PHY11T4DOR - WordPress.com

Dynamics of Rotation
Dynamics of Rotation
Torque (τ) :
The tendency or ability of a force to rotate an object about an
axis, fulcrum, or pivot. Just as a force is a push or a pull, a
torque can be thought of as a twist.
F
Axis
τ=Fxr
θ
Moment Arm (r)
Torque (τ)
τ is the Cross Product of the
Force vector & Moment Arm
Line of Force
Consider only the magnitude :
τ = Fr sinθ , τ is in N-m (MKS) & Dyne-cm (CGS)
Where
F – applied force
r – moment arm
θ – angle between F & r
Dynamics of Rotation
Torque (τ) : Could go in either a clockwise (CW) or counter-clockwise (CCW) in direction
F
Axis
CW Torque (τCW)
θ
Moment Arm (r)
Line of Force
Line of Force
Axis
CCW Torque (τCCW)
θ
Moment Arm (r)
F
Torque (τ) :
F1
r1
F3
r2
F4
r1
F2
r2
θ1
θ2
τ2 = (F2)(r2) sinθ2
F1
τ1 = (F1)(r1) sinθ1
but θ2 = 90°
τ2 = (F2)(r2)
F2
Torque (Γ) :
F1
r1
F3
r2
F2
F4
r2
F3
θ3
τ4 = (F4)(0) sinθ4
τ3 = (F3)(r2) sinθ3
but θ3 = 0°
τ3 = 0
τ4 = 0
F4
Dynamics of Rotation
MOMENT OF INERTIA (I) – It is the measure of the object’s resistance to
changes in rotational speed about an axis.
The greater the value of the mass moment of inertia, the smaller the
angular acceleration about an axis (slower rotation)
I of a point mass
General form of I
I = ∫r2 dm
Axis
m
Moment Arm (r)
I = mr2
Axis
Moment Arm (r)
I is in kg-m2 (MKS) & g-cm2 (CGS)
Dynamics of Rotation
Common Moments of Inertia
I = cMR2
c is a fraction
http://hyperphysics.phy-astr.gsu.edu
Dynamics of Rotation
Parallel Axes Theorem
-The moment of inertia can be calculated if moved to another axis as long as
that (new) axis is parallel to the (original) axis that passes through the center of
mass and the distance between them is known.
New Axis
Old Axis
d
IP = ICM + Md2
Where:
IP : Moment of Inertia about another axis that is parallel to the axis that goes
through the center of mass.
ICM : Moment of Inertia about an axis at the center of mass.
M : Mass of the object
d - distance between the center of mass and the point of the parallel axis.
Newton’s Second Law in Rotational Motion
a = Rα
τ = FR sinθ
for a point mass moving in a circle,
F is always perpendicular to the Radius thus
R
τ = FR
Fnet
If F is the net force (Fnet)
Applying NSLM
Fnet = Ma , where a is the tangential acceleration , a = Rα
τnet = (Ma)R = M(Rα)R
τnet = MR2α
But I =
M
MR2 (for a point mass)
where I is really I = cMR2
τnet = Iα
Rotation about a FIXED Axis
A bucket of water with a mass of 20kg is suspended by a rope wrapped around a
windlass in the form of a solid cylinder 0.4m in radius, also with a mass of 20kg. The
cylinder is pivoted on a frictionless axle through its center. The bucket is released from
rest at the top of a well and falls 20m to the water.
N
a) What is the tension in the rope while the bucket is falling?
a = Rα
b) With what speed does the bucket strike the water?
c) What is the time of fall?
R
R
M = 20 kg
FBD :
R = 0.4 m
T
Mg
m = 20 kg
T
h = 20 m
a
mg
Rotation about a FIXED Axis
A bucket of water with a mass of 20kg is suspended by a rope wrapped around a
windlass in the form of a solid cylinder 0.4m in radius, also with a mass of 20kg. The
cylinder is pivoted on a frictionless axle through its center. The bucket is released from
rest at the top of a well and falls 20m to the water.
a) What is the tension in the rope while the bucket is falling?
b) With what speed does the bucket strike the water?
c) What is the time of fall?
T
NSLM
ΣFy = ma ↑+
T – mg =−ma
T =mg −ma >eq-1
a
mg
Rotation about a FIXED Axis
A bucket of water with a mass of 20kg is suspended by a rope wrapped around a
windlass in the form of a solid cylinder 0.4m in diameter, also with a mass of 20kg. The
cylinder is pivoted on a frictionless axle through its center. The bucket is released from
rest at the top of a well and falls 20m to the water.
a) What is the tension in the rope while the bucket is falling?
b) With what speed does the bucket strike the water?
c) What is the time of fall?
N
a = Rα
τnet +
τ net = τT + τ N + τ Mg
τ net = − TR + 0 + 0
τ net = − TR >eq-2
τ net = Iα +
I = ½ MR2 (for solid cylinder)
α = a/R
τ net = − ½ MR2 (a/R)
T
τ net = − ½ MR a >eq-3
eq. 2 = ea. 3
R
Mg
R
Rotation about a FIXED Axis
A bucket of water with a mass of 20kg is suspended by a rope wrapped around a
windlass in the form of a solid cylinder 0.4m in radius, also with a mass of 20kg. The
cylinder is pivoted on a frictionless axle through its center. The bucket is released from
rest at the top of a well and falls 20m to the water.
a) What is the tension in the rope while the bucket is falling?
b) With what speed does the bucket strike the water?
c) What is the time of fall?
−TR = − ½ MRa
T = ½ Ma >eq. 4
a = (9.8m/s2) [20kg/(10kg + 20kg)]
eq. 4 = ea. 1
a = 6.533 m/s2
½ Ma = mg – ma
a in ea. 4
½ Ma + ma = mg
T = ½ (20kg)(6.533 m/s2)
a (½ M+m) = mg
T = 65.33 N
a = g [m/(0.5M + m)]
Rotation about a FIXED Axis
A bucket of water with a mass of 20kg is suspended by a rope wrapped around a
windlass in the form of a solid cylinder 0.4m in diameter, also with a mass of 20kg. The
cylinder is pivoted on a frictionless axle through its center. The bucket is released from
rest at the top of a well and falls 20m to the water.
a) What is the tension in the rope while the bucket is falling?
b) With what speed does the bucket strike the water?
c) What is the time of fall?
