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Solution datum at lowest point solving for the
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University of Ca… ENG 57 Homework Help nmrtinez29 8 싙 100% (1)
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Alice in The Sca
a
To Kill Wonderland rlet
gbird Letter
Mockin
At the instant the spring becomes undeformed, the center
of the 40-kg disk has a speed of 4 m>s . From this point
determine the distance d the disk moves down the plane
before momentarily stopping. The disk rolls without k = 200 N/m
slipping.
Literature Study
0.3 m
Guides SOLUTION
30°
Learn more about Datum at lowest point.
T1 + V1 = T 2 + V 2
characters, symbols,
1 1 4 2 1 1
c (40)(0.3)2 d a b + (40)(4)2 + 40(9.81)d sin 30° = 0 + (200)d 2
and themes in all your 2 2 0.3 2 2
100d 2 - 196.2d - 480 = 0
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d = 3.38 m Ans.
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ENG 57
12 pages
K i nemat i cs By
considering the
vertical motion of
the ball after the
University of Calif…
18–51.
ENG 57 - Fall 2015
A spring having a stiffness of k = 300 N>m is attached to
the end of the 15-kg rod, and it is unstretched when u = 0°.
If the rod is released from rest when u = 0° , determine its
angular velocity at the instant u = 30° . The motion is in the
vertical plane. B
11 pages u
k 300 N/m
SOLUTION 0.6 m
Potential Energy: With reference to the datum in Fig. a, the gravitational potential A
energy of the rod at positions (1) and (2) is
A Vg B 2 = -W(yG )2 = -15(9.81)(0.3 sin 30°)
Since the spring is initially unstretched,(Ve) 1 = 0. When u = 30°, the stretch of the
aGBnv2rGB3 spring is sP = 0.6 sin 30° = 0.3 m. Thus, the final elastic potential energy of the
spring is
2 07 a G OA n v 2
1 1
r G OA 3 2 03 27 AVe B2 = ks 2 = (300) A 0.32 B = 13.5 J
2 P 2
ms2aGBtarG Thus,
Ba V1 = (Vg ) 1 + (Ve)1 = 0 + 0 = 0
V2 = (Vg ) 2 + (Ve)2 = -22.0725 + 13.5 = -8.5725 J
University of Calif… Kinetic Energy: Since the rod is initially at rest, T1 = 0. From the geometry shown in
Fig. b, rG>IC = 0.3 m. Thus, (V G) 2 = v2rG >IC = v 2 (0.3). The mass moment of inertia
ENG 57 - Fall 2015 (15) A 0.62 B = 0.45 kg # m2 .Thus,
1 1
of the rod about its mass center is IG = ml2 =
12 12
the final kinetic energy of the rod is
16 pages 1 1
T2 = m(vG) 2 2 + IGv 2 2
2 2
1 1
= (15) C v2 A 0.3 B D 2 + A 0.45 B v 2 2
2 2
= 0.9v2 2
Conservation of Energy:
T1 + V 1 = T2 + V2
0 + 0 = 0.9v2 2 - 8.5725
u 15 r L u v2 = 3.09 rad>s Ans.
SOLUTION Free
Body D i a g ram
The free body
diagram of the
airplane
University of Calif…
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x 1 2 dx dA 2 y dx
2 2 2 x 1 2 dx y x 2
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TERM
Fall '15
University of Calif…
ENG 57 - Fall 2015 TAGS
Statics, Dynamics, Energy, Kinetic Energy, Mass, Potential Energy, Angular Velocity
15 pages
Report
aAvBvAABd
1237 As a body is
projected to a
high altitude
above the
University of Calif…
ENG 57 - Fall 2015
18 pages
The ride at an
amu s ement park
con s i s t s of a g
ondola which i s
lifted to
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