Angle Between Accel & Velocity: Solving Circular Motion Problems

[negation]

Homework Statement

A skater is gliding along the ice at 2.4 m/s, when she undergoes an acceleration of magnitude 1.1m/s^2 for 3.0s. At the end of that time she is moving at 5.7 m/s.

What must be the angle between the acceleration vector and the initial velocity vector?

The Attempt at a Solution

a = dv/dt

The issue I'm facing in circular motion is drawing a geometrical model. I don't know how to set it up geometrically. I can't progress further without being able to build a model and draw conclusions from there.

 

[collinsmark]

Homework Statement

A skater is gliding along the ice at 2.4 m/s, when she undergoes an acceleration of magnitude 1.1m/s^2 for 3.0s. At the end of that time she is moving at 5.7 m/s.

What must be the angle between the acceleration vector and the initial velocity vector?

The Attempt at a Solution

a = dv/dt

The issue I'm facing in circular motion is drawing a geometrical model. I don't know how to set it up geometrically. I can't progress further without being able to build a model and draw conclusions from there.

The skater is not moving in circular motion. I'm pretty sure you should assume that the acceleration vector has a constant direction.

The acceleration vector can have two components though: one component is parallel to the initial velocity vector and the other component is perpendicular to the initial velocity vector.

 

[negation]

The skater is not moving in circular motion. I'm pretty sure you should assume that the acceleration vector has a constant direction.

The acceleration vector can have two components though: one component is parallel to the initial velocity vector and the other component is perpendicular to the initial velocity vector.

Do you think I could have a leg-up with a geometrical model for me to make sense?

 

[negation]

The skater is not moving in circular motion. I'm pretty sure you should assume that the acceleration vector has a constant direction.

The acceleration vector can have two components though: one component is parallel to the initial velocity vector and the other component is perpendicular to the initial velocity vector.

Could you expound in the component being parallel to the initial velocity? Also, if the motion is non-circular, how could the acceleration be normal to the initial velocity?

 

[Simon Bridge]

There is a way to discover the geometry for yourself.

If the acceleration were at 0deg, what would the final speed have been?
If the acceleration were at 180deg, what would the final speed have been?
If the acceleration were at 90deg, what would the final speed have been?

What sort of geometry did you use to figure the answers?

 

[negation]

There is a way to discover the geometry for yourself.

If the acceleration were at 0deg, what would the final speed have been?
If the acceleration were at 180deg, what would the final speed have been?
If the acceleration were at 90deg, what would the final speed have been?

What sort of geometry did you use to figure the answers?

I haven't figure out any answer.

Let's see if I can proceed this procedurally.

On the assumption the motion is circular about the origin and counter-clockwise(Cartesian):

a(0π), velocity is in the +j direction.
a(π), velocity is in the +j direction
a(π/2), velocity is in the +i direction

 

[Simon Bridge]

OK - you think best in terms of axis:

put the initial velocity along the y axis.

Then $\vec v(t=0) = v_0\hat\jmath$ where v₀=2.4m/s

If the acceleration were at 0deg, then $\vec a = a\hat\jmath$

Where a=1.1m/s/s

- find the final speed after 3 seconds: $|\vec v(t=3)|$.

is that bigger than, smaller than, or equal to the 5.7m/s given in the problem?

 

[Chestermiller]

The motion is not circular. Consider very strongly using cartesian coordinates. The components of the acceleration vector in cartesian coordinates are going to be constant, independent of time. Consider aligning the initial velocity vector with the unit vector in the x-direction, i. The acceleration vector is going to have constant components in the x and y directions. Your job is to determine the these components, subject to the constraint that the overall acceleration vector has a magnitude of 1.1m/sec^2.

 

[negation]

OK - you think best in terms of axis:

put the initial velocity along the y axis.

Then $\vec v(t=0) = v_0\hat\jmath$ where v₀=2.4m/s

If the acceleration were at 0deg, then $\vec a = a\hat\jmath$

Where a=1.1m/s/s

- find the final speed after 3 seconds: $|\vec v(t=3)|$.

is that bigger than, smaller than, or equal to the 5.7m/s given in the problem?

does this make sense?

 

[negation]

The motion is not circular. Consider very strongly using cartesian coordinates. The components of the acceleration vector in cartesian coordinates are going to be constant, independent of time. Consider aligning the initial velocity vector with the unit vector in the x-direction, i. The acceleration vector is going to have constant components in the x and y directions. Your job is to determine the these components, subject to the constraint that the overall acceleration vector has a magnitude of 1.1m/sec^2.