From Kinematics
VF2 = VO2 + 2ah
a = 6.533 m/s2 h = 20 m
VF2 = (0)2 + 2(6.533)(20)
VF2 = 261.333 m2/s2
VF = 16.167 m/s
VO = 0
Rotation about a FIXED Axis
A bucket of water with a mass of 20kg is suspended by a rope wrapped around a
windlass in the form of a solid cylinder 0.4m in radius, also with a mass of 20kg. The
cylinder is pivoted on a frictionless axle through its center. The bucket is released from
rest at the top of a well and falls 20m to the water.
a) What is the tension in the rope while the bucket is falling?
b) With what speed does the bucket strike the water?
c) What is the time of fall?
From Kinematics
VF = VO + at
a = 6.533 m/s2
t = (VF – VO)/ a
t = (16.167 m/s – 0)/ 6.533 m/s2
t = 2.475 s
VF = 16.167 m/s
VO = 0
Work, Energy & Power in Rotational Motion
F
M
R
W = FS, but S = Rθ
W = FRθ , but τ = FR
W = τθ
K = ½ MV2 ,but ω = V/R and V = ωR
K = ½ M(ωR)2 = ½ MR2ω2 , but I = MR2
K=½
Iω2
Average Power (PAVE)
PAVE = W/t
P = (τθ)/t , but ωAVE = θ/t
PAVE = τωAVE
R
θ
S
M
Instantaneous Power
P(t) = dW/dt
P(t) = τ (dθ/dt) , but ω(t) = dθ/dt
P(t) = τω(t)
Rotation about a FIXED Axis
A playground merry-go-round has a radius of 4.4 m and a moment of inertia of 245 kg-m2
and turns with negligible friction about a vertical axle through its center.
A child applies a 25 N force tangentially to the edge of the merry-go-round for 20 seconds.
a)If the merry-go-round is initially at rest what is the angular velocity after this 20 second
interval?
b) How much work did the child do on the merry-go-round?
c) What is the average power supplied by the child?
R = 4.4 m
I = 245 kg-m2
F = 25 N
t = 20 s
Rotation about a FIXED Axis
A playground merry-go-round has a radius of 4.4 m and a moment of inertia of 245 kg-m2
and turns with negligible friction about a vertical axle through its center.
A child applies a 25 N force tangentially to the edge of the merry-go-round for 20 seconds.
a)If the merry-go-round is initially at rest what is the angular velocity after this 20 second
interval?
Using NSLM
b) How much work did the child do on the merry-go-round?
c) What is the average power supplied by the child?
τ net = Iα < eq -2
R = 4.4 m
eq-1 = eq-2
2
I = 245 kg-mFrom Kinematics
FR = Iα
ω F = ω O + αt
ω O = 0 t = 20 s α = ?
Using Torque
F = 25 N
t = 20 s
τ net = FnetR
= (F – f)R
= FR < eq -1
α = FR/I
α = (25N)(4.4m)/(245 kg-m2)
α = 0.45 rad/s2
ωF = 0 + (0.45 rad/s2)(20s)
ωF = 9 rad/s
Rotation about a FIXED Axis
A playground merry-go-round has a radius of 4.4 m and a moment of inertia of 245 kg-m2
and turns with negligible friction about a vertical axle through its center.
A child applies a 25 N force tangentially to the edge of the merry-go-round for 20 seconds.
a)If the merry-go-round is initially at rest what is the angular velocity after this 20 second
interval?
b) How much work did the child do on the merry-go-round?
c) What is the average power supplied by the child?
R = 4.4 m
Work
From Kinematics
I = 245 kg-m2
2
W=τθ
Using Torque
F = 25 N
t = 20 s
θ = ωOt + ½ αt
θ = (0)(20s) + ½ (0.45rad/s2)(20s)2
θ = 90 rad
τ net = FnetR
= (F – f)R
= FR
τ = (25N)(4.4m) = 110 Nm
W = (110 Nm)(90 rad)
W = 9,900 J = 9.9 KJ
Rotation about a FIXED Axis
A playground merry-go-round has a radius of 4.4 m and a moment of inertia of 245 kg-m2
and turns with negligible friction about a vertical axle through its center.
A child applies a 25 N force tangentially to the edge of the merry-go-round for 20 seconds.
a)If the merry-go-round is initially at rest what is the angular velocity after this 20 second
interval?
b) How much work did the child do on the merry-go-round?
c) What is the average power supplied by the child?
R = 4.4 m
From Average Power
I = 245 kg-m2
F = 25 N
t = 20 s
PAVE = τ ω AVE
Γ = 110 Nm θ = 90 rad
ωAVE = θ/t
ωAVE = 90 rad/20s = 4.5 rad/s
PAVE = (110 Nm)(4.5 rad/s)
PAVE = 495 Watts
Rotation about a MOVING Axis
A string wrapped several times around the rim of a small hoop with radius 0.08 m and a mass
of 0.28 kg (I = MR2). If the free end of the string is held in place and the hoop is released
from rest, calculate :
a) The tension in the string while the hoop descends as the string unwinds;
b) The time it takes the hoop to descend 0.5 m;
c) The angular velocity of the rotating hoop after it has descended 0.5 m.
Using Torque
h = 0.5 m
τ net = FnetR
= (T – f)R
= TR < eq -1
t=?
Using NSLM
T
T
ω=?
τ net = Iα
< eq -2
eq-1 = eq-2
TR = Iα
T = Iα/R
T = (MR2)(a/R)/R
T = Ma < eq-3
Rotation about a MOVING Axis
A string wrapped several times around the rim of a small hoop with radius 0.08 m and a mass
of 0.28 kg (I = MR2). If the free end of the string is held in place and the hoop is released
from rest, calculate :
a) The tension in the string while the hoop descends as the string unwinds;
b) The time it takes the hoop to descend 0.5 m;
c) The angular velocity of the rotating hoop after it has descended 0.5 m.
Using LCE
KE1 + PE1 = KE2 + PE2
0 + MgH = ½ Iω 2 + Mg(H-h)
2
h = 0.5 m 0 + MgH = ½ Iω + MgH – Mgh
2 – Mgh
0
=
½
Iω
t=?
½ Iω 2 = Mgh
T
T
H
ω=?
