr
If the moment M is constant, then
Because the forces are equal and in opposite
direction, the work done by the forces when the
body undergoes a translation cancels out.
U M = M (θ 2 − θ1 )
The Principle of Work and Energy
T1 + ∑U1−2 = T2
If the body undergoes a rotation dθ , each force
undergoes a displacement ds = (r / 2)dθ .
Hence the total work done by the forces is
⎛r
⎞
⎛r
⎞
dU M = F ⎜ dθ ⎟ + F ⎜ dθ ⎟ = F r dθ = M dθ
⎝2 ⎠
⎝2 ⎠
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Remains the same, but now the kinetic energy also contains
rotational energy, and the work includes the work done by
moments.
If we have several rigid bodies connected by pins, inextensible
cables or in mesh with each other, the principle applies to the
entire system, and the work done by internal forces cancels20out.
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