Ice skater gliding

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  • #26
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In your equation for vx, you left out the initial velocity of the skater. Otherwise, your first two equations are correct. Your second two equations (taking derivatives of the velocities) were not done correctly, and were not even necessary for solving this problem. Once you make the correction to the equation for vx, take the sum of the squares of the x and y components of velocity and set them equal to the magnitude of the final velocity you are seeking. This will give you an equation for solving for the angle θ (or sinθ or cosθ).

Chet
You're right I ought to have add the initial x-velocity to obtain the final x-velocity. Then determine the derivative (although as put it was irrelevant)

Let me work on your question.
 
  • #27
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these equations are not correct, but it won't take much to correct them.

The kinematic equations i was asking for can be written as:
[tex]\frac{dv_x}{dt}=(1.1) \cosθ[/tex]
[tex]\frac{dv_y}{dt}=(1.1) \sinθ[/tex]

the right hand sides of these equations are constant. They are the components of the acceleration. The angle θ is constant, and doesn't depend on time. Does this make sense so far?

From your problem statement, what are the initial values of vx and vy at time t = 0? Do you know how to integrate the above differential equations from time t = 0 to arbitrary time t, subject to these initial conditions? If yes, please do so and show us your results for vx(t) and vy(t).

Chet

Capture.JPG


I'm starting to see the big picture of this question.
If the above is true, I suppose the next step would be to set the antiderivative of vx = 2.4ms^-1?
and if the arbitrary time, t, = 3. Then vx(3) - vx(0) = 2.4 ms^-1, am I right?
 
  • #28
Simon Bridge
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Aside:

What must be the angle between the acceleration vector and the initial velocity vector?
It occurs to me that there is a judgement call here ... the question does not say that the acceleration maintains a constant angle to anything - though we cannot do it unless it maintains a constant angle to something.

So we take the acceleration vector as a constant magnitude and direction.
The fact the question says "angle to the initial velocity" suggests that the angle to the velocity will change as the velocity changes.

Basically this says that there is a constant force on the object.

We could treat it as an average acceleration - so we only need the difference between the final and initial velocities to come to the acceleration (times change in time). Could be - negation has had recent questions about average acceleration.

If the coursework has been around the ideas of circular motion - then we may infer that the acceleration maintains a constant angle to the velocity. But then why not say so hmmm?

negation has had a number of questions like this: similarly loosely worded.

I'm gonna leave you guys to it though ... see how it goes.
I can come back to this after if it turns out to be needed.
 
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  • #29
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View attachment 65622

I'm starting to see the big picture of this question.
If the above is true, I suppose the next step would be to set the antiderivative of vx = 2.4ms^-1?
and if the arbitrary time, t, = 3. Then vx(3) - vx(0) = 2.4 ms^-1, am I right?
The above is not done correctly. Sorry. θ is not a function of t, so sinθ and cosθ factor out of the integral.

If you integrate those equations, it goes like this:
[tex]v_x(3)-v_x(0)=\int_0^3{1.1\cosθdt}=1.1\cosθ\int_0^3{dt}=3.3\cosθ[/tex]
or
[tex]v_x(3)=2.4+3.3\cosθ[/tex]
Similarly,
[tex]v_y(3)=3.3\sinθ[/tex]
 
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  • #30
818
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The above is not done correctly. Sorry. θ is not a function of t, so sinθ and cosθ factor out of the integral.

If you integrate those equations, it goes like this:
[tex]v_x(3)-v_x(0)=\int_0^3{1.1\cosθdt}=1.1\cosθ\int_0^3{dt}=3.3\cosθ[/tex]
or
[tex]v_x(3)=2.4+3.3\cosθ[/tex]
Similarly,
[tex]v_y(3)=3.3\sinθ[/tex]
Sorry for that slip.
I'm attempting the method where I square vx and vy. I'll finish that first and once I get the answer, I'll do it via integral.
I get an intuition the angle is zero but we shall see.
 
  • #31
818
0
Aside:


It occurs to me that there is a judgement call here ... the question does not say that the acceleration maintains a constant angle to anything - though we cannot do it unless it maintains a constant angle to something.

So we take the acceleration vector as a constant magnitude and direction.
The fact the question says "angle to the initial velocity" suggests that the angle to the velocity will change as the velocity changes.

Basically this says that there is a constant force on the object.

We could treat it as an average acceleration - so we only need the difference between the final and initial velocities to come to the acceleration (times change in time). Could be - negation has had recent questions about average acceleration.

If the coursework has been around the ideas of circular motion - then we may infer that the acceleration maintains a constant angle to the velocity. But then why not say so hmmm?

negation has had a number of questions like this: similarly loosely worded.