ω 2 = 2Mgh/I
ω 2 = 2Mgh/(MR2)
ω 2 = 2gh/(R2) = 2(9.8)(0.5) / (0.08)2 = 1,531.25
ω = 39.13 rad/s
Rotation about a MOVING Axis
A string wrapped several times around the rim of a small hoop with radius 0.08 m and a mass
of 0.28 kg (I = MR2). If the free end of the string is held in place and the hoop is released
from rest, calculate :
a) The tension in the string while the hoop descends as the string unwinds;
b) The time it takes the hoop to descend 0.5 m;
c) The angular velocity of the rotating hoop after it has descended 0.5 m.
Using Kinematics
ω F = ω O + αt
T
ω F = ω = 39.13 rad/s ω O = 0
h = 0.5 m
T
t=?
ω=?
ω F = ω O + αt
39.13 = 0 + (122.5)t
α=?
ω F2 = ω O2+ 2αθ (solve first for α)
θ = s/r = h/r = 0.5 m / 0.08 m = 6.25 rad
1531.25 = 0 + 2α(6.25)
α = 122.5 rad/s2
1531.25 = 12.5α
t = 39.13 rad/s / (122.5 rad/s2)
t = 0.319 s
Rotation about a MOVING Axis
A string wrapped several times around the rim of a small hoop with radius 0.08 m and a mass
of 0.28 kg (I = MR2). If the free end of the string is held in place and the hoop is released
from rest, calculate :
a) The tension in the string while the hoop descends as the string unwinds;
b) The time it takes the hoop to descend 0.5 m;
c) The angular velocity of the rotating hoop after it has descended 0.5 m.
Using Torque
h = 0.5 m
τ net = FnetR
= (T – f)R
= TR < eq -1
t=?
Using NSLM
T
T
ω=?
τ net = Iα
< eq -2
eq-1 = eq-2
TR = Iα
T = Iα/R
T = (MR2)(a/R)/R
T = Ma < eq-3
a = αR
a = (122.5 rad/s2)(0.08m)
a = 9.8 m/s2
a = g = 9.8 m/s2
T = (0.28 kg)(9.8 m/s2)
T = 2.744 N
ROTATION ABOUT A MOVING AXIS : ROLLING OBJECT (Rolling without Slipping)
MS = 1 kg
RS = 0.2 m
h = 0.6 m
θ = 25°
1
2
3
s = 5.5 m
4
A sphere (as shown) starts from rest and rolls without slipping down an inclined surface that is 0.6 m above
the ground. Determine :
(a) Linear Acceleration of the sphere as it slides down the slope (a) & the frictional force (f)
(b) V3 = Velocity at pt. 3
(c) The μ of the flat surface for it to stop at pt. 4
In order that there will be NO SLIPPING. Friction on the slope is NOT zero.
f
θ
N
wS= MSg
τnet = ∑τ (CW+) = + τf
τnet = + (f)(RS) :eq. 1
eq.1 = eq.2
fRS = +(2/5)aMSRS
τnet = Iα (CW+) = +(2/5)MSRS2(a/RS)
τnet = +(2/5)aMSRS :eq. 2
f = (0.4)aMS : eq.3
ROTATION ABOUT A MOVING AXIS : ROLLING OBJECT (Rolling without Slipping)
MS = 1 kg
RS = 0.2 m
h = 0.6 m
θ = 25°
s = 5.5 m
1
2
3
4
A sphere (as shown) starts from rest and rolls without slipping down an inclined surface that is 0.6 m above
the ground. Determine
(a) Linear Acceleration of the sphere as it slides down the slope (a) & the frictional force (f)
N
∑Fx = ma (+→)
wSX – f = +Ms a
wSX = (MSg)sinθ
f = wSX – Msa : eq.4
a
Consider the forces acting on the surface of slope
f
f
eq.3 = eq.4
f
wS= MSg
N
θ
N
wS= MSg
θ
θ
wS= MSg
wSX = (MSg)cosθ
(0.4)aMS = wSX – MSa
(1.4)aMS = wSX
a = (MSgsinθ)/(1.4)MS
ROTATION ABOUT A MOVING AXIS : ROLLING OBJECT (Rolling without Slipping)
MS = 1 kg
RS = 0.2 m
h = 0.6 m
θ = 25°
s = 5.5 m
1
2
3
4
A sphere (as shown) starts from rest and rolls without slipping down an inclined surface that is 0.6 m above
the ground. Use LCE and Rotational Dynamics to determine
(a) Linear Acceleration of the sphere as it slides down the slope (a) & the frictional force (f)
a
a = (MSgsinθ)/(1.4)MS
N
f
wSX = (MSg)sinθ
a = (gsinθ)/(1.4)
a = (9.8m/s2)(sin25°)/(1.4)
wS= MSg
θ
wSX = (MSg)cosθ
a = 2.958
m/s2
f = (0.4)aMS : using eq.3
f = 0.4(2.958 m/s2)(1kg)
f = 1.1832 N
ROTATION ABOUT A MOVING AXIS : ROLLING OBJECT (Rolling without Slipping)
MS = 1 kg
RS = 0.2 m
h = 0.6 m
θ = 25°
s = 5.5 m
1
2
3
4
A sphere (as shown) starts from rest and rolls without slipping down an inclined surface that is 0.6 m above
the ground. Determine (b) V3 = Velocity at pt.3
WO = (K3 – K1) + (U3 – U1)
K1 = 0, since v1 = 0 (from rest)
U1 = MSgh = (1kg)(9.8m/s2)(0.6m) = 5.88 J
K3 = KLinear + KRotational = ½MSv32 + ½ Iω2 = 0.5MSv32 + 0.5[(2/5)MSRS2][v3/RS]2
U3 = 0, since h3 = 0
K3 = 0.5MSv32 + 0.2MSv32 = 0.7MSv32 = 0.7(1kg)v32
S1 = ?
