MECH1230_Dynamics_Lecture-IV_Lecture-Slides

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• Course Title & Information: MECH1230 Solid Mechanics Semester 2: Dynamics

  • Faculty: School of Mechanical Engineering, University of Leeds, UK

  • Instructor: Dr. I Shafagh (Email: I.Shafagh@leeds.ac.uk)

  • Copyright: 2025

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• Lecture Overview:

  • Recap of Lecture III

  • Specific Case: Constant Acceleration

  • Revisiting SUVAT Equations

  • Derivation of SUVAT Equations:

    • From Velocity-Time Graphs

    • Using 1D Particle Motion Analysis

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• Recap of Previous Concepts:

  • Displacement, velocity and acceleration relations introduced

  • Stated that in 1D, these are single-variable functions

  • Explained the use of the Chain Rule in 1D kinematics analysis

  • Discussed interpretation of d-t (displacement-time), v-t (velocity-time), and a-t (acceleration-time) graphs

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• Example Problem: Braking of a Dragster

  • Dragster reaches a max speed of 90 m/s, then deploys a parachute until speed reduces to 18 m/s

  • Acceleration with parachute:

    • a = -k1 * v^2 where k1 = 0.0022 m^-1

  • Speed between 18 m/s and rest (0 m/s) has constant acceleration of -3 m/s²

  • Problem Objective: Determine total distance travelled during deceleration.

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• Calculations from Example Problem:

  • Total time = 26.20 s

  • Use of relations:

    • v = dx/dt

    • a = dv/dt

    • Incorporate given acceleration equations to derive dx/dt.

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• Further Calculations Continues:

  • Initial conditions applied at t = 0 s when x = 0 m: Constant C = 0

  • At t = 20.20 s, x = 731.56 m

  • Acceleration calculation gives a = -3, which is consistent across time shifts between intervals

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• Final Calculations:

  • Update results to find x1 = 731.56 m

  • Final of t = 26.2 s leads to C1 = 244.1

  • Final distance x2 = 785.56 m indicating total distance travelled of 54 m during deceleration

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• The Chain Rule:

  • 1D displacement, velocity, and acceleration relations as single-variable functions

  • Chain rule for functions:

    • dy/dx = dy/du * du/dx (where u is another function of x)

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• Chain Rule Application in Calculus:

  • Example when a(x) instead of a(t)

  • Related acceleration to velocity derivatives across chain functions

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• Example 3: Particle Displacement Analysis in Resisting Medium

  • Displacement given by: x = 160(1 - e^(-0.25t)) - 20t

  • Tasks:

    • (a) Find velocity and acceleration at t = 0 s

    • (b) Determine total distance travelled before coming to rest

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• Velocity Calculation at t = 0 s:

  • v = dx/dt = evaluated resulting in v = 20 m/s

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• Acceleration Calculation at t = 0 s:

  • a = dv/dt = evaluated resulting in a = -10 m/s²

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• Total Distance Travelled before Rest:

  • Set velocity to zero, solve for t

  • Resultant t = 2.773 s

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• Substitute t Back into displacement equation for x

  • Resultant total distance travelled: x = 24.5 m

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• Constant Acceleration Overview:

  • Discussion on rare cases of uniform acceleration

  • Attention merited to near uniform acceleration in real-world scenarios

  • Includes gravitational constant: g = 9.81 m/s²

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• SUVAT Equations:

  • Gradient (dv/dt) represents acceleration

  • First equation derived: v = u + at

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• Area Under the Graph relates to average velocity

  • Derived area expression: s = (v + u)/2 * t

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• Another Area Equation for total distance:

  • s = ut + 1/2(v-u)t where v = u + at

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• Final SUVAT Expressions:

  • Transitioning from s = (v + u)/2 * t to v² = u² + 2as

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• Derivation from Acceleration Relation:

  • Expression: a = dv/dt leading to calculated relations

  • Proven correlation of constants during integration.

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• Further Derivation Towards Displacement:

  • x = ut + 1/2at², providing simplified motion relations in context

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• Velocity to Acceleration Relation:

  • General equations leading to nature of motion: v² = u² + 2as

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• Example 7: Car Accelerating from Rest

  • Calculate time taken to achieve a constant speed of 40 m/s

  • Determine the distance traveled during acceleration

  • Assess total time taken to reach destination

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• Calculations for Example 7:

  • Time to reach speed: t = (40-0)/2 = 20s

  • Total distance during acceleration = 400 m

  • Remaining travel calculation

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• Example 8 Overview:

  • Bus starts from rest & accelerates with parameters defined

  • Total journey from point A to D need to be computed

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• Diagram Sketch illustrating motion phases

  • Key points A, B, C, D relevant for calculations

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• Time from A to B:

  • Calculated using: t_AB = (vB – vA)/a_AB

  • Derived time = 12s

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• Distance from B to C:

  • Need to determine distance to apply proper time from point B

  • Total distance calculated for A to C routing

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• Exact Time from B to C:

  • Utilizes constant velocity for index, t_BC = s/v

  • Result yields time required

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• Time from C to D:

  • Calculate deceleration needed from velocity drop

  • Resulting constant yields time consistently

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• Final Summation of Journey Time:

  • Total time from A to D established

  • Total journey time = 29s

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• Lecture Summary:

  • Reviewed constant acceleration in particle motion

  • Derived significant equations of motion including SUVAT relations

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• Copyright Notice:

  • Protected by law and terms of restriction on reproduction and distribution.