What is the maximum speed of the piston and the maximum force acting on the piston?

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Homework Statement:
The motion of a piston in an auto engine is simple harmonic. If the piston travels back and forth over a distance of 10 cm, and the piston has a mass of 1.5 kg, what is the maximum speed of the piston and the maximum force acting on the piston when the engine is running at 4200 rpm?

Ansewer: 22 m/s, 14,500 N
Relevant Equations:
KEi + Ui = KEf + Uf
1/2 m (rw)^2 = 1/2 m( vf)2
rw =V = 0.05* ( 4200*2π /60 )= 22

1/2 mv^2 = 1/2 kx^2
1.5 * 22^2 = k * 0.05^2
k = 2.9 *10^5
F = kx = 2.9 *10^5 * 0.05 = 14,500
--------------------- (For the above calculation, I got the correct answer.)

I wonder whether we can use the method below to obtain force, which also leads me close to the correct answer?
a = rw^2 = 0.05* ( 4200*2π /60 )^2 = 9662.408
F = m* a = 1.5 * 9662.408 = 14,494
 

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  • #2
haruspex
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Homework Statement:: The motion of a piston in an auto engine is simple harmonic. If the piston travels back and forth over a distance of 10 cm, and the piston has a mass of 1.5 kg, what is the maximum speed of the piston and the maximum force acting on the piston when the engine is running at 4200 rpm?

Ansewer: 22 m/s, 14,500 N
Relevant Equations:: KEi + Ui = KEf + Uf

1/2 m (rw)^2 = 1/2 m( vf)2
rw =V = 0.05* ( 4200*2π /60 )= 22

1/2 mv^2 = 1/2 kx^2
1.5 * 22^2 = k * 0.05^2
k = 2.9 *10^5
F = kx = 2.9 *10^5 * 0.05 = 14,500
--------------------- (For the above calculation, I got the correct answer.)

I wonder whether we can use the method below to obtain force, which also leads me close to the correct answer?
a = rw^2 = 0.05* ( 4200*2π /60 )^2 = 9662.408
F = m* a = 1.5 * 9662.408 = 14,494
Yes, it is a valid method if supported by the right argument. But you have not provided any reasoning, so it is not clear whether it was just a lucky guess.

I believe the only reason you got a slightly different answer is that in the first method you rounded the speed to 22m/s. If yiu had kept more digits you would have got more like the second result.

Btw, it is not actually SHM. The connecting rod follows the hypotenuse, which changes the algebra a bit.
 
  • #3
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Yes, it is a valid method if supported by the right argument. But you have not provided any reasoning, so it is not clear whether it was just a lucky guess.

I believe the only reason you got a slightly different answer is that in the first method you rounded the speed to 22m/s. If yiu had kept more digits you would have got more like the second result.

Btw, it is not actually SHM. The connecting rod follows the hypotenuse, which changes the algebra a bit.
Thanks for your prompt response.
I think the question specifies that it is a simple harmonic motion in the beginning. If so, then both ways of calculating the force would be right?
 
  • #4
haruspex
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Thanks for your prompt response.
I think the question specifies that it is a simple harmonic motion in the beginning. If so, then both ways of calculating the force would be right?
Yes, I understand that the question tells you to treat it as SHM. Anyway, The questions asked relate to the extremes of motion, at which points it most closely approximates SHM. I was just pointing out that you should not be led to believing that the motion of a car piston really is SHM.

As I noted, whether your second method is valid depends on your reasoning for arriving at that equation. You have not provided that.
 
  • #5
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Yes, I understand that the question tells you to treat it as SHM. Anyway, The questions asked relate to the extremes of motion, at which points it most closely approximates SHM. I was just pointing out that you should not be led to believing that the motion of a car piston really is SHM.

As I noted, whether your second method is valid depends on your reasoning for arriving at that equation. You have not provided that.
My reasoning for that alternative way for calculating the force is:
Since v= rw and a = rw^2, I used the calculated acceleration to get the force, just as I used the calculated velocity in my original calculation. I know this reasoning is quite weak, but is there a more physic way of explaining why it is reasonable?
 
  • #6
haruspex
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My reasoning for that alternative way for calculating the force is:
Since v= rw and a = rw^2, I used the calculated acceleration to get the force, just as I used the calculated velocity in my original calculation. I know this reasoning is quite weak, but is there a more physic way of explaining why it is reasonable?
Yes, I suspected it was a bit of a shot in the dark.
the speed you calculated was of the piston when it was at the midpoint of its travel. The v you need for that equation is a tangential velocity of something rotating at rate w and radius r. Moreover, the acceleration you are trying to calculate will be at the extreme of the piston's motion, when its velocity is zero.
To understand how to justify your algeb, consider what the other end of the conrod is connected to.
 
  • #7
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Yes, I suspected it was a bit of a shot in the dark.
the speed you calculated was of the piston when it was at the midpoint of its travel. The v you need for that equation is a tangential velocity of something rotating at rate w and radius r. Moreover, the acceleration you are trying to calculate will be at the extreme of the piston's motion, when its velocity is zero.
To understand how to justify your algeb, consider what the other end of the conrod is connected to.
I guess I calculated the maximum of the speed, as I think of the equations for SHM.
x = A *sin(wt)
v = Aw *cos(wt), from this I obtain Vmax. Not sure if I blurred your focus.
 
  • #8
haruspex
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I guess I calculated the maximum of the speed, as I think of the equations for SHM.
x = A *sin(wt)
v = Aw *cos(wt), from this I obtain Vmax. Not sure if I blurred your focus.
I have no objection to the way you found Vmax. The issue is how to justify the equation you used for max acceleration.
To use ##r\omega^2## you need something of radius r rotating at rate ##\omega##. What is it, and how does it produce that acceleration for the piston?
 
  • #9
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I have no objection to the way you found Vmax. The issue is how to justify the equation you used for max acceleration.
To use ##r\omega^2## you need something of radius r rotating at rate ##\omega##. What is it, and how does it produce that acceleration for the piston?
I would just continue to derive: a = - Aw^2 *sin(wt)
Thus, I would then get the maximum force by F= m*a = m (Aw^2)
How about that?
 
  • #10
haruspex
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I would just continue to derive: a = - Aw^2 *sin(wt)
Thus, I would then get the maximum force by F= m*a = m (Aw^2)
How about that?
Yes, that works.
But you had rw^2, not Aw^2, implying some radius.
The answer I was looking for is that the piston is connected by the conrod to a crankshaft. The 10cm travel of the piston arises because the crankshaft turns the confod joint through a circle of radius 5cm. The max velocity is when when the conrod subtends a right angle at the centre of the crankshaft, but the max acceleration is when it is in line with that.
 
  • #11
Lnewqban
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Just a different point of view:
The ##V_{piston}=0## condition at the top and bottom dead centers of the engine does not depend on the angular velocity or radius of the crankshaft.
Being a reciprocating machine, it must be like that as much for fast engines as for slow rotating ones.

The force on the face of the piston determines the magnitude of the velocity of the piston at the middle point of the stroke, as well as the time that it will take reaching that velocity.
Therefore, the max acceleration should be greater for an engine of short stroke, if compared to another engine of long stroke, both being able to sustain the same angular velocity of the crankshaft.

Please, see:
https://en.m.wikipedia.org/wiki/Dead_centre_(engineering)

http://animatedengines.com/twostroke.html
 
  • #12
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Thank you all for your detailed/professional explanations!!
 

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