WO = WF + Wf = 0 + (– f)(s1) = – f (h/sinθ)
WO = (– 1.1832N)(0.6m/sin25°) = – 1.678 J
h = 0.6 m
θ = 25°
ROTATION ABOUT A MOVING AXIS : ROLLING OBJECT (Rolling without Slipping)
MS = 1 kg
RS = 0.2 m
h = 0.6 m
θ = 25°
s = 5.5 m
1
2
3
4
A sphere (as shown) starts from rest and rolls without slipping down an inclined surface that is 0.6 m above
the ground. Determine (b) V3 = Velocity at pt.3
WO = (K3 – K1) + (U3 – U1)
– 1.678 J = [(0.7)(1kg)(v32) – 0] + (0 – 5.88J)
v32 = 6 m2/s2
v3 = 2.45 m/s
ROTATION ABOUT A MOVING AXIS : ROLLING OBJECT (Rolling without Slipping)
MS = 1 kg
RS = 0.2 m
ΣFy = 0 (at flat surface)
N –WS = 0
h = 0.6 m
N = WS = MSg
θ = 25°
f2 = ? & μ = ?
s = 5.5 m
1
2
3
4
A sphere (as shown) starts from rest and rolls without slipping down an inclined surface that is 0.6 m above
the ground. Determine
(c) The μ of the flat surface for it to stop at pt.4
WO = (K4 – K3) + (U4 – U3)
K3 = 0.7 MSv32 = 0.7(1kg)(2.45m/s)2 = 4.2 J
K4 = 0, since v4 = 0 (sphere stops)
U3 = 0 , since h3 = 0
U4 = 0 , since h4 = 0
WO = WF + Wf = 0 + (– f2)(s) = (–μN)s
WO = –μMSgs = –μ(1kg)(9.8m/s2)(5.5m)
WO = –μ(53.9 J)
WO = (K4 – K3) + (U4 – U3)
– μ(53.9 J) = [ 0 – 4.2 J] + [0 – 0]
μ = 4.2 J /53.9 J
μ = 0.0779
Impulse & Momentum in Rotational Motion
Angular Momentum (L)
p = MV ,but ω = V/R and V = ωR
p = MωR
L = pR = MωR(R) = MωR2 but I = MR2
L = Iω
M
R
Angular Impulse (Jθ)
Jθ = τ Δt = ΔL
Jθ = τ Δt = Iω2 – Iω1
Conservation of Angular Momentum
“When two bodies interact with each other and nothing else (isolated
system), their total angular momentum is conserved or constant”
ΣLbefore = ΣLafter
Iaωa + Ibωb = (Ia + Ib)ω
V
Impulse & Momentum in Rotational Motion
A solid wood door 1 m wide and 2 m high is hinged along one side (2m side) and has a
total mass of 35 kg. Initially open and at rest, the door is struck with a hammer at its
center. During the blow, an average force of 2,000 N acts for 5 ms. Find the angular
velocity of the door after the impact
F = 2,000 N
R
ωO = 0
F = 2,000 N
Δt = 5x10-3 s
R
ωA= ?
Angular Impulse (Jθ)
Jθ = τ Δt = Iω2 – Iω1
τ = FR = (2,000N)[½ (1 m)]
τ = 1,000 Nm
Jθ = (1,000 Nm)(5x10-3s) = 5 Nms
Impulse & Momentum in Rotational Motion
b
b
a
a
Inertia of a rectangular
plane – axis at its geometric
center
I=
(1 /
12
)M(a2 +
b2)
Inertia of a rectangular
plane – axis at one edge
I = (1/3)Ma2
Impulse & Momentum in Rotational Motion
A solid wood door 1 m wide and 2 m high is hinged along one side (2m side) and has a
total mass of 35 kg. Initially open and at rest, the door is struck with a hammer at its
center. During the blow, an average force of 2,000 N acts for 5 ms. Find the angular
velocity of the door after the impact
M = 35 kg
F = 2,000 N
R
R
ωA= ?
ωO = 0
F = 2,000 N
Δt = 5x10-3 s
Jθ = 5 Nms = Iω2 – Iω1
I = (1/3)Ma2
I = (1/3)(35 kg)(1 m)2
I = 11.667 kg-m2
ω1 = ωO = 0
ω2 = ωA = ?
Impulse & Momentum in Rotational Motion
A solid wood door 1 m wide and 2 m high is hinged along one side (2m side) and has a
total mass of 35 kg. Initially open and at rest, the door is struck with a hammer at its
center. During the blow, an average force of 2,000 N acts for 5 ms. Find the angular
velocity of the door after the impact
M = 35 kg
F = 2,000 N
R
R
ωO = 0
F = 2,000 N
Δt = 5x10-3 s
5 Nms = IωA – IωO
5 Nms = (11.667 kg-m2)ωA
ωA = 5 Nms / 11.667 kg-m2
ωA = 0.429 rad/s
ωA= ?
Conservation of Angular Momentum
A large turntable rotates about a fixed vertical axis, making 1 revolution in 6 seconds.
The moment of inertia about this axis is 1,200 kg-m2. A man with a mass of 80 kg, then
steps into the turntable and stops at a 2m distance from the axis. What is the new
angular velocity of the turntable due to the presence of the man at that distance? (Treat
the man as a point mass when solving for his inertia)
ωO = 1 rev / 6 sec = 0.1667 rev/s
ωN = ?
r=2m
M = 80 kg
I T = 1,200 kg-m2
I T = 1,200 kg-m2
ωO = (0.1667 rev/s) (2π rad / 1 rev) = 1.047 rad/s
I M = MR2 = (80kg)(2m)2 = 320 kg-m2
Conservation of Angular Momentum
ωO = 1 rev / 6 sec = 0.1667 rev/s
ωN = ?
I M = 320 kg-m2
r=2m
ωO = 1.047 rad/s
M = 80 kg
I T = 1,200 kg-m2
I T = 1,200 kg-m2
Lbefore = Lafter
Iaωa + Ibωb = (Ia + Ib)ω
ITωOT + IMωOM = (IT + IM)ωN
(1,200)(1.047) + 0 = (1,200+320)ωN
1,256.4 kg-m2/s= (1,520 kg-m2)ωN
ωN = (1,256.4 kg-m2/s) / (1,520 kg-m2)
ωN = 0.827 rad/s
A circular disc rotates on a thin air film with a period of 0.3s. Its moment of inertia about its
axis of rotation is 0.06 kg m2. A small mass is dropped onto the disc and rotates with it. The
moment of inertia of the mass about the axis of rotation is 0.04 kg m2. Determine the final
period of the rotating disc and mass
T1 = 0.3 s
T2 = ?
IM = 0.04 kg-m2
M
ID = 0.06 kg-m2
ID = 0.06 kg-m2
I1 ω1 = I2 ω2
ID ω1 = (ID + IM)ω2
(0.06)(2π/0.3) = (0.06 + 0.04)( 2π/T2)
(0.06/0.3) = (0.1/T2)
0.2 = 0.1/T2
T2 = 0.5 sec
Given : BEFORE
ω2 = 20 rad/s
ω=?
