Is the Magnitude of the Fly's Acceleration Constant in a Helical Path?

• Ravenatic20
In summary: \mathbf{a}(t)=[-b\omega^2\sin(\omega t)\mathbf{i}-b\omega^2\cos(\omega t)\mathbf{j}+2c\mathbf{k}*\mathbf{b}cos(\omega t)]^3, where \mathbf{b}cos(\omega t) is x.
Ravenatic20
I hope I posted in the right place. Sorry in advanced.

Homework Statement

A buzzing fly moves in a helical path given by the equation
r(t) = ib sin $$\omega$$t + jb cos $$\omega$$t + kct$$^{2}$$
Show that the magnitude of the acceleration of the fly is constant, provided b, $$\omega$$, and c are constant.

The Attempt at a Solution

x = b sin $$\omega$$t
y = b cos $$\omega$$t
z = ct$$^{2}$$

In class we did a similar problem, but in that problem we had to find the trajectory in space. I'm a little slow, but it's just not helping me with this one. Same with the textbook. I'll go to my teacher if I have to.

I'm not asking someone to do the problem, just get me started. Once that happens, I'll try to go over it here in case I have more questions. Thank you!

The acceleration is the second derivative of r(t) with respect to t. What is that?

Dick said:
The acceleration is the second derivative of r(t) with respect to t. What is that?

The first derivative:
ib$$\omega$$ cos $$\omega$$t - jb$$\omega$$ sin $$\omega$$t + 2kct

Second derivative:
-ib$$\omega$$$$^{2}$$ sin $$\omega$$t - jb$$\omega$$$$^{2}$$ cos $$\omega$$t + 2kc

Is that right?

Now find the magnitude...

Ravenatic20 said:
Second derivative:
-ib$$\omega$$$$^{2}$$ sin $$\omega$$t - jb$$\omega$$$$^{2}$$ cos $$\omega$$t + 2kc

Is that right?

Sure; that;s correct ...But very ugly! Try writing the entire equation inside the [ tex] or [ itex] tags instead:

$$\mathbf{a}(t)=-b\omega^2\sin(\omega t)\mathbf{i}-b\omega^2\cos(\omega t)\mathbf{j}+2c\mathbf{k}$$

(You can click on the above equation to see the code that generated it)

Now, as Nabeshin said, calculate the magnitude

Thank you guys :)

Now what is the first step in calculating the magnitude? I am used to plugging in numbers to do that.

You know what the x,y, and z-components of a are, so square them, add the squares, and take the square root as per usual.

$$||\mathbf{a}||=\sqrt{a_x^2+a_y^2+a_y^2}$$

So I take $$\mathbf{a}$$ (the second derivative above), and factor in x for the first part. So it would look like this:

$$\mathbf{a_x}(t)=[-b\omega^2\sin(\omega t)\mathbf{i}-b\omega^2\cos(\omega t)\mathbf{j}+2c\mathbf{k}*\mathbf{b}sin(\omega t)]^2$$, where $$\mathbf{b}sin(\omega t)$$ is x.

Then do the same thing for y and z, add the terms up, and take the square root. Am I on the right track? Then I simplify as much as possible?

Ravenatic20 said:
So I take $$\mathbf{a}$$ (the second derivative above), and factor in x for the first part. So it would look like this:

$$\mathbf{a_x}(t)=[-b\omega^2\sin(\omega t)\mathbf{i}-b\omega^2\cos(\omega t)\mathbf{j}+2c\mathbf{k}*\mathbf{b}sin(\omega t)]^2$$, where $$\mathbf{b}sin(\omega t)$$ is x.

Then do the same thing for y and z, add the terms up, and take the square root. Am I on the right track? Then I simplify as much as possible?

Huh?!

No! $a_x$ is the x-component of a...that's just $-b\omega^2\sin(\omega t)$...what are
$a_y$ and $a_z$?

gabbagabbahey said:
Huh?!

No! $a_x$ is the x-component of a...that's just $-b\omega^2\sin(\omega t)$...what are
$a_y$ and $a_z$?
Thank you. I always over complicate things. I believe I know how to do it now.

To answer your question, y is $$-b\omega^2\cos(\omega t)$$ and z is $$2c$$
Now I square them, add them up, and take the square root of that sum (this:$$||\mathbf{a}||=\sqrt{a_x^2+a_y^2+a_y^2}$$). Correct?

Correct. So what is the result?

What is acceleration?

Acceleration is the rate at which an object's velocity changes over time. It is a vector quantity, meaning it has both magnitude and direction.

How is acceleration measured?

Acceleration is typically measured in units of meters per second squared (m/s²) or feet per second squared (ft/s²). This represents the change in velocity (meters or feet per second) over the change in time (seconds).

What does it mean for acceleration to be constant?

If an object's acceleration is constant, it means that its velocity is changing at a constant rate. This can either be a constant increase or decrease in velocity, or a constant velocity in a certain direction.

How is constant acceleration shown in an experiment?

In an experiment, constant acceleration can be shown by measuring the object's velocity at equal time intervals and observing that it changes at a consistent rate. This can be done using tools such as a stopwatch, ruler, and motion sensor.

What are some real-world examples of constant acceleration?

Some examples of constant acceleration in the real world include a car accelerating at a steady rate, a pendulum swinging back and forth, and a falling object in a vacuum. These all exhibit a constant change in velocity over time.

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