Repeat Problem 1.110 if the initial displacement is θ0=π/ 2...

Key Concepts

Simple Pendulum Mechanics

The simple pendulum is a classical example in mechanics that illustrates oscillatory motion under gravity. Its dynamics are governed by a nonlinear differential equation that relates the angular displacement to the restoring gravitational force, showing how energy is exchanged between potential and kinetic forms during motion.

Nonlinear Oscillations

For small displacements, the simple pendulum can be approximated by linear equations, but when the displacement is large, as in the case of an initial displacement of ?/2 radians, the full nonlinear behavior must be considered. This entails solving the nonlinear differential equation without the small-angle approximation, which leads to significant differences in oscillatory behavior and period.

Elliptic Integrals

When addressing nonlinear oscillations in a pendulum with large initial displacements, the period of oscillation is often found by evaluating an elliptic integral. Unlike elementary functions, elliptic integrals arise naturally in the analysis of systems where the restoring force is not proportional to the displacement, capturing the exact dynamics over a larger range of motion.

Initial Conditions

Initial conditions play a crucial role in determining the subsequent motion of any dynamical system. In the context of the pendulum, changing the initial displacement from a small angle to ?/2 radians significantly alters the energy of the system, thereby affecting the period and amplitude of the oscillations due to the nonlinear terms in the motion equation.

Transcript

00:01 Hello friends here it is given the engine system in which rod vd of length 250mm and mass 1 .2 kg connected to the piston of mass 1 .8 kg and with the crank av of length 100mm as given in the diagram.

00:38 The crank is rotating in clockwise direction with 600 rpm, that is 62 .832, radiant per second.

01:01 No force is applied by the piston.

01:08 Calculate force exerted on the connecting rod b and d for theta 2v 90 degrees.

01:44 Let us see here.

01:45 First we will analyze the geometry.

01:46 We will draw the mathematical figure of it first here this is the piston p this is d point dv this is the crank this is l this is v so here value of x this distance you can calculate by pythagoras syrup and x will be root of l is square minus b square l is 0 .25 minus b square l is 0 .25 minus 0 .1.

03:16 So value of x you will get 0 .2291 meter position vector of v with respect to a.

03:27 So rva can be written as 0 .1 meter j cap and r dv vector can be written as minus 0 .2291 i cap minus 0 .1 j cap meter.

03:59 Now angular velocity of crank ab is given we are writing here in vector form minus 62 .823 radiant per second k cap.

04:27 The velocity of v point will be cross product of angular velocity of av and rba vector substituting the value minus 62 .832 i cap cross r b a that is 0 .1 j cap so it will be 6 .2832 meter per second along i cap now velocity of d can be written as velocity of b point plus velocity of d with respect to b so it will be velocity of v plus omega vd cross r dv vector substituting the value velocity of b we have obtained 6 .2832 meter per second i cap omega vd k cap cross r d is minus 0 .229291 i cap minus 0 .2 .1 i cap minus 0 .1 so velocity of d along x direction you may write 6 .2832 icap plus 0 .1 omega vd j cap minus 0 .2291 omega vd i cap equating its components i and jcap component you will get velocity of d to be 6 .2832 meter per second and angular velocity of vd to be 0 acceleration angular acceleration of av is 0 angular acceleration of vd rod alpha vd k cap and acceleration of d is ad i cap so acceleration of v you can write alpha a v cross r v a vector minus omega square a v r b a vector alpha a v is 0 minus 62 .832 square into rva vector that is 0 .1 j cap so you will get acceleration of v .2v minus 394 .78 j cap meter per second square and acceleration of d is defined as acceleration of v plus alpha vd cross r vd vector that is tangential acceleration minus normal acceleration omega square vd or dv vector substituting the value acceleration of d is along x -axis so a d a d i cap a v is already we have calculated 394 .78 j cap plus alpha a v a cap cross r bd vector which is minus 0 .2291 i cap minus 0 .21 j cap minus 0 .1 j cap minus 0 .1 jcap minus 0.

11:09 On solving it above equation can be written as 394 .78 icap minus 0 .2291 alp alpha a .v j cap plus 0 .1 alpha a .v i.

11:40 Equating the components along yx and jcap.

11:45 So y components, you may write 0 is called 2 .2, minus 3 94 .78 this is j cap minus 0 .2291 alpha a v so from here angular acceleration of a b you will get minus 1723 radian per second square and x component you will get acceleration of b point will be 0 .1 into minus 1723, that is minus 172 .3 meter per second square.

12:52 So here we can write angular acceleration of av in vector form, which is minus 1732, sorry, 1723 k -cap, radiant per second square, and acceleration of d, is minus of 172 .3 meter per second square i cap.

13:39 Acceleration of center of mass of bar v .d.

13:46 Acceleration of center of mass g of bar vd.

13:57 So position vector of g with respect to d will be half of 0 .2291i cap plus point 1 j cap so you can write point 1145 i cap plus 0 .05 .0 .05.

14:23 J cap meter.

14:27 So acceleration of center of mass will be acceleration of d plus angular acceleration of vd cross position vector of g with respect to d minus omega square vd in to position vector of g with respect to d.

14:59 Substituting the value, acceleration of b is minus 172 .3 icap, alpha vd we have calculated minus 172 .323k cap cross.

15:38 .11455 i cap plus 0 .0505 j cap minus omega is 3.

15:46 Square which is omega vd is zero so this term becomes zero so acceleration of center of mass of vd rod will be minus 86 .15 i cap minus 197 .34 j cap meter per second square.

16:19 Now we have to draw the free body diagram of bar vd at p fvd fvd of fvd of a fvd of rod vd at p that is piston so this is the piston here it will be normal reaction this distance already we have major x here b y and b x this is point bx it having mass of vd mass of vd into acceleration of g this is center of mass and it is rotating so it having moment of units of the vd into alpha vd...

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