AFTER
ω1 = 20 rad/s
M2 = 5 kg
M1 = 2 kg
R2 = 0.6 m
R1 = 0.2 m
ω =−15.294 /s
ΣLbefore = ΣLafter
+
I1ω1 + I2ω2 =(I1+I2)ω
(−52kg·m2/s) = (3.4kg·m2)ω
ω = (−52kg·m2/s)/(3.4kg·m2)
I = ½ MR2
=−15.294 rad/s
ω =15.2954 rad/s ,CCW
½M1R12ω1 + ½M2R22ω2 =(½M1R12+ ½M2R22)ω
½(M1R12ω1 +M2R22ω2) = ½(M1R12+ M2R22)ω
M1R12ω1 +M2R22ω2 = (M1R12+ M2R22)ω
(2kg)(0.2m)2(+20rad/s) +(5kg)(0.6m) 2(−20rad/s) = [(2kg)(0.2m)2+(5kg)(0.6m)2]ω
(8kg·m2/s)+(−60kg·m2/s) = (3.4kg·m2)ω
SECOND CONDITION OF EQUILIBRIUM
Recall :
FIRST CONDITION :
“At Equilibrium the vector sum of a force, must be zero”
ΣFx = 0 , ΣFy = 0 , Fnet = R = 0
SECOND CONDITION :
“The sum of all Torques of all forces acting on the body with
respect to ANY specified axis, must be zero”
Στ = 0 (about any axis) , τnet = 0
Or
Total CCW Torque equals Total CW Torque
ΣτCCW = ΣτCW
SECOND CONDITION OF EQUILIBRIUM
CENTER OF GRAVITY (C.G.)
This is the point on the body that the entire force of gravity is
concentrated.
For a body of definite shape, it is found at its geometric center. For a
composite body the center of gravity can be calculated using
torques.
Center of Gravity of Common Shapes
Circle
Semi-Circle
Y-axis
Y-axis
XCG = r
YCG = r
XCG = r
(0.0)
X-axis
(0.0)
YCG = 4r/3π = 0.424 r
X-axis
Note : All shapes are made from the same material and uniform thickness
Center of Gravity of Common Shapes
Rectangles
Square
Y-axis
Height
(H)
Y-axis
XCG = B/2
side (s)
YCG = s/2
YCG = H/2
(0.0)
X-axis
Base (B)
XCG = s/2
(0.0)
X-axis
side (s)
Note : All shapes are made from the same material and uniform thickness
Center of Gravity of Common Shapes
Right Triangle
Triangles
Isosceles and Equilateral Triangle
Y-axis
Height
(H)
Y-axis
XCG = B/3
Height
(H)
YCG = H/3
(0.0)
YCG = H/3
X-axis
Base (B)
XCG = B/2
(0.0)
X-axis
Base (B)
Note : All shapes are made from the same material and uniform thickness
Center of Gravity of Common Shapes
Generally all CG of triangles are determined by
½c
½A
C
A
C.G
½B
B
Note : All shapes are made from the same material and uniform thickness
Center of Gravity of Common Shapes
The Center of Gravity is NOT always INSIDE the object
Ring
Y-axis
XCG = r
YCG = r
(0.0)
X-axis
Note : All shapes are made from the same material and uniform thickness
Center of Gravity of Composite Shapes
H= 9.5 c m
Given : Find the CG of this figure of uniform density & thickness.
B = 10 c m
Center of Gravity of Composite Shapes
1st Step : Place the figure in an x-y plane, and identify the basic shape it contains.
Use the formulas given earlier to determine the INDIVIDUAL C.G.’s
Y-axis
X1 = r
Y2 = H/3
H= 9.5 c m
X2 = B/2
Y1 = 0.424r
B = 10 c m
(0.0)
X-axis
Center of Gravity of Composite Shapes
2nd Step : Make a tabulation for your measurements and calculations. The number
of rows depends on the number of basic shapes
Y-axis
Shape
Semicircle
Triangle
Total
Area
xCG from origin
yCG from origin
Ax
Ay
Center of Gravity of Composite Shapes
3rd Step : Compute for the individual areas and also the total area
Y-axis
Shape
Area
Semicircle
A1
Triangle
A2
Total
xCG from origin
yCG from origin
Ax
Atotal
A1 = ½ πr2 : Area of a Semicircle
A2 = ½BH : Area of a Triangle
A1 = ½ π(5cm)2
A2= ½ (10cm)(9.5cm)
A1 = 39.27 cm2
A2 = 47.5 cm2
AT = 39.27 cm2 + 47.5 cm2
AT = 86.77 cm2
Ay
Center of Gravity of Composite Shapes
4th Step : Measure the x-coordinate of the individual shape FROM the origin
Y-axis
Shape
Area
xCG from origin
Semicircle
39.27
X1
Triangle
47.5
X2
Total
86.77
x1 = ½ B = ½(10 cm) = 5 cm
x2 = ½ B = 5 cm
yCG from origin
Ax
Ay
Center of Gravity of Composite Shapes
5th Step : Measure the y-coordinate of the individual shape FROM the origin
Y-axis
Shape
Area
xCG from origin
yCG from origin
Semicircle
39.27
5
Y1
Triangle
47.5
5
Y2
Total
86.77
y1 = (4r/3π) + H = (0.424r)+H
y1 = 0.424(5 cm) + 9.5 cm = 11.62 cm
y2 = (2/3) H = (2/3)(9.5cm) = 6.33 cm
Ax
Ay
Center of Gravity of Composite Shapes
6th Step : Compute the area “torque”
Y-axis
Shape
Area
xCG from origin
yCG from origin
Ax
Semicircle
39.27
5
11.62
A1x1
A1y1
Triangle
47.5
5
6.33
A2x2
A2y2
Total
86.77
A1x1 = (39.27cm2)(5cm) = 196.35 cm3
A1y1 = (39.27cm2)(11.62cm) = 456.32 cm3
A2x2 = (47.5cm2)(5cm) = 237.5 cm3
A2y2 = (47.5cm2)(6.33cm) = 300.675 cm3
Ay
Center of Gravity of Composite Shapes
7th Step : Compute the total “torque” per axis
Y-axis
Shape
Area
xCG from origin
yCG from origin
Ax
Semicircle
39.27
5
11.62
196.35
456.32
Triangle
47.5
5
6.33
237.5
300.675
Total
86.77
ΣAx
ΣAy
ΣAx = 196.35 cm3 + 237.5 cm3 = 433.85 cm3
ΣAy = 456.32 cm3 + 300.675 cm3 = 757 cm3
Ay
Center of Gravity of Composite Shapes
8th Step : Compute the composite’s center of gravity coordinates:
Y-axis
Shape
Area
xCG from origin
yCG from origin
Ax
Ay
Semicircle
39.27
5
11.62
196.35
456.32
Triangle
47.5
5
6.33
237.5
300.675
Total
86.77
433.85
757
XCG = ΣAx/Atotal = ( 433.85 cm3 ) / (86.77cm2) = 5 cm
YCG = ΣAy/Atotal = ( 757 cm3 ) / (86.77cm2) = 8.724 cm
Center of Gravity of Composite Shapes
last Step : Locate the C.G. on your composite figure
XCG = 5 cm
YCG = 8.724 cm
(0.0)
Center of Gravity of Composite Shapes
r = 1 cm
H= 6 c m
3cm
Given : Find the CG of this figure of uniform density & thickness.
