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

To estimate the speed of an orangutan moving by brachiation, we can calculate the angular velocity and convert it to linear speed. Assuming the orangutan swings back and forth in a 20° angle with a time period of 2 seconds, the linear speed is approximately 0.349 m/s.

Step-by-step explanation:

To estimate the speed at which an orangutan is moving while brachiating (swinging) from one branch to another, we can use the concept of angular velocity. Angular velocity is the rate at which an object rotates or moves in a circular path. In this case, the orangutan swings back and forth, creating a circular motion with a small amplitude of 20 degrees.

To find the angular velocity in radians per second (rad/s), we need to convert the swing angle from degrees to radians. The conversion formula is: 1 radian = 180/pi degrees. So, 20 degrees is equal to (20 * pi)/180 radians.

Next, we need to find the time it takes for the orangutan to complete one swing. Let's assume it takes 1 second for the orangutan to swing from one handhold to the next. Therefore, the time period (T) for one complete swing is 2 seconds (since the orangutan swings back and forth).

Now, we can calculate the angular velocity:

Angular velocity (ω) = Angle (in radians) / Time period (T)

ω = ((20 * pi)/180) / 2 rad/s

Simplifying the equation gives:

ω = (pi/9) rad/s

Finally, to find the linear speed of the orangutan, we multiply the angular velocity by the distance between handholds. Let's assume the distance between handholds is 1 meter. Therefore, the linear speed (v) of the orangutan is:

Linear speed (v) = Angular velocity (ω) x Distance between handholds (d)

v = (pi/9) rad/s x 1 m

v ≈ 0.349 m/s

Answer 2

how fast it is moving in m/s? 0.65 m/s.

Step-by-step explanation:

given information:

the length of the arm, L - 0.9 m

angle, θ = 20°

First we calculate the distance in horizontal motion

s = 2 L sin θ

= 2 (0.9) sin 20°

= 0.62 m

now calculate the time

t/2 = 2π√(L/g)

t = π√(0.9/9.8)

= 0.95 s

the speed is

v = s/t

= 0.62/0.95

= 0.65 m/s