Sallie Banks

2022-01-07

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 ${20}^{\circ }$ angle, taking one swing after another, estimate its speed of forward motion in m/s. While this is somewhat beyond the range of validity of the small-angle approximation, the standard results for a pendulum are adequate for making an estimate.

otoplilp1

Expert

Step 1
1- The period T of a simple pendulum of length L is given by
$T=2\pi \sqrt{\frac{L}{g}}$
where g is the acceleration due to gravity. Keep in mind that a simple pendulum's period is solely dependent on its length and the gravitational constant's strength. It does not depend on the mass of the object hanging at its end or the amplitude of vibration.
2- The ratio of the total distance a particle travels to the total time it takes to cover that distance is its average speed
${v}_{avg}=\frac{d}{\mathrm{△}t}$ (2)
Step 2
- The length of the orangutan arms is: $L=0.90m$
- The angle of swinging of the orangutan is: $\theta ={20}^{\circ }$
- The orangutan is modeled as a pendulum.
Step 3
We are asked to estimate the orangutans

Corgnatiui

Expert

Step 1
Write the formula for the time period of the pendulum.
$T=2\pi \sqrt{\frac{l}{g}}$
Step 2
Substitute the known values and solve.
$T=2\pi \sqrt{\frac{0.90m}{9.8\frac{m}{{s}^{2}}}}$
$=1.904s$
Step 3
The time taken by Orangutans to move from one end to another end will be half of the time period. Therefore,
$t=\frac{T}{2}$
$=\frac{1.904s}{2}$
$=0.952s$
Step 4
Draw the displacement diagram of the Orangutans when it moves from one end to another.
Step 5
From the diagram, the distance from one end to another end is:
$d=2l{\mathrm{sin}20}^{\circ }$
Step 6
Therefore, the speed of the pendulum will approximately:
$v=\frac{d}{t}$
$=\frac{2l{\mathrm{sin}20}^{\circ }}{0.952s}$
$=\frac{2\left(0.9m\right){\mathrm{sin}20}^{\circ }}{0.952s}$
$=0.647\frac{m}{s}$
$\approx 0.65\frac{m}{s}$

karton

Expert

We should really be treating the orangutan as a physical pendulum not a simple pendulum, but there is not enough information given to do so, so it seems we will have to treat then as a simple pendulum
We can calculate easily the period of such a pendulum from
$P=2\pi \sqrt{\frac{L}{g}}=6.28$
$\sqrt{\frac{0.9m}{9.8m.s.s}}=1.90s$
The time from the lowest point to the next branch is $\frac{1}{4}$ of period, so this time is 0.48s
If they mean forward horizontal velocity, draw a triangle representing the situation to see that the horizontal distance covered is given by
hor distance $=L\mathrm{sin}20=0.9\mathrm{sin}20=0.31m$, and the forward speed is
$\frac{0.31m}{0.48s}=0.65m/s$

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