B= 4cm
Center of Gravity of Composite Shapes
1st Step : Place the figure in an x-y plane, and identify the basic shape it contains.
Use the formulas given earlier to determine the INDIVIDUAL C.G.’s
Let : Object 1 : Rectangle
X 2= B/2
X 1= B/2
B= 4cm
(0.0)
r = 1 cm
r = 1 cm
H= 6 c m
3cm
Object 2 : Circle (Hole)
Y 2= r
Y 1= H/2
X-axis
Center of Gravity of Composite Shapes
2nd Step : Make a tabulation for your measurements and calculations. The number
of rows depends on the number of basic shapes
Y-axis
Shape
Rectangle
Circle
Total
Area
xCG from origin
yCG from origin
Ax
Ay
Center of Gravity of Composite Shapes
3rd Step : Compute for the individual areas and also the total area
Y-axis
Shape
Area
Rectangle
A1
Circle
A2
Total
xCG from origin
yCG from origin
Ax
Atotal
A1 = BH : Area of a rectangle
A2 = πr2 : Area of a Circle
A1 = (4cm)(6cm)
A2= π(1cm)2
A1 = 24 cm2
A2 = 3.1416 cm2
AT = 24 cm2 + ( – 3.1416 cm2)
AT = 20.858 cm2
Ay
Center of Gravity of Composite Shapes
4th Step : Measure the x-coordinate of the individual shape FROM the origin
Y-axis
Shape
Area
xCG from origin
Rectangle
24
X1
Cicle
-3.1416
X2
Total
20.858
x1 = ½ B = ½(4 cm) = 2 cm
x2 = ½ B = 2 cm
yCG from origin
Ax
Ay
Center of Gravity of Composite Shapes
5th Step : Measure the y-coordinate of the individual shape FROM the origin
Y-axis
Shape
Area
xCG from origin
yCG from origin
Rectangle
24
2
Y1
Cicle
-3.1416
2
Y2
Total
20.858
y1 = H/2 = (6cm/2) = 3 cm
y2 = (H/2) + r = 3 cm + 1 cm = 4 cm
Ax
Ay
Center of Gravity of Composite Shapes
6th Step : Compute the area “torque”
Y-axis
Shape
Area
xCG from origin
yCG from origin
Ax
Rectangle
24
2
3
A1x1
A1y1
Cicle
-3.1416
2
4
A2x2
A2y2
Total
20.858
A1x1 = (24cm2)(2cm) = 48 cm3
A1y1 = (24cm2)(3cm) = 72 cm3
A2x2 = (-3.1416cm2)(2cm) = – 6.2832 cm3
A2y2 = (-3.1416cm2)(4cm) = 300.675 – 12.5664 cm3
Ay
Center of Gravity of Composite Shapes
7th Step : Compute the total “torque” per axis
Y-axis
Shape
Area
xCG from origin
yCG from origin
Ax
Rectangle
24
2
3
48
72
Cicle
-3.1416
2
4
– 6.2832
– 12.5664
Total
20.858
ΣAx
ΣAy
ΣAx = 48 cm3 – 6.2832 cm3 = 41.7168 cm3
ΣAy = 72 cm3 – 12.5664 cm3 = 59.4336 cm3
Ay
Center of Gravity of Composite Shapes
8th Step : Compute the composite’s center of gravity coordinates:
Y-axis
Shape
Area
xCG from origin
yCG from origin
Ax
Ay
Rectangle
24
2
3
48
72
Cicle
-3.1416
2
4
– 6.2832
– 12.5664
Total
20.858
41.7168
59.4336
XCG = ΣAx/Atotal = ( 41.7168cm3 ) / (20.858cm2) = 2 cm
YCG = ΣAy/Atotal = ( 59.4336 cm3 ) / (20.858cm2) = 2.85 cm
Center of Gravity of Composite Shapes
last Step : Locate the C.G. on your composite figure
r = 1 cm
H= 6 c m
3cm
B = 3 cm
XCG
YCG
(0.0)
B= 4cm
XCG = 2 cm
YCG = 2.85 cm
SECOND CONDITION OF EQUILIBRIUM
Calculation of Center of Gravity using Weights
ΣτX = w1x1 + w2x2 + . . . .
ΣτX = Σwx
ΣτY = w1y1 + w2y2 + . . . .
ΣτY = Σwy
XCG = Σwx / ΣW
YCG = Σwy / ΣW
Where
w1 , w2 - weight of individual element
x1, x2 – lever arm of individual element with
respect to x-axis
y1, y2 – lever arm of individual element with
respect to y-axis
ΣW – total weight of composite body
X,Y - coordinate of the center of gravity
For simplicity : One of the axes can be graphically set to zero, IF the composite body
is PURELY horizontal or vertical in structure, such that we only need to solve for one
coordinate
SECOND CONDITION OF EQUILIBRIUM
Calculation of Center of Gravity using Weights
Given :
A 14 kg beam is hinged at one end. A 6 kg triangular object and a 7.5 kg object are positioned
as shown. Dots indicate the individual centers of gravity of the beam and the two objects. What
is the distance from the axis of rotation to the center of gravity for this system?
