Q46P - University Physics with Modern Physics

Question

Birds of prey typically rise upward on thermals. The paths these birds take may be spiral-like. You can model the spiral motion as uniform circular motion combined with a constant upward velocity. Assume that a bird completes a circle of radius 6.00 m every 5.00 s at a constant rate of 3.00 m/s . 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.

Step-by-Step Solution

Answer

a) The bird’s speed relative to the ground is 8.11 m/s

b) the bird’s acceleration is $9.48m/s^2$ and the direction of the acceleration will be towards the center of the circle.

c) The angle between the bird’s velocity vector and the horizontal is $21.7°$.

Step 1: Introduction

The velocity contains two components, one is the horizontal component and another one is vertical. 

 

The vertical component of the acceleration will be zero and the horizontal component will be equal to the square of velocity divided by distance.

 

$a_{rad}=\frac{v_x^2}{r}$

Step 2: Given data

• Radius, $r=6.00 m$ .

 

• Time, $t=5.00 s$ .

 

• Velocity in the circle, $v_y=3.00 m/s$ .

 

Step 3: (a) The bird’s speed relative to the ground

The tangential velocity of the bird will be,

$$\begin{gathered} v_x=\frac{circumference}{time} \\ {v}_x=\frac{2\mathrm{\pi }\times 6.00 \mathrm{m}}{5.00 s} \\ {v}_x=7.54 m/s \end{gathered}$$ 

Another velocity component is $v_y=3.00m/s$ . Now the speed of the bird relative to the ground will be,

$$\begin{gathered} v=\sqrt{v_x^2+v_y^2} \\ v=\sqrt{{\left(7.54 m/s\right)}^2+{\left(3.00m/s\right)}^2} \\ v=8.11 m/s \end{gathered}$$                                                                                         

Thus, the bird’s speed relative to the ground is 8.11 m/s

Step 4: (b) The bird’s acceleration

The magnitude of the acceleration,

 $$\begin{gathered} a_{rad}=\frac{v_x^2}{r} \\ {a}_{rad}=\frac{{\left(7.54m/s\right)}^2}{6.00 m} \\ {a}_{rad}=9.48 m/s^2 \end{gathered}$$

The speed of the bird is constant at every place so the direction of the acceleration will be towards the center of the circle, and the magnitude of the acceleration is $9.48 m/s^2$ .

Step 5: (c) The angle between the bird’s velocity vector and the horizontal

The angle between the bird’s velocity vector and the horizontal,

 $$\begin{gathered} \tan \theta =\frac{vertical velocity component}{horizontal velocity component} \\ \tan \theta =\frac{3.00 m/s}{7.54 m/s} \\ \theta =21.7° \end{gathered}$$                                                                                      

The angle between the bird’s velocity vector and the horizontal is $21.7°$ .