Kinematics motion - Given position, find velocity and acceleration

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Homework Help Overview

The discussion revolves around kinematics, specifically analyzing the motion of a particle whose position is defined by the equation xf = xi . sin(wt), where xi and w are constants. Participants are tasked with finding expressions for velocity and acceleration based on this position function.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the differentiation of the position function to derive velocity and acceleration, with some questioning the clarity of notation and the simplification of expressions. There are inquiries about the implications of constants in differentiation and the relevance of jerk in the context of maximum acceleration.

Discussion Status

The discussion is ongoing, with participants providing feedback on each other's attempts and suggesting simplifications. There is a mix of agreement and confusion regarding the steps taken, particularly in relation to the treatment of constants and the use of derivatives to find maximum values.

Contextual Notes

Some participants express uncertainty about the notation used and the process of utilizing LaTeX for clearer communication. There is also a focus on ensuring that the correct derivatives are set to zero to find maximum values, with some participants reflecting on their previous reasoning.

negation
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Kinematics motion -- Given position, find velocity and acceleration

Homework Statement



A particle's position as a function of time is given by xf = xi . sin wt, where xi and w are constants.

a) Find expressions for velocity and acceleration.
b) What are the maximum values of velocity and acceleration.

Homework Equations


xf = xi. sin(wt)

The Attempt at a Solution



a) xf = xi. sin(wt)

velocity v = xf' = d/dt xi sin(wt) + xi d/dt sin(w.t) . d/dt w.t

xf' = xi cos(wt).w = xi.w.cos(wt)

acceleration = v' or xf " = [d/dtxi.w + xi d/dt w]cos(w.t) + xi.w d/dt (cos(w.t) . d/dt (w.t)

xf " = x dw/dt cos (wt) - xw^2 sin(w.t)

Can someone verify the above steps for me before I proceed to part (b)?
 
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negation said:
xf " = x. dw/dt cos (wt) - x.w^2 sin (w.t)
Looks OK, but use the fact that ω is constant to simplify that expression. (Your notation is difficult to read. Consider using Latex in the future.)
 
Doc Al said:
Looks OK, but use the fact that ω is constant to simplify that expression. (Your notation is difficult to read. Consider using Latex in the future.)
Is it possible to simplify further?

On a tangent, how do I utilize Latex? Matlab?
 
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negation said:
On a tangent, how do I utilize Latex? Matlab?
On the advanced editor of physicsforums,you can use latex by pressing the top right button.
 
adjacent said:
On the advanced editor of physicsforums,you can use latex by pressing the top right button.

I just realized. But the nitty gritty details would have to wait.
 
negation said:
Is it possible to simplify further?
Yes.
 
doc al said:
yes.

https://www.physicsforums.com/attachments/64882
 
dω/dt = ?
 
doc al said:
dω/dt = ?

Capture.JPG


Why do I have to express the equation as dω/dt?
 
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  • #10
negation said:
Why do I have to express the equation as dω/dt?
Huh? Your expression for the acceleration contains dω/dt.

negation said:
A particle's position as a function of time is given by xf = xi . sin wt, where xi and w are constants.
 
  • #11
Capture.JPG
 
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  • #12
negation said:
View attachment 64885
That is incorrect. Your first post was correct, just not completed.
 
  • #13
Doc Al said:
That is incorrect. Your first post was correct, just not completed.

I believe I'm stuck. Even if ω were a constant, hasn't the equation been differentiated to it's second derivative? I could factor out some variables in the second derivative but I don't think that's what you were going at.
 
  • #14
negation said:
I believe I'm stuck. Even if ω were a constant, hasn't the equation been differentiated to it's second derivative? I could factor out some variables in the second derivative but I don't think that's what you were going at.
What do you mean "even if"? You are told that ω is a constant.

Answer this. What's the derivative of a constant?
 
  • #15
Another hint: The point I am making is so trivial you will slap yourself once you realize it.
 
  • #16
  • #17
Doc Al said:
What do you mean "even if"? You are told that ω is a constant.

