CP SHM in a Car Engine. The motion of the piston of an...

CP SHM in a Car Engine. The motion of the piston of an automobile engine is approximately simple harmonic. (a) If the stroke of an engine (twice the amplitude) is 0.100 $\mathrm{m}$ and the engine runs at 4500 $\mathrm{rev} / \mathrm{min}$ , compute the acceleration of the piston at the endpoint of its stroke. (b) If the piston has mass $0.450 \mathrm{kg},$ what net force must be exerted on it at this point? (c) What are the speed and kinetic energy of the piston at the mid- point of its stroke? (d) What average power is required to accelerate the piston from rest to the speed found in part (c)? (e) If the engine runs at 7000 rev/min, what are the answers to parts (b), (c), and (d)?

Key Concepts

Kinetic Energy and Speed

Kinetic energy in the context of SHM is related to the velocity of the oscillating object and is calculated using the formula KE = 1/2 mv². At the midpoint of its path, an object in SHM has its maximum speed, and understanding this concept is vital for relating the energy and motion dynamics of the system.

Average Power

Average power involves the rate at which work is performed or energy is transferred in a system. In the context of SHM, calculating the average power needed to accelerate a piston involves understanding the work done to reach a certain kinetic energy and linking it to the time over which this acceleration occurs. This concept is key for quantifying the energy demands of mechanical systems like an engine.

Newton's Second Law

Newton’s Second Law of Motion states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration (F = ma). This principle is used to compute the net force required to accelerate the piston when its mass and the maximum acceleration from SHM are known.

Acceleration in SHM

The acceleration in SHM is given by the equation a = -?²x, where x is the displacement from equilibrium. At the endpoints of the motion, where x equals the amplitude, the acceleration reaches its maximum magnitude. This concept is essential when determining the forces acting at these extreme positions in oscillatory systems such as a car engine piston.

Amplitude and Stroke

Amplitude in SHM represents the maximum displacement of the oscillating object from its equilibrium position. The stroke of an engine, which is twice the amplitude, indicates the full range of motion of the piston and is a critical parameter in determining the extreme positions where acceleration reaches its maximum values.

Angular Frequency

Angular frequency, denoted by ?, is a measure of how rapidly the oscillations occur in SHM and is related to the period or frequency of the motion. It is crucial for converting rotational measures, such as revolutions per minute, into radians per second, which are used in key SHM formulae for acceleration and energy calculations.

Simple Harmonic Motion (SHM)

Simple Harmonic Motion is a type of periodic oscillatory motion where the restoring force is directly proportional to the displacement and acts in the opposite direction. It is characterized by sinusoidal time dependence, which allows for straightforward analysis of velocity, acceleration, and energy within the system.

Transcript

00:01 We know that force can be written as mass into acceleration.

00:06 Also force can be written as minus k of x.

00:09 So equating both these equations, we will get m -a -x equal to minus k -x.

00:17 So from this equation, we can also find out a -max, that is, k -by -m into a, that is, omega -square.

00:33 Now, this is the acceleration when x equal to plus or minus a.

00:42 Okay? so, this is the acceleration when x is equal to plus or minus a.

00:48 So, from this we can find out v max.

00:52 What is the value of v max from this? root of k upon m into a, which is equal to omega a.

01:01 And power equal to work upon time that is equal to delta k upon t so these are the basic equations that we require to solve our questions so in this now we have to see what are the values given a equal to 0 .0 5000 meter this is the amplitude omega is given as 4500 r p m which is also equal to when we convert 471 .24 brad per second so these are the things that we have now we are going to the first part the first part we have to find out in maximum acceleration we write it at a max that we have omega square a from where did we get we have already solid about so that is equal to 471 .24 rad a second the whole square into 0 .5 .0 .0 .0 .05.

02:19 We'll be very careful.

02:21 Okay.

02:22 These small mistakes can make a lot of difference.

02:25 0 .0500 meter which is equal to 1 .11 into 10 raised to 4 .4 .4 .5 .0 meter which is equal to 1 .1 into 10 raised to 4.

02:36 Meter per second square okay so because it is acceleration it will be meter per second square so this is the answer for the first part it gives you the maximum acceleration understood the second part what we have to find the part is maximum force experienced maximum force experience is m a x so put the values that is 0 .455 0 kg into acceleration we have here 1 .11 into 10 raised to 4 meter per second square when we multiply it we are going to get the answer as 5 .00 into 10 raised to 3 newton okay so this is our maximum force remember to write every answer in in the standard form okay so now let's move on to this c part okay let's move on to the c part c part is asking for maximum velocity is v max that is omega a so just put the values now 471 .24 rad a second into 0 .050m .0 meter so the answer for this will be 23 .6 meter per second.

04:14 So here we have got in all the cases, you have got here v max also.

04:19 So because of this v max, you can find out the maximum kinetic energy, that is k max.

04:24 What's a formula for that? half into m into z max square.

04:33 So we are putting the values simply, simply just put the values half into 0 .4.

04:41 450 into 23 .6 we have already found out here.

04:46 So the answer would be 125 jutes.

04:53 So in order to find out the kinetic energy, first we have to find out the maximum velocity.

05:03 Okay, maximum velocity we have found out.

05:06 Now, the last, the d part of it.

05:11 So, d, we have to find out.

05:12 So, d, we have to find out.

05:14 Out of p, right? so that is k max upon t...