Running head: ENGINEERING DYNAMICS 0
ENGINEERING DYNAMICS
Name of Student
Institution Affiliation
QUESTION 1
Part 1
Given that the particle has a mass of m=0.6kg
, ENGINEERING DYNAMICS 2
Angular velocity w=2.4rad/s
Ѳ=320
Ѳ´= 2.4rad/s
0.6 0.6 0.6
r´ = cos θ = cos 32 = 0.8480 = 0.7075 m
Calculate the linear velocity of the rod
r´ = 0.6 secѲ ( (tanѲ )( Ѳ´))
= 0.6sec(32)(tan(32)(2.4)
=0.6(1.11791)(0.6248(2.4))
=1.06085m/s
Calculate the linear acceleration of the rod
r´´ 0.6secѲ (tanѲ(Ѳ´)) tanѲ +secѲ(sec2Ѳ´(Ѳ) Ѳ´+sec(tanѲѲ´´ )
(0.6secѲ´×tan2Ѳ×Ѳ´)+ (secѲ×sec2Ѳ×Ѳ2) +0
= (0.6 ×1.1791× 0.6242 ×2.4) + (1.1791×1.3902×5.76)
, ENGINEERING DYNAMICS 3
= (0.6611) + (9.4417) = 10.1028 m/s2
Determine the radial acceleration
ar =r´´ - rѲ2
= 10.1028-(0.7075(2.4)2
=10.1028 – 4.0752
= 6.0276 m/s 2
Angular acceleration
aѲ= rѲ´´ +2r´´ Ѳ
=(0.7075×0 + (2×1.0608×2.4)
=5.09184m/s2
Now calculate the force due to the linear acceleration
mar= Fr
mar=µcos Ѳ- mgcos Ѳ
(0.6 )(6.0276)= µ cos 32 – ( 0.6 ×9.81×cos 32)
3.6156=0.8480N-4.9916
0.8480N=8.6081
N= 10.1511N
Part 2
ENGINEERING DYNAMICS
Name of Student
Institution Affiliation
QUESTION 1
Part 1
Given that the particle has a mass of m=0.6kg
, ENGINEERING DYNAMICS 2
Angular velocity w=2.4rad/s
Ѳ=320
Ѳ´= 2.4rad/s
0.6 0.6 0.6
r´ = cos θ = cos 32 = 0.8480 = 0.7075 m
Calculate the linear velocity of the rod
r´ = 0.6 secѲ ( (tanѲ )( Ѳ´))
= 0.6sec(32)(tan(32)(2.4)
=0.6(1.11791)(0.6248(2.4))
=1.06085m/s
Calculate the linear acceleration of the rod
r´´ 0.6secѲ (tanѲ(Ѳ´)) tanѲ +secѲ(sec2Ѳ´(Ѳ) Ѳ´+sec(tanѲѲ´´ )
(0.6secѲ´×tan2Ѳ×Ѳ´)+ (secѲ×sec2Ѳ×Ѳ2) +0
= (0.6 ×1.1791× 0.6242 ×2.4) + (1.1791×1.3902×5.76)
, ENGINEERING DYNAMICS 3
= (0.6611) + (9.4417) = 10.1028 m/s2
Determine the radial acceleration
ar =r´´ - rѲ2
= 10.1028-(0.7075(2.4)2
=10.1028 – 4.0752
= 6.0276 m/s 2
Angular acceleration
aѲ= rѲ´´ +2r´´ Ѳ
=(0.7075×0 + (2×1.0608×2.4)
=5.09184m/s2
Now calculate the force due to the linear acceleration
mar= Fr
mar=µcos Ѳ- mgcos Ѳ
(0.6 )(6.0276)= µ cos 32 – ( 0.6 ×9.81×cos 32)
3.6156=0.8480N-4.9916
0.8480N=8.6081
N= 10.1511N
Part 2