I'm using Cartesian but I'm still unsure as to how the geometrical model should be built

 

[negation]

I'm having a really big problem with this. My conceptual understanding is pretty shaky given I'm doing a self-study before the semester opens.

 

[haruspex]

I'm using Cartesian but I'm still unsure as to how the geometrical model should be built

Take the initial motion as being in the x direction.
Create unknowns to represent the acceleration components in the x and y directions.
You know the magnitude of the acceleration. What equation does that give you?
You know the duration of the acceleration. What will be the velocity changes in the x and y directions?
What are the new velocities in the x and y directions?
You know the final speed. What equation does that give you?

 

[Chestermiller]

I'm using Cartesian but I'm still unsure as to how the geometrical model should be built

Are you saying you don't know what equations to use? What is the relationship between velocity and acceleration (in vector form)? What is the relationship between velocity and acceleration in cartesian component form?

Chet

 

[negation]

Take the initial motion as being in the x direction.
Create unknowns to represent the acceleration components in the x and y directions.
You know the magnitude of the acceleration. What equation does that give you?
You know the duration of the acceleration. What will be the velocity changes in the x and y directions?
What are the new velocities in the x and y directions?
You know the final speed. What equation does that give you?

x^2 + y^2 = |a|?

I'm really clueless.

 

[negation]

Are you saying you don't know what equations to use? What is the relationship between velocity and acceleration (in vector form)? What is the relationship between velocity and acceleration in cartesian component form?

Chet

Acceleration is the derivative of velocity.
I'm saying how should I intepret the question geometrically. My books is pretty much worthless at only 2 pages for circular motion.

 

[haruspex]

x^2 + y^2 = |a|?

I'm really clueless.

To avoid confusion, let's label the x and y components of acceleration a_x, a_y. Your equation is wrong. On the left hand side you have squares of accelerations; on the right hand side you have an acceleration that's not squared. Equations should always have the same sort of thing on each side.
Please try to correct the equation and attempt my next question:

You know the duration of the acceleration. What will be the velocity changes in the x and y directions?

 

[Chestermiller]

Acceleration is the derivative of velocity.
I'm saying how should I intepret the question geometrically. My books is pretty much worthless at only 2 pages for circular motion.

I'm not sure I understand what you mean by interpreting the question geometrically. But, OK, here goes. Initially the skater is moving in a straight line in the x directions. The skater's direction starts changing as soon as she starts her consstant acceleration, oriented at an angle of θ to the x direction. After accelerating for a long time, her velocity and trajectory will eventually be a straight line at the angle θ to the x axis.

You already stated that the derivative of the velocity is equal to the acceleration. All you need to do now is to translate this statement into the language of mathematics with an equation. First write it down in terms of the vectors. Then write it down as two equations, in terms of the x and y components of the vectors. If the acceleration is 1.1 m/sec^2 and it is oriented at an angle θ to the x axis, you can resolve the acceleration into its components in the x and y directions. In terms of θ, what are these components?

Chet

 

[negation]

To avoid confusion, let's label the x and y components of acceleration a_x, a_y. Your equation is wrong. On the left hand side you have squares of accelerations; on the right hand side you have an acceleration that's not squared. Equations should always have the same sort of thing on each side.
Please try to correct the equation and attempt my next question:

What is the mathematical reasoning for having squares of acceleration on the left hand and non-squared acceleration on the right?

 

[negation]

I'm not sure I understand what you mean by interpreting the question numerically. But, OK, here goes. Initially the skater is moving in a straight line in the x directions. The skater's direction starts changing as soon as she starts her consstant acceleration, oriented at an angle of θ to the x direction. After accelerating for a long time, her velocity and trajectory will eventually be a straight line at the angle θ to the x axis.

You already stated that the derivative of the velocity is equal to the acceleration. All you need to do now is to translate this statement into the language of mathematics with an equation. First write it down in terms of the vectors. Then write it down as two equations, in terms of the x and y components of the vectors. If the acceleration is 1.1 m/sec^2 and it is oriented at an angle θ to the x axis, you can resolve the acceleration into its components in the x and y directions. In terms of θ, what are these components?

Chet

X-component: 1.1m/s^2 cos theta
Y-component: 1.1m/s^2 sin theta

 

[Chestermiller]

X-component: 1.1m/s^2 cos theta
Y-component: 1.1m/s^2 sin theta

Excellent. Now write down the equations I asked for in component form, in terms of θ. What we are looking for is:
dv_x/dt=?
dv_y/dt=?

