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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 6.00m every 5.00s and rises vertically at a rate of 3.00m/s.

1. Find the speed of the bird relative to the ground.
2. Find the magnitude of the bird's acceleration.
3. Find the direction of the bird's acceleration.
4. Find the angle between the bird's velocity vector and the horizontal.

Respuesta :

skyluke89

1. 8.11 m/s

In order to find the speed of the bird, we need to combine both the horizontal and the vertical component of its velocity.

The horizontal component is equal to the tangential velocity of the circular motion of the bird. We have:

r = 6.00 m radius of the circle

T = 5.00 s period of the circular motion

So the tangential velocity (=horizontal component of the velocity) is

[tex]v_x=\frac{2\pi r}{T}=\frac{2\pi (6.00 m)}{5.00 s}=7.54 m/s[/tex]

The vertical component of the velocity is given by the problem

[tex]v_y = 3.00 m/s[/tex]

So the speed of the bird is the magnitude of its velocity:

[tex]v=\sqrt{v_x^2 + v_y^2}=\sqrt{(7.54 m/s)^2+(3.00 m/s)^2}=8.11 m/s[/tex]

2. 9.48 m/s^2

The only component of the acceleration of the bird is the one that determine the circular motion in the horizontal plane - so it is the centripetal acceleration.

The centripetal acceleration of the bird is given by:

[tex]a=\frac{v_x^2}{r}[/tex]

where

[tex]v_x = 7.54 m/s[/tex] is the tangential velocity of the bird

r = 6.00 m is the radius of the circular trajectory

Substituting,

[tex]a=\frac{(7.54 m/s)^2}{6.00 m}=9.48 m/s^2[/tex]

3. Towards the centre of the circle

The bird is moving upward at constant velocity and in a straight line, so there is no component of the acceleration in the vertical direction.

In the horizontal plane, however, the bird is moving in a uniform circular motion: this means that the tangential acceleration is zero, while there is a centripetal acceleration (calculated in the previous part of the exercise). The direction of the centripetal acceleration is always towards the centre of the circular trajectory: so, the acceleration of the bird has exactly the same direction as the centripetal acceleration, i.e. towards the centre of the circle.

4. [tex]21.7^{\circ}[/tex]

We said that the components of the velocity of the bird are:

[tex]v_x = 7.54 m/s[/tex] in the horizontal direction

[tex]v_y = 3.00 m/s[/tex] in the vertical direction

So we can find the angle of the velocity with the horizontal by using the equation:

[tex]\theta = tan^{-1} (\frac{v_y}{v_x})[/tex]

Substituting the values in, we find

[tex]\theta = tan^{-1} (\frac{3.00 m/s}{7.54 m/s})=21.7^{\circ}[/tex]

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