Engineering Dynamics Solutions

ENGINEERING DYNAMICS 1
[Author Name(s), First M. Last, Omit Titles and Degrees]
Author Note
[Include any grant/funding information and a complete correspondence address.]

2
.
4. Letting the cylindrical co-ordinates be:
Er is the unit vector in xy plane in radial direction.
Eꬾ is the unit vector in xy plane perpendicular to Er in the direction of increasing ꬾ.
K is the unit vector in z. Thus,
R = (x2 + y2)1/2 ; ꬾ = tan-1(y/x).
The position is: rop = x(t)i + y(t)j +z(t)k
= r(ꬾ)Er + z(t)k
Er = cosꬾi+sinꬾj= ∂ r
∂ r / ∂ r
∂ r
Eꬾ=
∂ r
∂ ꬾ /∂ r
∂ ꬾ
but ∂ r
∂ ꬾ = r(-sinꬾi+cosꬾj)
Velocity thus = vp = drop/dt=rer+rder/dt+zk+zdk/dt
When z-direction(k) fixed = dk/dt = 0
In figure 4.4, calculating the pressure force and coordinate x.
= mgsinꬾ
= 0.09×0.9×sin60
=0.070151
Also,
=0.01 × 0.1 ×sin 60
= 0.000866
X= x0 + v0t +1/2gsinꬾt2
0.07015= 0.2 + ½ sin60t2 + 4πt
0.07015 – 0.2 = 0.433t2 + 12.57t

3
- 0.12985 = 0.433t2 + 12.57t
Also,
0.000866 = 0.5 + 1/2sin60t2 + πt
0.000866 – 0.5 = 0.433t2 + 3.14t
-0.49914 = 0.433t2 + 3.14t
- 0.12985 = 0.433t2 + 12.57t
0.36929 = 9.43t
T=0.0392
Solution to figure 4.15
Er is the unit vector in xy plane in radial direction.
Eꬾ is the unit vector in xy plane perpendicular to Er in the direction of increasing ꬾ.
K is the unit vector in z. Thus,
R = (x2 + y2)1/2 ; ꬾ = tan-1(y/x).
The position is: rop = x(t)i + y(t)j +z(t)k
= r(ꬾ)Er + z(t)k
Er = cosꬾi+sinꬾj= ∂ r
∂ r / ∂ r
∂ r
Eꬾ=
∂ r
∂ ꬾ /∂ r
∂ ꬾ
but ∂ r
∂ ꬾ = r(-sinꬾi+cosꬾj)
Velocity thus = vp = drop/dt=rer+rder/dt+zk+zdk/dt
When z-direction(k) fixed = dk/dt = 0
In figure 4.15, calculating the pressure force and coordinate x.
= mgsinꬾ
= 0.01×0.1×sin60

4
=0.000866
X= x0 + v0t +1/2gsinꬾt2
Also,
0.000866 = 0.0005 + πt
0.000866 – 0.0005 = 3.14t
0.000366 = 3.14t
T= 0.0001166
5. Determine the velocity at positions B and C and get the pressure force on the
particle.
V = 4Q/πD2
But Q = ∫
0
D 2
u 2 πdr
Q = 2π ∫
0
D 2
urdr
= 2π × 0.25
=1.571
(4 × 1.571)/ π × 0.1×0.1
Velocity at B= 200.02m/s
Velocity at C
V = 4Q/πD2