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Prof. Dr. Ersel Canyurt 6 - 1
Dynamics R.C. Hibbeler Ferdinand P. Beer E. Russell Johnston, Jr. Lecture Notes: Prof. Dr. O. Ersel Canyurt Gazi University Prof. Dr. Ersel Canyurt 6 - 2 Mechanical Engineering Second Moment of Area x y Ix = (^) ʃ y^2 dA A Iy = ʃ^ x^2 dA A Iz = (^) ʃ (x^2 + y^2 ) dA A y x Unit m^4
Prof. Dr. Ersel Canyurt 6 - 3 x y
b a G
Rectangle x y
r G
Circle = Iy Second Moment of Area (Alan Atalet Momenti) Second Moment of Area according to x- axes Second Moment of Area according to y- axes Prof. Dr. Ersel Canyurt 6 - 4 Mechanical Engineering Second Moment of Area according to CG Second Moment of Area Parallel axes Theorem: Move Center of Gravitation (CG) Point to another axes z y G z’ d A
B
Second Moment of Area according to
Unit m^4 B
G z
Prof. Dr. Ersel Canyurt 6 - 7
2
m
Moment of Inertia Integrating over the entire region of the cylinder yields = ^2 ^ h^ ʃ^ r^ 3
0 R Im = h R^4 2 Mass of Cylinder m = ^ ^ R^2 h Im = 1 2 m • R^2 Moment of Inertia according to axes that passes through CG of Cylinder Prof. Dr. Ersel Canyurt 6 - 8 Mechanical Engineering Moment of Inertia for A Rod (Çubuk)
O
Moment of Inertia for A Rod (OA) according to CG
G
G m O A l Moment of Inertia for A Rod according to Point O or Point A
Prof. Dr. Ersel Canyurt 6 - 9 Radius of gyration, (k) (Eylemsizlik Moment Yarıçapı)
I k m The radius of the moment of inertia of a body based on a specified axis The mass of body m is given Moment of Inertia m Prof. Dr. Ersel Canyurt 6 - 10 Mechanical Engineering Parallel Axis Theorem Unit z
A Z X Y dm
A
m Moment of Inertia according to CG Moment of Inertia according to z axis that passes through point A m d^ G
Prof. Dr. Ersel Canyurt 6 - 13 Example 17. O B A C 1 m The pendulum is suspended from the pin at O and consists of two thin rods, each having a weight of 50 N. Determine the moment of inertia of the pendulum about an axis passing through (a) point 0, (b) the mass center G of the pendulum. 0.5 m 0.5 m Prof. Dr. Ersel Canyurt 6 - 14 Mechanical Engineering
O
m=W/g = 50/9.81^ = 5.09 [kg] Rod OA, Rod 1.
G 1 m +^ m^ ^ d 1 2
2
O
G 1
G 1 m O B A C d 1
O
3
Prof. Dr. Ersel Canyurt 6 - 15
O
Rod BC, Rod 2.
G 2 m +^ m^ ^ d 2 2
O
G 2 m G 2 O B A C G (^1) d 2 Prof. Dr. Ersel Canyurt 6 - 16 Mechanical Engineering
O
O m
O
G 2 O B A C G 1 y 2 =
O m
y 1 =0. i i
y
O
Go m+ mT ^ d 2 Go
O
Go m =^ – mT ^ d 2
Go
Prof. Dr. Ersel Canyurt 6 - 19 Example 17. When the forward speed of 1000 kg truck shown was 10 m/s, the brakes were suddenly applied, causing all four wheels to stop rotating. It was observed that the truck skidded (kayarak) to rest in 8m. Determine the magnitude of the normal reaction and of the friction force at each wheel as the truck skidded to rest. 2 2
2 0 10 2 aG (^8) aG= – 6.25 m/s^2 1.2 m 1.5 m 2.1^ m Prof. Dr. Ersel Canyurt 6 - 20 Mechanical Engineering x y O
W NA FA
No vertical motion 1 NB FB^ A m. aGx 1.2 m 1.5 m 2.1 m
(– 6.25) (^) μ k = 0.