M2 = 7.5 kg
M1 = 6 kg
MB = 14 kg
Solution :
Since the body is purely horizontal. We can set the Y coordinate for the C.G. to zero and solve only for X
coordinate. Place the composite body at the center of the Y= 0 and one end of it at X=0 :
Hence YCG = 0
SECOND CONDITION OF EQUILIBRIUM
Calculation of Center of Gravity using Weights
M2 = 7.5 kg
M1 = 6 kg
MB = 14 kg
Using the origin as axis
x1 = 0.36m
x2 =0.36 m + 0.53m = 0.89m
x3 = 0.89m + 1.92m = 2.81m
Weights :
w1 = m1g = (6kg)(9.8m/s2) = 58.8 N
wB = mBg = (14kg)(9.8m/s2) = 137.2 N
w2 = m2g = (7.5kg)(9.8m/s2) = 73.5 N
WT = w1 + wB + w2
WT = 58.8N + 137.2N + 73.5N = 269.5 N
Στx = Σwx = w1x1 + wBx2 + w2x3
Σwx = (58.8N)(0.36m)+(137.2N)(0.89m)+(73.5N)(2.81m)
Σwx = 349.811 Nm
XCG = Σwx / W = (349.811Nm) / (269.5N)
XCG = 1.298 m
SECOND CONDITION OF EQUILIBRIUM
Calculation of Center of Gravity using Weights
Given :
A 14 kg beam is hinged at one end. A 6 kg triangular object and a 7.5 kg object are positioned
as shown. Dots indicate the individual centers of gravity of the beam and the two objects. What
is the distance from the axis of rotation to the center of gravity for this system?
M2 = 7.5 kg
M1 = 6 kg
MB = 14 kg
XCG = 1.298m
SECOND CONDITION OF EQUILIBRIUM
Given :
An 80 kg man balances (system is at equilibrium) the boy on a teeter-totter (similar to a seesaw) as shown. Note : Ignore the weight of the board they are standing on.
a)What is the approximate mass of the boy
b)What, approximately, is the magnitude of the downward force exerted on the fulcrum?
m=?
M = 80 kg
4m
fulcrum
1m
SECOND CONDITION OF EQUILIBRIUM
Given :
An 80 kg man balances (system is at equilibrium) the boy on a teeter-totter (similar to a seesaw) as shown. Note : Ignore the weight of the board they are standing on.
a)What is the approximate mass of the boy
b)What, approximately, is the magnitude of the downward force exerted on the fulcrum?
m=?
M = 80 kg
L1 =4 m
fulcrum
L2 =1 m
a) Using the fulcrum as axis
ΣτCCW = ΣτCW
mgL1=MgL2
mg(4m) =(80kg)g(1m)
m = 20 kg
SECOND CONDITION OF EQUILIBRIUM
Given :
An 80 kg man balances (system is at equilibrium) the boy on a teeter-totter (similar to a seesaw) as shown. Note : Ignore the weight of the board they are standing on.
a)What is the approximate mass of the boy
b)What, approximately, is the magnitude of the downward force exerted on the fulcrum?
F=?
m=?
M = 80 kg
L1 =4 m
fulcrum
L2 =1 m
WB
WM
b) Using the first condition of equilibrium
ΣFY = 0 +↑
– F – WBm – WM = 0
F = – (WB + WM) = – (mg + Mg) = –[(20 kg)(9.8m/s2)+(80kg)(9.8m/s2)] = – 980N
Considering only the magnitude
|F | = 980N
SECOND CONDITION OF EQUILIBRIUM
SCE on the ladder using red dot ( pt.A) as axis
A ladder 10m long and has a weight of 400N, with
the center of gravity at its center. It leans in
equilibrium against a vertical frictionless wall.
Dimensions are provided in the figure. Find the
magnitudes forces C, N & f.
ΣτA CCW = ΣτA CW
ΣτA CCW = (C)(8m) = C(8m)
ΣτA CW = (400N)(3m) = 1,200 Nm
ΣτA CCW = ΣτA CW
C(8m) = 1,200 Nm
C
3m
C = (1,200 Nm)/(8m)
C = 150 N
8m
W = 400N
f
6m
N
SECOND CONDITION OF EQUILIBRIUM
SCE on the wall corner using blue dot ( pt.B)
as axis
ΣτB CCW = ΣτB CW
A ladder 10m long and has a weight of 400N, with
the center of gravity at its center. It leans in
ΣτB CCW = (C)(8m) + (400N)(3m)
equilibrium against a vertical frictionless wall.
ΣτB CCW = (150N)(8m) + 1200Nm = 2400 Nm
Dimensions are provided in the figure. Find the
magnitudes forces C, N & f.
ΣτB CW = (N)(6m) = N(6m)
ΣτB CCW = ΣτB CW
2400 Nm = N(6m)
3m
N= 400 N
C
3m
SCE on the wall and ladder contact point using
green dot ( pt.D) as axis
ΣτD CCW = ΣτD CW
8m
ΣτD CCW = (400N)(3m) + (f)(8m)
ΣτD CCW = 1200Nm + f(8m)
ΣτD CW = (N)(6m) = (400N)(6m) = 2400Nm
W = 400N
f
6m
N
ΣτD CCW = ΣτD CW
1200 Nm + f(8m) = 2400 Nm
f = 150 N
SECOND CONDITION OF EQUILIBRIUMAlternative Solution : for f & N
SCE on the ladder using red dot ( pt.A) as axis
A ladder 10m long and has a weight of 400N, with
the center of gravity at its center. It leans in
equilibrium against a vertical frictionless wall.
Dimensions are provided in the figure. Find the
magnitudes forces C, N & f.
ΣτA CCW = ΣτA CW
ΣτA CCW = (C)(8m) = C(8m)
ΣτA CW = (400N)(3m)
ΣτA CW = 1,200 Nm
ΣτA CCW = ΣτA CW
C(8m) = 1,200 Nm
C
3m
C = (1,200 Nm)/(8m)
C = 150 N
8m
W = 400N
f
6m
N
Using FCE : ΣFx = 0 & ΣFy = 0
ΣFx = 0 (+ )
+f – C = 0
f=C
ΣFx = 0 (+ ↑)
+N – W = 0
N=W
f = 150 N
N = 400 N
SECOND CONDITION OF EQUILIBRIUM
ΣτA CCW = ΣτA CW
Two men are carrying a uniform ladder that is 5m
long & weighs 100N. If one man applies an
upward (lifting) force equal to 42N at one end.
