Download presentation
Presentation is loading. Please wait.
1
Modeling & Simulation of Dynamic Systems
Lecture-5 Modeling Examples Dr. Imtiaz Hussain URL :
2
Rotational and Translational Motion (Example-1)
Consider the system shown below. The disc is of radius ‘R’ and has a moment of inertia ‘J’. There is friction between the disc and the block of mass ‘m’.
3
Rotational and Translational Motion
Free body diagram 𝑚 𝑥 𝐹=𝑚 𝑥 +𝑘𝑥+𝑐 𝑥 + 𝐹 𝑓 (1)
4
Rotational and Translational Motion
Free body diagram of disc Since 𝑥=𝑅𝜃 𝑇=𝐽 𝜃 𝐹 𝑓 𝑅=𝐽 𝜃 𝐹 𝑓 = 𝐽 𝑅 2 𝑥
5
Rotational and Translational Motion
Substituting the value of 𝐹 𝑓 in equation (1) Taking Laplace Transform of the equation 𝐹=𝑚 𝑥 +𝑘𝑥+𝑐 𝑥 + 𝐽 𝑅 2 𝑥 𝐹 𝑠 =𝑚 𝑠 2 𝑋(𝑠) +𝑘𝑋(𝑠)+𝑐𝑠𝑋(𝑠) + 𝐽 𝑅 2 𝑠 2 𝑋(𝑠) 𝑋(𝑠) 𝐹(𝑠) = 1 𝑚+ 𝐽 𝑅 2 𝑠 2 +𝑐𝑠+𝑘
6
Example-2 Consider the system shown below (Assume no friction).
The relation between the rotation of the disc and the linear displacement moved by the disc is given by x = Rθ
7
Example-2 Free body diagram of the disc Where 𝑥=𝑅𝜃 0=𝑚 𝑥 +2𝑘𝑥
8
Example-3
9
Example-3 Free Body diagram of Mass m 𝑚 𝑦 𝑚 𝑦 +𝑐 𝑦 −𝑅 𝜃 +𝑘 𝑦− 𝑅 2 𝜃 =0
10
Example-3 Free body diagram of drum
𝑇=𝐽 𝜃 +𝑐 𝑅 𝜃 − 𝑦 𝑅+𝑘 𝑅 2 𝜃−𝑦 𝑅 2 +𝑘𝜃𝑅 𝑅 +𝑐 𝑅 2 𝜃 ( 𝑅 2 )
11
Example-4: Elevator
12
Example-4: Elevator The cage of an elevator is hoisted by a long cable wound over a drum driven through a gear-set by an electric motor. The motor is relay-operated (i.e., either on or off)
13
Example-4: Elevator 𝒏 𝒈𝒆𝒂𝒓 𝜽 𝒓𝒊𝒎 𝒓 𝒅𝒓𝒖𝒎 𝜽 𝒎𝒐𝒕𝒐𝒓 𝒆 𝒎𝒐𝒕𝒐𝒓 𝒌 𝒄𝒂𝒃𝒍𝒆 𝒙 𝒄𝒂𝒈𝒆
𝒎 𝒄𝒂𝒈𝒆
14
Example-4: Elevator Differential Equation
𝒎 𝒄𝒂𝒈𝒆 𝑭 𝒎 𝒄𝒂𝒈𝒆 𝑭 𝒌 𝒄𝒂𝒃𝒍𝒆 𝑭 𝒈 𝑚 𝑐𝑎𝑔𝑒 𝑥 𝑐𝑎𝑔𝑒+ 𝑘 𝑐𝑎𝑏𝑙𝑒 𝑥 𝑐𝑎𝑔𝑒 − 𝑥 𝑟𝑖𝑚 + 𝑚 𝑐𝑎𝑔𝑒 𝑔=0
15
Example-4: Elevator Transmission 𝑥 𝑟𝑖𝑚 =𝑟 𝑑𝑟𝑢𝑚 𝜃 𝑑𝑟𝑢𝑚
𝑥 𝑟𝑖𝑚 =𝑟 𝑑𝑟𝑢𝑚 𝜃 𝑑𝑟𝑢𝑚 𝜃 𝑑𝑟𝑢𝑚 =𝑛 𝑔𝑒𝑎𝑟 𝜃 𝑚𝑜𝑡𝑜𝑟 𝜃 𝑚𝑜𝑡𝑜𝑟 =𝑒 𝑚𝑜𝑡𝑜𝑟 𝐾 𝑡 / 𝑅 𝑎 𝑠[𝐽𝑠+(𝐵+ 𝐾 𝑡 𝐾 𝑏 / 𝑅 𝑎 )]
16
Example-5: Tap Drive
17
Example-5: Tap Drive
18
Example-5: Tap Drive
19
End of Lecture-5 To download this lecture visit
End of Lecture-5
Similar presentations
© 2025 SlidePlayer.com. Inc.
All rights reserved.