Prof. Dr. Ersel Canyurt 6 - 21 x y O
W=9810 N NA FA
NB FB^ A m. aGx 1.2 m 1.5 m 2.1 m NB =
W = m g = 1000 9. = 9810 [N]
Prof. Dr. Ersel Canyurt 6 - 25 aBn= ^2 · AB = 62 · (0.5) aBn= 18 [m/s^2 ] B n t aBn aBt Rod BD Translation B G D aG = aB B G D Acting Forces TB TD W Effective Forces aBn Wt Wn aBt B 30 o Wn= W · cos 30 Wn= 849.6 [N] Wt= W · sin 30 Wt= 490.5 [N]
Wt = 100 aGt 490.5 = 100 aGt aGt = 4.9 [m/s^2 ] Prof. Dr. Ersel Canyurt 6 - 26 Mechanical Engineering B n t B G^ D B G D Acting Forces TB TD W Effective Forces aBn Wt Wn aBt 30 o Wn= W · cos 30 Wn= 849.6 [N] D
TB + TD – Wn = 100 (18) TB + TD = 2649.
Prof. Dr. Ersel Canyurt 6 - 27
Fn = m aGn Ft = m^ aGt MO Equations of Motion – ROTATION about a Fixed Axis When body rotates about a Fixed Axis Equations of Motion Dönme aGt aGn
aGn is ALWAYS directed through the center of rotation. Prof. Dr. Ersel Canyurt 6 - 28 Mechanical Engineering Example 30 kg disc rotates about pin O. A cord of negligible mass is wrapped around the disc. Cord Force of F= 10 N and M= 5 N•m Moment was applied to the disc. If it is released from rest, determine a) Required cycle when disc reaches 20 rad/s angular velocity b) The horizontal and vertical components of reaction forces at the pin O.
Prof. Dr. Ersel Canyurt 6 - 31 Example 17. O (^60) Nm 3 m = 5 rad/s At the instant shown in Figure, the 20 kg slender rod has an angular velocity of = 5 rad/s. Determine a) the angular acceleration and b) the horizontal and vertical components of reaction of the pin on the rod at this instant? Prof. Dr. Ersel Canyurt 6 - 32 Mechanical Engineering O 60 Nm 1.5 m G W = O 1.5 m G mat man
60 + W (1.5) = IG + m at (1.5) m l 2
= 15 [kgm^2 ] at = (1.5) an =^2 r
= 5.9 [rad/s^2 ]
Prof. Dr. Ersel Canyurt 6 - 33 O 60 Nm 1.5 m G W =
m l 2
60 + 20 9.81 (1.5) = 60 ^ = 60^ [kgm^2 ] 354.3 = 60 = 5.9 [rad/s^2 ]
Prof. Dr. Ersel Canyurt 6 - 34 Mechanical Engineering O 60 Nm 1.5 m G W = O 1.5 m G mat man
On Ot Reaction Forces an =^2 r
On = 20 (37.5)
aGn = 37.5 [m/s^2 ] Coordinate System On = 750 [N] n t
Prof. Dr. Ersel Canyurt 6 - 37
Block Fy = m ay W – T = 20^ ^ a x y G T =196.2 – 20^ ^ a
Prof. Dr. Ersel Canyurt 6 - 38 Mechanical Engineering
O
=
x y O
O = 11.29 [rad/s^2 ] T =196.2 – 8
Prof. Dr. Ersel Canyurt 6 - 39
x y (^) • Forces acting on periphery =
O = 11.29 [rad/s^2 ]
W (0.4) =^ IO + (m^ ^ a)^ ^ (0.4) = 3.75 78.48 = 3.2^ ^ + 3.75
Prof. Dr. Ersel Canyurt 6 - 40 Mechanical Engineering
The spool (Makara) has a mass of 8 kg and a radius of gyration of kG = 0.35 m. The cords of negligible mass are wrapped around its inner hub and outer rim. Determine the angular acceleration of the spool in Figure?
