A person is to be released from rest on a swing pulled up from thevertical by 20
ID: 1678882 • Letter: A
Question
A person is to be released from rest on a swing pulled up from thevertical by 20 degrees. The Two ropes of the swing are 2.75 longand will break if the tension is wither is them is greater than355N. a. What is the maximum weight the person can have so that theropes will not break? b. If the angle is increased does this weight increase,decrease or stay the same?From what I gather Tension when swinging is T= mg= mv^2/r r= 2.75 mass has to be less than 37
change in potential energy from where the swing was releasedto when it reaches straight perpendicular. to find the change in height =L-L (cos20)
a. What is the maximum weight the person can have so that theropes will not break? b. If the angle is increased does this weight increase,decrease or stay the same?
From what I gather Tension when swinging is T= mg= mv^2/r r= 2.75 mass has to be less than 37
change in potential energy from where the swing was releasedto when it reaches straight perpendicular. to find the change in height =L-L (cos20)
Explanation / Answer
h = L ( 1 - cos)= 2.75 * ( 1 -cos20)
= 0.1658 meters Now when the swing is released, the child speeds up as theydrop toward the lowest point. By cons of energy, all the PE becomesKE so (1/2) m v2 = m g h v2 = 2 g h
= 2 * 9.8 * 0.1658
= 3.25m2/s2 Consider the forces acting on the swing at its lowest point...there are two ropes pulling up, gravity pulling down. So the sum ofthe forces is T + T - mg and since the swingmoves in circ motion, the sum of the forces must equal mv2 / r so 2T - mg = mv2 / r
2T = mg + mv2 / r 2T = m ( g + v2/r)
2*355 = m * ( 9.8 + 3.25/2.75) 710 = m * 10.98
m= 64.66 kg
(b) yes, the higher the swing is pulled back, thefaster the swing will go and the greater the tension will be. So tokeep it below the max, the person will have to be lighter.
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