For the system to be at equilibrium:
a) at what point should the other man lift?
b) what should be the magnitude of F?
(100N)(x) = (42N)(2.5m+x)
5m
b) SCE on the ladder using blue dot ( pt.B) as axis
2.5 m
42N
100N
x
(100N)X = 105Nm + (42N)X
(58N)X = 105 Nm
X = 1.81 m (from the center)
ΣτB CCW = ΣτB CW
F
ΣτB CCW = (F)(2.5m+x)
ΣτB CCW = (F)(2.5m+1.81m) = F(4.31m)
A
B
a) SCE on the ladder using red dot ( pt.A) as axis
ΣτA CCW = ΣτA CW
ΣτA CCW = (100N)(x)
ΣτA CW = (42N)(2.5m+x)
ΣτB CW = (100N)(2.5m) = 250 Nm
ΣτB CCW = ΣτB CW
F(4.31m) = 250 Nm
F = 58 N
SECOND CONDITION OF EQUILIBRIUM
a) SCE on the beam using red dot ( pt.A) as axis
ΣτA CCW = ΣτA CW
The horizontal beam weighs 200N and its C.G. is at
its center. Find:
a. The tension of the cable
b. The horizontal (C) & vertical (f) components of
the forces exerted on the beam by the wall.
ΣτA CCW = (T)(4m)sin θ
ΣτA CCW = (T)(4m)sin 36.87°
ΣτA CCW = T(2.4m)
ΣτA CW = (200N)(2m) + (300 N)(4m)
FBD
ΣτA CW = 1,600 Nm
T
T
3m
3m
ΣτA CCW = ΣτA CW
5m
5m
T(2.4m) = 1,600 Nm
T = 666.667 N
θ
C
2m
4m
2m
f
300N
200N
300N
5m
3m
θ = tan -1 ( 3m/4m) = 36.87°
θ
4m
SECOND CONDITION OF EQUILIBRIUM
b) SCE on the beam using blue dot
( pt.B) as axis
The horizontal beam weighs 200N and its C.G. is at
ΣτB CCW = ΣτB CW
its center. Find:
ΣτB CCW = (200N)(2m) = 400 Nm
a. The tension of the cable
ΣτB CW = (f)(4m) = f(4m)
b. The horizontal (C) & vertical (f) components of
the forces exerted on the beam by the wall.
Στ
= Στ
B CCW
400Nm = f(4m)
FBD
f = 100 N
T
5m
5m
We have no more valid axis. Hence we
use FCE on the entire beam to get the
value of C
T
3m
3m
Tx
C
2m
4m
Using FCE : ΣFx = 0 & ΣFy = 0
θ
2m
f
300N
200N
300N
ΣFx = 0 (+ )
+C – Tx = 0
C = Tx = Tcosθ
C = (666.667N)( cos 36.87°)
5m
3m
θ = tan -1 ( 3m/4m) = 36.87°
θ
4m
B CW
C = 533.333 N
SECOND CONDITION OF EQUILIBRIUM
Alternative Solution for friction (f) using FCE
The horizontal beam weighs 200N and its C.G. is at
its center. Find:
a. The tension of the cable
b. The horizontal (C) & vertical (f) components of
the forces exerted on the beam by the wall.
Using FCE : ΣFx = 0 & ΣFy = 0
ΣFy = 0 (+ )
+f – 200N – 300N + Ty = 0
f = 500N – Ty = 500N – Tsinθ
FBD
T
5m
5m
f = 100 N
T
3m
3m
f = 500N – (666.667N)( sin 36.87°)
Ty
θ
C
2m
4m
2m
f
300N
200N
300N
5m
3m
θ = tan -1 ( 3m/4m) = 36.87°
θ
4m
COUPLES
Two forces with equal magnitude and opposite direction with a line of action that are
parallel but does not coincide.
F
0
X1
L
F
X2
τCouple = FL
Στo +
Στo = −Fx1+ Fx2 , But x2 = x1 + L
Στo = −Fx1 + F(x1 + L)
Στo = −Fx1 + Fx1 + FL
Στo = + FL
Στo = FL (CW)
Couple principle : At any point at
the plane the couple produces the
SAME torque.
COUPLES
1. Two equal parallel forces F1 = F2 = 5N are applied to a rod as shown :
a. Calculate the net torque about pt. O due to these two forces by calculating the
torque of each separate force.
b. Calculate the net torque about pt. P due to these two forces by calculating the
torque of each separate force.
c. Compare your results using the couple equation.
F1
F2
3m
37°
37°
O
P
1.5m
1.5m
a) Getting net torque on the beam using green dot( pt. O) as axis
ΣτO = τO net +
ΣτO = -(F1)(3m)sin(180°-37°) + (F2)(3m+1.5m) sin(180°-37°)
ΣτO = -(5N)(1.8054m) + (5N)(2.7082m)
τO net = + 4.514 Nm = 4.514 Nm , CW
COUPLES
1. Two equal parallel forces F1 = F2 = 5N are applied to a rod as shown :
a. Calculate the net torque about pt. O due to these two forces by calculating the
torque of each separate force.
b. Calculate the net torque about pt. P due to these two forces by calculating the
torque of each separate force.
c. Compare your results using the couple equation.
F1
F2
3m
37°
37°
O
P
1.5m
1.5m
b) Getting net torque on the beam using red dot( pt. P) as axis
ΣτP = τP net +
ΣτP = +(F1)(3m)sin(37°) – (F2)(1.5m) sin(37°)
ΣτP = +(5N)(1.8054m) – (5N)(0.9027m)
τP net = + 4.514 Nm = 4.514 Nm , CW
COUPLES
1. Two equal parallel forces F1 = F2 = 5N are applied to a rod as shown :
a. Calculate the net torque about pt. O due to these two forces by calculating the
torque of each separate force.
b. Calculate the net torque about pt. P due to these two forces by calculating the
torque of each separate force.
c. Compare your results using the couple equation.
F1
F2
3m
37°
37°
37°
O
P
1.5m
sin 37° = L/1.5 m
L = 1.5m(sin 37°)
1.5 m
L = 0.9027 m
1.5m
c) Using couple equation
τCouple = FL
τCouple = (5N)(0.9027m)
τCouple = 4.514Nm, CW
All the values are the same. Thus
proving the couple principle. That
at any point at the plane the couple
produces the SAME torque.

Download Report