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 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.

Question

otoplilp1 · Accepted Answer

Step 1&amp;nbsp;1- The period T of a simple pendulum of length L is given by&amp;nbsp;T=2πLg&amp;nbsp;where g is the acceleration due to gravity. Keep in mind that a simple pendulum&#039;s period is solely dependent on its length and the gravitational constant&#039;s strength. It does not depend on the mass of the object hanging at its end or the amplitude of vibration.&amp;nbsp;2- The ratio of the total distance a particle travels to the total time it takes to cover that distance is its average speedvavg=d△t&amp;nbsp;(2)&amp;nbsp;Step 2&amp;nbsp;- The length of the orangutan arms is:&amp;nbsp;L=0.90m.&amp;nbsp;- The angle of swinging of the orangutan is:&amp;nbsp;θ=20∘.&amp;nbsp;- The orangutan is modeled as a pendulum.&amp;nbsp;Step 3&amp;nbsp;We are asked to estimate the orangutans

Corgnatiui · Accepted Answer

Step 1  Write the formula for the time period of the pendulum.  T=2πlg  Step 2  Substitute the known values and solve.  T=2π0.90m9.8ms2  =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=T2  =1.904s2  =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=2lsin⁡20∘  Step 6  Therefore, the speed of the pendulum will approximately:  v=dt  =2lsin⁡20∘0.952s  =2(0.9m)sin⁡20∘0.952s  =0.647ms  ≈0.65ms

karton · Accepted Answer

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=2πLg=6.28 0.9m9.8m.s.s=1.90s The time from the lowest point to the next branch is 14 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 =Lsin⁡20=0.9sin⁡20=0.31m, and the forward speed is 0.31m0.48s=0.65m/s