Question

It is common to see birds of prey rising upward on thermals. The paths they take may be spiral-like. You can model the spiral motion as uniform circular motion combined with a constant upward velocity. Assume a bird completes a circle of radius 6.00 m every 5.00 s and rises vertically at a rate of 3.00 m/s.

Answer 1

Answer:

a) $8.115 m/s$

b) $9.472 m/s^{2}$

c) $21.7\°$

Explanation:

The rest of the question is written below:

Determine (a) the bird’s speed relative to the ground; (b) the bird’s acceleration (magnitude and direction); and (c) the angle between the bird’s velocity vector and the horizontal.

And we have the following data:

$r=6 m$  is the radius of the circular path in the x-axis

$T=5 s$ the period of the circular motion of the bird's path

$V_{y}=3 m/s$ the vertical component of the bird's velocity, which is directed upward and is constant.

Now let's begin with the answers:

a) Bird’s speed relative to the ground

In order to find this speed, we have to calculate the magnitude of the bird's velocity vector:

$V=sqrt{{V_{x}}^{2} + {V_{y}}^{2}$ (1)

We already know the value of $V_{y}$. So, we have to find $V_{x}$.

Since the bird is describing a circular path in the x-axis, $V_{x}$ will be its tangential velocity:

$V_{x}=\omega r=\frac{2 \pi}{T} r$ (2)

Where $\omega=\frac{2 \pi}{T}$ is the birds angular velocity

$V_{x}=\frac{2 \pi}{5 s} 6 m$ (3)

$V_{x}=7.539 m/s$ (4)

Substituting (4) in (1):

$V=sqrt{(7.539 m/s)}^{2} + (3 m/s)^{2}$ (5)

$V=8.1147 m/s \approx 8.115 m/s$ (6) This is the bird's speed relative to the ground

b) Bird’s acceleration (magnitude and direction)

Since the vertical component of the bird's velocity is constant, the vertical component of its acceleration is zero:

$a_{y}=0 m/s^{2}$

However, the bird has radial acceleration $a_{r}=a_{x}$ that results from its rotation on the circular path horizontally:

$a_{x}=\frac{{V_{x}}^{2}}{r}$ (7)

$a_{x}=\frac{(7.539 m/s)^{2}}{6 m}$ (8)

$a_{x}=9.472 m/s^{2}$ (9) This is the magnitude of the bird's acceleration, which is directed to the center of the circular path the bird describes while it is moving upwards in the spiral.

c) Angle between the bird’s velocity vector and the horizontal

In order to find the direction of the bird's velocity vector with the horizontal, we have to find the angle between the horizontal and the vertical component of this velocity:

$\theta=tan^{-1}(\frac{V_{y}}{V_{x}})$ (10)

$\theta=tan^{-1}(\frac{3 m/s}{7.539 m/s})$ (11)

Finally:

$\theta=21.69\° \approx 21.7 \°$ (12)