A child is playing on a swing. He swings down from an angle of 53 degree. The ma

Question

A child is playing on a swing. He swings down from an angle of 53 degree. The mass of the child is 40 kg and the seat of the swing has a mass of 5 kg. The length of the swing is 5 m. At the bottom of the swing the boy quickly picks up his 5-kg dog from rest and continues to swing up with it on his lap. a) How far will the child and dog swing up before coming to rest? b) Instead of picking up the dog, the child slips off the swing at the bottom with zero speed relative to the swing. How far will the swing rise all by itself? c) In a third version of this problem, the child slips of the swing and the swing rises to an angle of 37 degree all by itself. What was the speed of the child relative to the ground when he slipped off? .. .relative to the swing? d) In which part of this problem is energy conserved? ... is momentum conserved?

Explanation / Answer

a) Let us consider the zero level of potential enery at the bottom of the swing.

Initial potential energy is PE1 = mgL(1-cos530) = (40kg + 5kg)gL(1-cos530)

Initial kinetic energy is zero.

When the swing reaches bottom let the speed be v, then kinetic energy = (1/2)(40kg + 5kg)v2

potenital energy will be zero at bottom

Applying consevration of mehcanical energy we have:

(40kg + 5kg)gL(1-cos530) = (1/2)(40kg + 5kg)v2

or v2 = 2gL(1-cos530) = 2(9.8m/s2)(5m)(1-cos530)

v = 6.246m/s

At the bottom, the child picks up dog of 5 kg. Let the speed of child and dog be v after this, then as per conservation of momentum;

45kg(6.246m/s) = (50kg)v

or v = 5.62m/s

So, the kinetic energy of the dog+child+swing system is KE = 1/2(50kg)(5.62m/s)2 = 789.61J

Let the swing rises to a point where the angle made by the swing's length with the vertical is before the swing comes to rest.

Then applying conservation of energy we have;

1/2(50kg)(5.62m/s)2 = (50kg)(9.8m/s2)(5m)(1- cos)

or = 47.330

So, it rises to = 47.330, or a heigth h = (5m)(1- cos) = 1.611m.

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b) At the bottom of the swing the child+swing seat has speed v = 6.246m/s. Let the swing's speed be v' after the child slips off with zero speed relative to the swing. Applying conservation of momentum, we have;

(40kg+5kg)(6.246m/s) = 40kg(6.246m/s)+ (5kg)v'

or v' = 6.246m/s

Let the swing rises to a point where the angle made by the swing's length with the vertical is before the swing comes to rest. b conservation of energy we have

1/2(5kg)(6.246m/s)2 = (5kg)(9.8m/s2)(5m)(1 - cos)

or = 530 , so the swing rises to same height where it was at the start of motion.

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At the bottom of the swing the child+swing seat has speed v = 6.246m/s. let the swing's speed be v' after child slips off. Let the speed of child relative to the ground be v. Then aplying conservation of momentum we have;

45kg(6.246m/s) = (40kg)v + (5kg)v' --------- (1)

Now the KE of swing after the child slipps off is KE = 1/2(5kg)v'2

It reaches to an angle 370, so by conservation of energy we have;

1/2(5kg)v'2 = (5kg)(9.8m/s2)(5m)(1-cos370)

v' = 4.44m/s, puttin this in equation (1) above we get;

45kg(6.246m/s) = (40kg)v + (5kg)(4.44m/s)

or v = 6.47m/s was the speed of child relative to the ground.

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d) Momentum is conserved every where in the problem but only immediately before and after the child picks up the dog or slips off the swing. After a sufficient time, the force of gravity changes the momentum of the system.

Energy is conserved only after the child picks up the dog to the point where child+dog+swing rise to rest together.

Again Energy is conserved only after the child slips off the swing to the point where swing rises to rest together.

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This concludes the answers. Check the answer and let me know if it's correct. If you need anymore clarification or correction I will be happy to oblige....

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