Answer this. What's the derivative of a constant?

it's zero

y" = [d/dt(xw) + (x)d/dt(w)]cos(wt) + (xw) d/dt cos(wt) d/dt(wt)
=
dw/dt cos(wt) - (xw)sin(wt)w
=
dw/dt cos(wt) - xw^2 sin(wt)

Sent from iPhone
 
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  • #18
Doc Al said:
Another hint: The point I am making is so trivial you will slap yourself once you realize it.

dw/dt cos (wt) - xw sin(wt)w
dw/dt =0
So,
y" = -xw^2 sin(wt)
Is the workings above correctly reduced?
 
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  • #19
I would suggest you that while calculating derivative of an function if you see any constant in that equation ,you can substitute 0 directly which makes your equation simple.
In your equations
substitute 0 in place of dw/dt of your acceleration equation .
 
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  • #20
negation said:
it's zero
Of course!

y" = [d/dt(xw) + (x)d/dt(w)]cos(wt) + (xw) d/dt cos(wt) d/dt(wt)
=
dw/dt cos(wt) - (xw)sin(wt)w
=
dw/dt cos(wt) - xw^2 sin(wt)
Keep going.

negation said:
dw/dt cos (wt) - xw sin(wt)w
dw/dt =0
So,
y" = -xw^2 sin(wt)
Is the workings above correctly reduced?
Exactly.
 
  • #21
Doc Al said:
Of course!Keep going.Exactly.

Part (b)

In find maximum velocity:

y' = xw cos (wt)
xw cos (wt) = 0
Cos(wt) = 0
Cos(pi/2) = 0 A cos(3pi/2) = 0
wt = pi/2
t = pi/2w

wt = 3pi/2
t = 3pi/2w

y"(pi/2w)= -xw^2 sin(wt)
= -xw^2 sin(pi/2)
= xw^2 < 0 local max

y"(3pi/2w) = -xw^2
= xw^2 > 0 local min

Max velocity occurs at t = pi/2wIn finding maximum acceleration:

To determine maximum acceleration, we first determine the rate of change of jerk.

Jerk is defined as the 3rd derivative of displacement with respect to time, t.

y'' = -xw^2 sin(wt)

y"' = [d/dt(-xw^2) + (-x)d/dt(w^2)] sin(wt) + (-xw^2) d/dt sin(wt) d/dt(wt)

= [0 + 0 ]sin(wt) - (xw^2) cos(wt)w

= -(xw^3)cos(wt)

y" = 0 = -xw^2 sin(wt)

-xw^2 sin(wt)= 0

sin(wt) = 0

sin(0) = 0 and sin(pi)= 0

wt = 0 and wt = pi

t = 0 and t = pi/w

y"'(0) = -xw^3 cos(w.0)
= -xw^3(1)
= -xw^3y"'(pi/w) = -xw^3 cos(w.pi/w)
= -xw^3 cos(pi)
= -xw^3(-1)
= xw^3

y"'(0) = -xw^3 < 0
local max

therefore, maximum acceleration occurs at t = 0
 
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  • #22
negation said:
Part (b)

In find maximum velocity:

y' = xw cos (wt)
xw cos (wt) = 0
Why are you setting that to zero?

Cos(wt) = 0
Cos(pi/2) = 0 A cos(3pi/2) = 0
wt = pi/2
t = pi/2w

wt = 3pi/2
t = 3pi/2w

y"(pi/2w)= -xw^2 sin(wt)
= -xw^2 sin(pi/2)
= xw^2 < 0 local max

y"(3pi/2w) = -xw^2
= xw^2 > 0 local min

Max velocity occurs at t = pi/2w
No.

In finding maximum acceleration:

To determine maximum acceleration, we first determine the rate of change of jerk.

Jerk is defined as the 3rd derivative of displacement with respect to time, t.
OK.

y'' = -xw^2 sin(wt)

y"' = [d/dt(-xw^2) + (-x)d/dt(w^2)] sin(wt) + (-xw^2) d/dt sin(wt) d/dt(wt)

= [0 + 0 ]sin(wt) - (xw^2) cos(wt)w

= -(xw^3)cos(wt)
OK.

y" = 0 = -xw^2 sin(wt)
Why are you setting the acceleration = zero? You calculated the jerk, so use it!

