Youngs Modulus and Centripetal Acceleration

AI Thread Summary
The discussion revolves around two physics problems involving Young's Modulus and centripetal acceleration. In the first question, participants calculate the net force on a bungee rider using the properties of rubber tubes, arriving at a force of approximately 4417.5 N after correcting earlier miscalculations. For the maximum height of the rider, they explore energy conservation principles, debating how to determine the velocity upon release from the slingshot. In the second question, confusion arises regarding the angular acceleration of a door being pulled shut by a rope, with participants sharing their approaches and calculations. Overall, the thread highlights the complexities of applying physics concepts to real-world scenarios and the collaborative effort to clarify misunderstandings.
zainy199
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Question #1)
In a new bungee sport a 70 kg rider is projected from the shore, out into a lake for a splash with a human scale slingshot. The stretched slingshot is made with a set of 50 cylindrical rubber tubes 1) stretched at a 45 degree angle from the horizontal. The tubes are cylindrical, having a 1cm outside diameter and a 0.5cm inside diameter. They are stretched from a 10 meter loose length back by an additional 15 meters and released accelerating the rider. Y of rubber = (1.0x10^6 pa) . Consider the accelerated rider.

a) Find the net force on the rider upon release.

b) Find the maximum height of the rider in his flight out into the water.

Question #2)
A 14 Kg door is pulled shut using tension in a rope. The rope is fixed to the door, 10cm from the hinge. The rope passes over a pulley and extends down a 2Kg hanging mass. The mass falls at a nearly constant speed, falling a total of 10cm during the 5 seconds to pull the door through a 90 degree rotation from fully open to fully shut. Find the angular acceleration of the door when the rope is at a 45 degree angle from the doors surface. You may consider the shutting door as a mass of 14Kg pivoting 60 cm from the hinge
 
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Sorry for the repost.
This is what I've done for question 1a)

F/A = (y x delta L) / L

delta L = 25-10 = 15 meters
A = A(outer) - A (inner)
A = 50pi(7.5x10^-5)
=0.0118 m^2

F = ((0.0118)(1x10^-6)(15))/(10)
F = 17,700 N

For Part B)
Im not exactly sure how to do this but I know that K.E initial = P.E final.
We need to calculate the velocity as he leaves the slingshot but I have no idea how.

I thought maybe this could work. F=ma
a= 17700/70
a= 253 m/s^2

v^2/r = a
v = sqrt (253x25)
v = 80 m/s

The I did 0.5mv^2= mgh
h = 322.6 meters

I have no clue how to do question number 2.
 
zainy199 said:
I thought maybe this could work. F=ma
a= 17700/70
a= 253 m/s^2

v^2/r = a
Where did you get that equation from? I believe it to be right - just don't know on what basis you justify it.
The usual way is conservation of energy. How much energy is stored in a spring?
v = sqrt (253x25)
v = 80 m/s
The I did 0.5mv^2= mgh
So all the KE goes into PE?
 
Well I assumed that the net force is the one we calculate using the young's modulus equation. Once i get acceleration I use a= v^2/r. Assuming that a is the centripetal acceleration. The radius is 25 because he's being stretched 25m from the slingshot at a 45 degree angle, hence creating a circle.

It says what is the maximum height reached, and this is only when the K.E energy is 0, and P.E is maximum.
Yes, I assumed that all of it gets transferred into P.E. We can ignore non-conservative forces like air resistance etc.
 
zainy199 said:
Well I assumed that the net force is the one we calculate using the young's modulus equation. Once i get acceleration I use a= v^2/r. Assuming that a is the centripetal acceleration. The radius is 25 because he's being stretched 25m from the slingshot at a 45 degree angle, hence creating a circle.

It says what is the maximum height reached, and this is only when the K.E energy is 0, and P.E is maximum.
Yes, I assumed that all of it gets transferred into P.E. We can ignore non-conservative forces like air resistance etc.
We seem to be interpreting the set-up quite differently. You mention centripetal acceleration and a circle... what circle? The slingshot is what in England I would have called a catapult. The hand-held version is a Y-shaped stick with elastic stretched across the prongs. You put the payload in the middle of the elastic, pull and release.
In the present case, the initial trajectory is at 45 degrees to horizontal (thereby maximising range). The trajectory as a whole will be a parabola.
 
Oh, now I see the problem. It's not accelerating in a circle, so there is no centripetal acceleration. But then how do I calculate the velocity with which it leaves the slingshot?
 
zainy199 said:
how do I calculate the velocity with which it leaves the slingshot?
Conservation of energy.
 
for the first question a) how did you calculate the area?
 
helpneed said:
for the first question a) how did you calculate the area?
Forget about the area. That was based on a misreading of the problem.
(Edit: ignore that comment)
 
Last edited:
  • #10
haruspex said:
Forget about the area. That was based on a misreading of the problem.

so why is he multiplying by 50. is it because of the number of tubes. I have the same question and i a little confused with the first one . i got my force to be 88.35N and if i multiply it by 50 i get 4417.5N. i don't know if I am correct.
 
  • #11
helpneed said:
so why is he multiplying by 50. is it because of the number of tubes. I have the same question and i a little confused with the first one . i got my force to be 88.35N and if i multiply it by 50 i get 4417.5N. i don't know if I am correct.

Yeh sorry, you're right. The answer if 4417.5 N. I was using the diameter, not the radius lol. Still don't know what to do for the second part.
 
  • #12
zainy199 said:
Yeh sorry, you're right. The answer if 4417.5 N. I was using the diameter, not the radius lol. Still don't know what to do for the second part.

yea cody pretty much killed us this time
 
  • #13
I don't really like this system of taking tests back home and bringing them next class.
He should just make the questions on the tests easier, so we don't have to take them home.
 
  • #14
zainy199 said:
I don't really like this system of taking tests back home and bringing them next class.
He should just make the questions on the tests easier, so we don't have to take them home.

exactly, i sent him an email complaining on how we can't even do the questions at home with all the resources, because they are too damn hard.
 
  • #15
A = A(outer) - A (inner)
A = 50pi(7.5x10^-5)
=0.0118 m^2

I don't get that part, how did you get the actual area and where did you get 7.5x10^-5 from?
could you plase explain that part to me
 
  • #16
Yes can someone please explain how they are getting there area. Because when i do my area, i am not getting the same value for my F as you guys are. My value is extremely low and I am confused on what I am doing wrong. my radius i was using was 7.5x10^-03.
 
  • #17
A= 50pi( (5x10^-3)^2-(2.5x10^-3)^2)
= 2.94x10^-3 m^2
 
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