Question

Orangutans can move by brachiation, swinging like a pendulum beneath successive handholds. If an orangutan has arms that are 0.90 m long and repeatedly swings to a 20° angle, taking one swing immediately after another, estimate how fast it is moving in m/s.

Answer 1

Answer:

 v = 1.03 m / s

Explanation:

For this case we must use energy conservation. Let's define an energy system at the lowest point of the movement.

Initial energy lowest point

     Em₀ = K = ½ m v²

Final energy highest point

   $Em_{f}$ = U = m g y

Energy is conserved

    Em₀ =   $Em_{f}$

    ½ m v² = m g y

    v = √ (2g y)

Let's use trigonometry to find the height

    cos 20 = L’/ L

    L’= L cos 20

The height of the displacement is the total length minus L ’

    y = L - L’

    y = L - L cos 20

    y = L (1- cos 20)

Let's replace

    v = √ (2g L (1-cos 20))

Let's calculate

   v =√(2 9.8 0.90 (1- cos 20))

   v = 1.03 m / s