I'm gonna leave you guys to it though ... see how it goes.
I can come back to this after if it turns out to be needed.
I copied the question word for word from the book. There are many times I have no idea what is expected.
 
  • #32
818
0
in your equation for vx, you left out the initial velocity of the skater. Otherwise, your first two equations are correct. Your second two equations (taking derivatives of the velocities) were not done correctly, and were not even necessary for solving this problem. Once you make the correction to the equation for vx, take the sum of the squares of the x and y components of velocity and set them equal to the magnitude of the final velocity you are seeking. This will give you an equation for solving for the angle θ (or sinθ or cosθ).

Chet


Capture.JPG


EDIT: Sub 5.7 into |v|
 
  • #33
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Sorry for that slip.
I'm attempting the method where I square vx and vy. I'll finish that first and once I get the answer, I'll do it via integral.
I get an intuition the angle is zero but we shall see.
There is no need for doing any more integration. You have your two components of the velocity. All you need to do now is solve for θ under the condition that:

[tex]\sqrt{(v_x(3))^2+(v_y(3))^2}=5.7[/tex]
 
  • #34
818
0
There is no need for doing any more integration. You have your two components of the velocity. All you need to do now is solve for θ under the condition that:

[tex]\sqrt{(v_x(3))^2+(v_y(3))^2}=5.7[/tex]

x is undefined so x is either π/2 or 3π/2. But since (i,j), therefore, x = π/2
 
  • #35
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Sorry for that slip.
I'm attempting the method where I square vx and vy. I'll finish that first and once I get the answer, I'll do it via integral.
I get an intuition the angle is zero but we shall see.
Your intuition was correct. 2.4 + 3.3 = 5.7
The angle is zero.
 
  • #36
818
0
Your intuition was correct. 2.4 + 3.3 = 5.7
The angle is zero.
check post#34, something could be wrong
 
  • #37
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check post#34, something could be wrong
No way. It's definitely zero.

(2.4 + 3.3 cosθ)2+(3.3sinθ)2=5.72

(5.76+ 10.89cos2θ+15.84cosθ)+10.89sin2θ)=32.49

5.76+10.89(cos2θ+sin2θ)+15.84cosθ=32.49

5.76+10.89 +15.84 cosθ = 32.49

15.84 cosθ = 15.84

cosθ = 1

θ = 0
 
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  • #38
818
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No way. It's definitely zero.

(2.4 + 3.3 cosθ)2+(3.3sinθ)2=5.72

(5.76+ 10.89cos2θ+15.84cosθ)+10.89sin2θ)=32.49

5.76+10.89(cos2θ+sin2θ)+15.84cosθ=32.49

5.76+10.89 +15.84 cosθ = 32.49

15.84 cosθ = 15.84

cosθ = 1

θ = 0
I had an arithmetic error.
Θ = 0 is correct.
Thanks and I shall now review the question again and try to sketch a mental model so as to better cement my understanding.
 
  • #39
818
0
No way. It's definitely zero.

(2.4 + 3.3 cosθ)2+(3.3sinθ)2=5.72

(5.76+ 10.89cos2θ+15.84cosθ)+10.89sin2θ)=32.49

5.76+10.89(cos2θ+sin2θ)+15.84cosθ=32.49

5.76+10.89 +15.84 cosθ = 32.49

15.84 cosθ = 15.84

cosθ = 1

θ = 0

Would I be right in sketching
On the x-axis Vx=3.3cos(theta) +2.4
And
On the y-axis Vy=3.3sin(theta)
With the hypothenus being 5.7?
 
  • #40
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Would I be right in sketching
On the x-axis Vx=3.3cos(theta) +2.4
And
On the y-axis Vy=3.3sin(theta)
With the hypothenus being 5.7?
Yes, as long as it's considered schematic. But, in the actual case with θ=0, the triangle is going to be degenerate.
 
  • #41
818
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The above is not done correctly. Sorry. θ is not a function of t, so sinθ and cosθ factor out of the integral.

If you integrate those equations, it goes like this:
[tex]v_x(3)-v_x(0)=\int_0^3{1.1\cosθdt}=1.1\cosθ\int_0^3{dt}=3.3\cosθ[/tex]
or
[tex]v_x(3)=2.4+3.3\cosθ[/tex]
Similarly,
[tex]v_y(3)=3.3\sinθ[/tex]

Strange.
If one integrates 1.1cos(x) shouldn't one obtain 1.1sin(x)? The antiderivative of cos is sin.
 
  • #42
haruspex
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Strange.
If one integrates 1.1cos(x) shouldn't one obtain 1.1sin(x)?
Only if you integrate with respect to x. If you integrate with respect to something unrelated to x, x staying constant, then cos(x) is a constant. Why should it change to sin(x)?
 
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