dynamics Retired engineer Donald and his wife Betty are contestants in the reali

Question

dynamics

Retired engineer Donald and his wife Betty are contestants in the reality television show "The Amazingly Elderly Race.** They have been doing pretty well until recently when Donald started acting strangely and decided to "go native" and wear only a loincloth. Betty suspects that he has not been taking his blood pressure medication on a regular basis. They were traveling down river in their hand-made boat through the tropical section of the racc when during the night Donald just disappeared. Betty spent the better pari of the morning trying to find Donald, but she finally decided he had just taken this, "going native" idea a bit too far. Suddenly, the boat struck a submerged log and began to sink. Unfortunately, this section of the river is infested with crocs and Betty is in a very unpleasant situation! Fortunately, Donald was nearby (having dccidcd to improve his vine swinging techniques during the morning) and spotted Betty's, situation. Using his vast knowledge of rigid-body dynamics, Donald decided that he must swing down and grab her in mid-flight and release his grip from the vine at just the right moment to allow he and Betty to sail over the crocs. Donald weighs 210 lbs and he guesses that Betty weighs (always a dangerous thing for a husband to do) 125 lbs. He quickly sketches out his plan and the local geometry as shown below. He needs to know the following information. If lie starts from rest at point A. then what is his velocity just before he grabs Betty at point B? What is the velocity of Donald and Betty after he grabs her at point B? What is their velocity at point C when he releases the vine? If the horizontal distance from pint C to D is 30 feet, then will they make it to point D and safety?

Explanation / Answer

1) Potential energy loss from point A to B = m*g*50(cos20-cos60) = Kinetic energy = 1/2 m v^2

v^2=100*g(cos20-cos60) (g=32ft/s^2) v=37.51 ft/s

2) by conservation of momentum m1v1 = m2v2

210*37.51 = (210+125)*v2 v2=23.51 ft/s

3) Potential energy gain from point B to C = m2*g*50(cos20-cos35) = Kinetic energy loss = 1/2 m2 (v2^2 - vc^2)

vc^2=v2^2 - 100*g(cos20-cos35) v=12.92 ft/s

4) time taken to fall to the ground s = ut -1/2 gt^2 = 12.92sin35 * t -16t^2 =-9.04 ft = 7.41t-16t^2

t= 1.02s

forward distance covered in that time = 12.92cos35 * t < 30ft

so they cannot make it till D

5)max tension will be at bottommost point

T = mg + mv^2/r = (335 lb) (32+v^2/50)

v^2 can be got by energy difference as before = 3174

T = 335*(32+63.5)=32000 lb ft/s^2

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