Q1(a) A car in Figure Q1(a) travels in a straight line with acceleration shown in the graph. The car starts from the origin with vo = -18 m/s. By assume t' = 8.4s, sketch the v-t and s-t graphs. a (m/s*) 6. 3 t (s) 4 12 -5
Q1(a) A car in Figure Q1(a) travels in a straight line with acceleration shown in the graph. The car starts from the origin with vo = -18 m/s. By assume t' = 8.4s, sketch the v-t and s-t graphs. a (m/s*) 6. 3 t (s) 4 12 -5
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
Related questions
Question
Use t'=20.4 instead of 8.4 and use the formulas attached
Transcribed Image Text:Q1(a) A car in Figure Q1(a) travels in a straight line with acceleration shown in the graph.
The car starts from the origin with vo = -18 m/s. By assume t' = 8.4s, sketch the
v-t and s-t graphs.
a (m/s")
3
t (s)
4
12
-5
Figure Q1(a)
(b)
The rod OA in Figure Q1(b) rotates clockwise with a constant angular velocity. Two-
pin-connected slider blocks, located at B, moves freely on OA and the curved rod is
describe by the equation r = 100(1 – cos 8) mm.
(i)
Determine the speed of the slider blocks at e = 120°
(11)
Calculate its magnitude of acceleration when e = 120°
6 rad/s
400 mm
600 mm
200 mm
Figure Q1(b)
![Fundamental Equation of Dynamics
KINAMATICS
Equations of Motion:
Particle Rectilinear motion:
Constant a = de
EF = ma
EF, = m(ag),
EF, = m(ag),
Variable a
Particle
dv
ds
v = vo + at
1
Rigid Body
(Plane Motion)
dt
s = , + vạt +at
12 = vỷ + 2a.(s - 50)
ads = vdv
EMG = Iça a EMp = E(M)p
Particle Curvilinear Motion:
Principle of Work and Energy:
T, + U,-2 = T,
Cartesian Coordinates (1,y,z)
V = * a, = *
Vy = ý ay = jỹ
Kinetic Energy
Particle
T =
: T=mvå + lgw?²
or T= 1,w?
Polar Coordinates (r,8,z)
Rigid Body
a, = # - re?
ag = rê + 2rẻ
(Plane Motion)
V = ré
Work
Uş = SF cos e de
U, = (F cos 8)As
Uw = -WAy
Variable Force :
Normal-Tangential Coordinates (n,t,b)
Constant Force:
a = i = v,
Weight
U, = -(kei -kei)
Spring
[1+ (dy/dx)*]a/2
|d²y/dx*|
Couple of Moment: UM = M AO
Where p=
Power and Efficiency
du
Uout
Pout
P = = Fv, E=
Pin
Conservation of Energy Theorem
!!
dt
Relative Motion
VB = VA + VB/A
T +V = T2 + V2
Potential Energy
Rigid Body Motion About a Fixed Axis
V = V, + V.
Where:
V = ± Wy , V, = +ks?
Иariable a
Constant a = de
Principle of linear Impulse and Momentum:
dw
w = w, +at
a =
dt
1
e = 0, + wat +
w? = w3 + 2a.(0 - 0.)
mv, +ES Fåt = mvz
Rigid Body: m(vc)ı +ES Fdt = m(va)a
Particle
de
dt
wdw = ade
For Point P
Conservation of Linear Momentum:
s= er, v= wr. a = ar, an = w'r
Σ(mv), Σ (mv),
Relative General Plane Motion-Translating
Coefficient of Restitution
(v3), – (v)2
Axes
Vg = VA + vn/Acpin) B = Ga+ aB/Apin)
Principle of Angular Impulse and Momentun:
: (H,), +ES M,dt = (H,)2
Where H, = (d)(mv)
Relative General Plane Motion-Tran. And
Rot. Axis
Particle
Vg = VA + w X TB/a + (VB/A)vz
ag = an + i x P/a +w x (w x rayA)
+ 2w x (V8/A)v
Rigid Body : (H.h +ES M,dt = (H)2
(Plane Motion)
+
Where He = Igw
Kineties
(H.)1 +
M,dt = (H,)2
Mass Moment of Inertia
Parallel-Axis Theory
1= [r*dm
1 = lg + m d?
Where H, = 1,w
Conservation of Angular Momentum
Radius of Gyration
k =
Σ(Η), Σ (Η),](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F33488049-1422-455e-b50d-4d67f5321948%2Ff34983fb-c2e0-4bd4-9a2e-34a5d2cac732%2Fajhlmxn_processed.png&w=3840&q=75)
Transcribed Image Text:Fundamental Equation of Dynamics
KINAMATICS
Equations of Motion:
Particle Rectilinear motion:
Constant a = de
EF = ma
EF, = m(ag),
EF, = m(ag),
Variable a
Particle
dv
ds
v = vo + at
1
Rigid Body
(Plane Motion)
dt
s = , + vạt +at
12 = vỷ + 2a.(s - 50)
ads = vdv
EMG = Iça a EMp = E(M)p
Particle Curvilinear Motion:
Principle of Work and Energy:
T, + U,-2 = T,
Cartesian Coordinates (1,y,z)
V = * a, = *
Vy = ý ay = jỹ
Kinetic Energy
Particle
T =
: T=mvå + lgw?²
or T= 1,w?
Polar Coordinates (r,8,z)
Rigid Body
a, = # - re?
ag = rê + 2rẻ
(Plane Motion)
V = ré
Work
Uş = SF cos e de
U, = (F cos 8)As
Uw = -WAy
Variable Force :
Normal-Tangential Coordinates (n,t,b)
Constant Force:
a = i = v,
Weight
U, = -(kei -kei)
Spring
[1+ (dy/dx)*]a/2
|d²y/dx*|
Couple of Moment: UM = M AO
Where p=
Power and Efficiency
du
Uout
Pout
P = = Fv, E=
Pin
Conservation of Energy Theorem
!!
dt
Relative Motion
VB = VA + VB/A
T +V = T2 + V2
Potential Energy
Rigid Body Motion About a Fixed Axis
V = V, + V.
Where:
V = ± Wy , V, = +ks?
Иariable a
Constant a = de
Principle of linear Impulse and Momentum:
dw
w = w, +at
a =
dt
1
e = 0, + wat +
w? = w3 + 2a.(0 - 0.)
mv, +ES Fåt = mvz
Rigid Body: m(vc)ı +ES Fdt = m(va)a
Particle
de
dt
wdw = ade
For Point P
Conservation of Linear Momentum:
s= er, v= wr. a = ar, an = w'r
Σ(mv), Σ (mv),
Relative General Plane Motion-Translating
Coefficient of Restitution
(v3), – (v)2
Axes
Vg = VA + vn/Acpin) B = Ga+ aB/Apin)
Principle of Angular Impulse and Momentun:
: (H,), +ES M,dt = (H,)2
Where H, = (d)(mv)
Relative General Plane Motion-Tran. And
Rot. Axis
Particle
Vg = VA + w X TB/a + (VB/A)vz
ag = an + i x P/a +w x (w x rayA)
+ 2w x (V8/A)v
Rigid Body : (H.h +ES M,dt = (H)2
(Plane Motion)
+
Where He = Igw
Kineties
(H.)1 +
M,dt = (H,)2
Mass Moment of Inertia
Parallel-Axis Theory
1= [r*dm
1 = lg + m d?
Where H, = 1,w
Conservation of Angular Momentum
Radius of Gyration
k =
Σ(Η), Σ (Η),
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