Chet

 

[negation]

Excellent. Now write down the equations I asked for in component form, in terms of θ. What we are looking for is:
dv_x/dt=?
dv_y/dt=?

Chet

v_x = [1.1m/s^2 cosΘ]3s
v_y = [1.1m/s^2 sinΘ]3s

dv_x/dt = 3.3ms^-1 d/dt cos Θ .dΘ/dt = -3.3ms^-1 sinΘ .ω
dv_y/dt = 3.3ms^-1 d/dt cos Θ .dΘ/dt = 3.3ms^-1 cosΘ .ω

 

[Chestermiller]

v_x = [1.1m/s^2 cosΘ]3s
v_y = [1.1m/s^2 sinΘ]3s

dv_x/dt = 3.3ms^-1 d/dt cos Θ .dΘ/dt = -3.3ms^-1 sinΘ .ω
dv_y/dt = 3.3ms^-1 d/dt cos Θ .dΘ/dt = 3.3ms^-1 cosΘ .ω

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:
\frac{dv_x}{dt}=(1.1) \cosθ
\frac{dv_y}{dt}=(1.1) \sinθ

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 v_x and v_y 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 v_x(t) and v_y(t).

Chet

 

[haruspex]

What is the mathematical reasoning for having squares of acceleration on the left hand and non-squared acceleration on the right?

You wrote

x^2 + y^2 = |a|

Since that was in response to my question about x and y components of acceleration, I presume that's what the x and y refer to there. On the left hand side you have x^2 and y^2, so those are each the square of an acceleration. In dimensional terms, L²T^-4 (length squared over time to the fourth). On the right hand side you have just an acceleration, LT^-2. That cannot be right.
This is a really useful test to apply to your equations. If they are inconsistent dimensionally then they must be wrong.

 

[negation]

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:
\frac{dv_x}{dt}=(1.1) \cosθ
\frac{dv_y}{dt}=(1.1) \sinθ

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 v_x and v_y 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 v_x(t) and v_y(t).

Chet

Yes, your equation make sense. However, why doesn't mine? in taking a_x and multiplying by time, I obtain velocity.

The initial velocity is (2.4ms^-1, 0ms^-1)

Yes I do know how to integrate dvx/dt and dvy/dt

 

[Chestermiller]

Yes, your equation make sense. However, why doesn't mine? in taking a_x and multiplying by time, I obtain velocity.

In your equation for v_x, 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 v_x, 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

 

[negation]

In your equation for v_x, 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 v_x, 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.

 

[negation]

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:
\frac{dv_x}{dt}=(1.1) \cosθ
\frac{dv_y}{dt}=(1.1) \sinθ

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 v_x and v_y 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 v_x(t) and v_y(t).

Chet

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?

 

[Simon Bridge]

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 going to leave you guys to it though ... see how it goes.
I can come back to this after if it turns out to be needed.

 

[Chestermiller]

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:
v_x(3)-v_x(0)=\int_0^3{1.1\cosθdt}=1.1\cosθ\int_0^3{dt}=3.3\cosθ
or
v_x(3)=2.4+3.3\cosθ
Similarly,
v_y(3)=3.3\sinθ

 

[negation]

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:
v_x(3)-v_x(0)=\int_0^3{1.1\cosθdt}=1.1\cosθ\int_0^3{dt}=3.3\cosθ
or
v_x(3)=2.4+3.3\cosθ
Similarly,
v_y(3)=3.3\sinθ

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.

 

[negation]

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 going to 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.

 

[negation]

in your equation for v_x, 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 v_x, 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

EDIT: Sub 5.7 into |v|

 

[Chestermiller]

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:

\sqrt{(v_x(3))^2+(v_y(3))^2}=5.7

 

[negation]

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:

\sqrt{(v_x(3))^2+(v_y(3))^2}=5.7

x is undefined so x is either π/2 or 3π/2. But since (i,j), therefore, x = π/2

 

[Chestermiller]

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.

 

[negation]

Your intuition was correct. 2.4 + 3.3 = 5.7
The angle is zero.

check post#34, something could be wrong

 

[Chestermiller]

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

 

[negation]

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.

 

[negation]

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?

 

[Chestermiller]

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.

 

[negation]

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:
v_x(3)-v_x(0)=\int_0^3{1.1\cosθdt}=1.1\cosθ\int_0^3{dt}=3.3\cosθ
or
v_x(3)=2.4+3.3\cosθ
Similarly,
v_y(3)=3.3\sinθ

Strange.
If one integrates 1.1cos(x) shouldn't one obtain 1.1sin(x)? The antiderivative of cos is sin.

 

[haruspex]

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)?