Download - LEC 14 (review 12_13)
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King Fahd University ofPetroleum & Minerals
Mechanical Engineering
Dynamics ME 201
BY
Dr. Meyassar N. Al-HaddadLecture # 14
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Projectile Determine the speed at which
the basketball at A must be
thrown at the angle of 30oso
that it makes it to the basket at
B. At what speed does it pass
through the hoop?
tss
Horizontal
oo
toA 30cos010
2
2
1tatss
Vertical
coo
2)81.9(2
130sin5.13 ttoA
933.0tsmA /4.12
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A ball traveling at 25 m/s drive off of the edgeof a cliff 50 m high. Where do they land?
25 m/s
Vertically
v = v0-gt
y = y0+ v0t + 1/2gt2.
v2= v02- 2g(y-y0).
Horizontally
x = x0 + (v0)x t x = 25 *3.19 = 79.8 m
79.8 m
Initial Conditions
vx= 25m/s
vy0= 0m/s
a =- 9.8 m/s
2
t = 0
y0= 0 m
y =- 50 m
x0=0 m
V2= -2(9.8
)(-50-0)V = 31.3 m/s
-50 = 0+0+1/2(-9.8)t2 t = 3.19 s
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Example 13-4
m A= 3 kg
m B= 5 kg
From rest
vB= ? In 2 second
yy maF AaT 32981
Block A
yy maF BaT 52.196
Block B
lss BA 2 02 BA 02 BA aa
taBo
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VA= ?
2Sc+ S
A= L
2ac+ aA= 0
aA= -2ac
ca
ccx
y
amgTNmaF
NNNF
20030sin25.0:
1699030cos81.9*200:0
Problem
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Relative motion At the instant shown, cars A and
B are traveling at the speeds
shown. If B is accelerating at
1200 km/h2while A maintains a
constant speed, determine the
velocity and acceleration of Awith respect to B.
BABA /vvv
BA
oo
/vi65j45sin20i45cos20
j14.14i14.79v / BA
hkmBA /4.80)14.14()14.79(v 22
/ BABA /aaa
BA
oo
/
22
ai1200j45sin1.0
)20(i45cos
1.0
)20(
j2828i1628a / BA
222
/ /3260)2828()1628( hkma BA
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Polar and Cylindrical The slotted fork is rotating about O at a
constant rate of 3 rad/s. Determine the radial
and transverse components of velocity and
acceleration of the pin A at the instant q= 360o.
The path is defined by the spiral groove r =(5+q/p)in., where qis in radians.
pq
pq p
q
p
q
rrr 752
032360
qqpq
o
sinrvr /955.03
p
sinrv /21)3(7 qq
222 /63)3(70 sinrrar q
2/73.5)3)(3(202 sinrra p
qqq
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Example 12-20
qq
q
q
andfind
180atft/s30
ft/s4
ft)cos1(5.0
o2
a
v
r
)cos1(5.0 qr
qq )sin(5.0rqqqqq )sin(5.0)()(cos5.0 r
2
o
500ft1
180at
.-rrr
q
rad/s44)1()0()()( 2222 qqq rr
30)]4)(0(21[])4(1)4(5.0[)2()( 2222222 qqqq rrrra
2rad/s18q
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Problem 13-89Smooth
Horizontal
m = 0.5 kg
r = (0.5 q)
q = 0.5 t2rad
Applied force
Normal force
t = 2 s.
q
r
qqq
5.05.05.0 rrr15.0 2 qqq tt
2
2
/5.0/11
/1/26.1142
sec2
smrsmrmr
sradsradrad
tat
o
qqq
o
radddr
r
43.63
5.0
5.0
/tan
2
q
q
q
222 /5.3)2(15.0 sftrrar q2/5)2)(1(2)1(12 sftrra qqq
NFFmaF
NNNmaF ccrr
62.1)5(5.043.63cos96.1
96.1)5.3(5.043.63sin
qq
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Problem
m=2 kg
Smooth horizontal
r = 0.4 qPtangent=?
N=?
At q = 45o
qqq 4.04.04.0 rrr
0/64
45 qqp
q srado
0/4.21.0 rsmrr p
222
/31.11)6()1.0(0 smrrar pq
2/8.28)6)(4.2(202 smrra qqq
o
ddr
r15.38
44.0
4.0
/tan
pq
q
q
)31.11(215.38sin15.38cos; NPmaF rr )8.28(215.38cos15.38cos; NPmaF qq
NPNN 8.173.59
PN
r
q
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