-xw^2 sin(wt)= 0

sin(wt) = 0

sin(0) = 0 and sin(pi)= 0

wt = 0 and wt = pi

t = 0 and t = pi/w

y"'(0) = -xw^3 cos(w.0)
= -xw^3(1)
= -xw^3


y"'(pi/w) = -xw^3 cos(w.pi/w)
= -xw^3 cos(pi)
= -xw^3(-1)
= xw^3

y"'(0) = -xw^3 < 0
local max

therefore, maximum acceleration occurs at t = 0
No.

Try once more.
 
  • #23
Doc Al said:
Why are you setting that to zero?

What was I thinking. I was rushing to type these.

Will retry
 
  • #24
Doc Al said:
Why are you setting the acceleration = zero? You calculated the jerk, so use it!

But in finding maximum velocity, shouldn't I find the derivative of velocity, which gives me acceleration, then set it to zero to determine the roots(max velocity implies y"(root)<0 )?
or are you implying that because I have derived the function for jerk, I should work on jerk instead?
 
  • #25
Doc Al said:
Why are you setting the acceleration = zero? You calculated the jerk, so use it!

In finding maximum velocity:

y''= 0 = -xw^2 sin (wt)
sin(wt) = 0 if sin(0) or sin(pi)

wt = 0
t = 0
and
wt = pi
t = pi/w

roots are t = 0 and t = pi/w

y'(0) = xw cos(wt) = xw cos(0) = xw
y''(pi/w) = xw cos(w. pi/w) = xw cos(pi) = -xw

max velocity occurs at t = 0

In finding max acceleration:

y''' = -xw^3 cos(wt) = 0
cos(wt) = 0 if cos(pi/2) and cos(3pi/2)

wt = pi/2
t = pi/2w
and
wt = 3pi/2
t = 3pi/2w

y''(pi/2w) = -xw^2 sin (w . pi/2w)
= -xw^2 sin (pi/2)
= -xw^2

y''(3pi/2w) = -xw^2 sin(w. 3pi/2w)
= -xw^2 sin(3pi/2)
xw^2

max acceleration occurs at t = 3pi/2w
 
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  • #26
negation said:
But in finding maximum velocity, shouldn't I find the derivative of velocity, which gives me acceleration, then set it to zero to determine the roots(max velocity implies y"(root)<0 )?
That's what you should have done, but not what you did. You set the velocity equal to zero, not the acceleration.

or are you implying that because I have derived the function for jerk, I should work on jerk instead?
Jerk can be used when calculating the maximum acceleration.
 
  • #27
Doc Al said:
That's what you should have done, but not what you did. You set the velocity equal to zero, not the acceleration.


Jerk can be used when calculating the maximum acceleration.

Is my workings now correct?
 
  • #28
negation said:
In finding maximum velocity:

y''= 0 = -xw^2 sin (wt)
sin(wt) = 0 if sin(0) or sin(pi)

wt = 0
t = 0
and
wt = pi
t = pi/w

roots are t = 0 and t = pi/w

y'(0) = xw cos(wt) = xw cos(0) = xw
y''(pi/w) = xw cos(w. pi/w) = xw cos(pi) = -xw

max velocity occurs at t = 0
Good. Max velocity is ±xω. That occurs when ωt = 0, π, ...


In finding max acceleration:

y''' = -xw^3 cos(wt) = 0
cos(wt) = 0 if cos(pi/2) and cos(3pi/2)

wt = pi/2
t = pi/2w
and
wt = 3pi/2
t = 3pi/2w

y''(pi/2w) = -xw^2 sin (w . pi/2w)
= -xw^2 sin (pi/2)
= -xw^2

y''(3pi/2w) = -xw^2 sin(w. 3pi/2w)
= -xw^2 sin(3pi/2)
xw^2

max acceleration occurs at t = 3pi/2w
Good. Max acceleration is ±xω^2. That occurs when ωt = π/2, 3π/2, ...

negation said:
Is my workings now correct?
Yes, much better! :thumbs:
 

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