Find velocity given y and angle

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Homework Help Overview

The problem involves Tarzan swinging on a vine, with the goal of determining his speed at the bottom of the swing under two different initial conditions. The context includes concepts from mechanics, specifically energy conservation and forces acting on a swinging object.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss using conservation of energy, drawing free-body diagrams, and applying Newton's Second Law. There are questions about the role of the angle in calculations and the correct interpretation of height changes. Some participants suggest different methods for calculating velocity, while others express confusion about the initial conditions and the mass of Tarzan.

Discussion Status

There is ongoing exploration of different approaches to the problem, with some participants providing guidance on using energy conservation principles. Multiple interpretations of the setup and calculations are being discussed, but no consensus has been reached on the final answers.

Contextual Notes

Participants note the importance of accurately determining the change in height based on the angle of the vine and the initial position of Tarzan. There is also mention of constraints related to homework rules and the need for careful consideration of the problem's parameters.

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Homework Statement


Tarzan swings on a 34.0 m long vine initially inclined at an angle of 38.0° with the vertical.
(a) What is his speed at the bottom of the swing if he starts from rest?
(b) What is his speed at the bottom of the swing if he pushes off with a speed of 2.00 m/s?


The Attempt at a Solution


I honestly do not even know where to begin.
 
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Use conservation of energy, you know the height at which he starts, you know the height at the bottom of the swing. You know the initial speed for both cases. This should get you started. KEi + PEi = KEf + PEf
 
mandy9008 said:

Homework Statement


Tarzan swings on a 34.0 m long vine initially inclined at an angle of 38.0° with the vertical.
(a) What is his speed at the bottom of the swing if he starts from rest?
(b) What is his speed at the bottom of the swing if he pushes off with a speed of 2.00 m/s?


The Attempt at a Solution


I honestly do not even know where to begin.

(1)Firstly draw freebody diagram. Consider how many forces acting on Tarzan.
(2) Use Newton's Second law then find acceleration.
(3) Find the length between tarzan initial and final condition.
(4) Find the velocity at the bottom.
 
KE + PE = KE + PE
mgh = 1/2 mv2
v2 = gh/2
v2 = (9.8 m/s2)(34m) / 2
v= 12.91 m/s

where does the angle come into play?
when i entered this answer, it told me that I was within 10% of the correct answer
 
See figure attached.
 

Attachments

  • angle.JPG
    angle.JPG
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basically find the change in height using trig... the length of the vine * cos(angle)

then you take the length and subtract what you found..
next find (mass)(9.81)(change in height)

equate that to 1/2mv^2

so take the answer, multiply by 2, divide by mass and then get the squareroot..

thats the velocity..

all this is is mgh=1/2mv^2 (Eg=Ek)
 
I don't know the mass though
 
mandy, if you write out KEi + PEi = KEf + PEf, you will notice a common term which can be canceled.
 
i just did the question.. ill post the solution.. scanning it atm..

mass cancels out.. are the answers 16 and 16.1m/s?
 
  • #10
I did cancel out the mass when I solved this equation. Is this not the correct equation?

mandy9008 said:
KE + PE = KE + PE
mgh = 1/2 mv2
v2 = gh/2
v2 = (9.8 m/s2)(34m) / 2
v= 12.91 m/s
 
  • #11
its not bc you're dividing by 2 when you should be multiplying.. you're also ignoring the angle and delta heighthttp://yfrog.com/6bscan0001yj
 
  • #12
http://img227.imageshack.us/img227/5659/scan0001y.jpg

this is my solution..
 
Last edited by a moderator:
  • #13
The equation is correct, and you canceled out the mass, but the height is not correct. Instead use delta h in the figure which i attached. Tarzan did not release from 0 degrees, therefore take this into account.
 
  • #14
joshmdmd, you must be rather careful here. If the 38 degrees are measured from the x-axis like I have drawn then in order to obtain the y-component you must take the sine of the angle.
 
  • #15
your eq'n should be v^2=2gh
 
  • #16
wwshr87 said:
joshmdmd, you must be rather careful here. If the 38 degrees are measured from the x-axis like I have drawn then in order to obtain the y-component you must take the sine of the angle.

mmm

careful?

cos(90-38) = sin38
 
  • #17
joshmdmd said:
mmm

careful?

cos(90-38) = sin38

oops.. i took theta in your drawing as 38.. =] you messed me up
 
  • #18
Sorry for the confusion, just making sure you know what's going on.
 
  • #19
here we go.. corrections to answer..

v=(SQRT)2gh
v=11.886m/s

b)
v=12.22
 
  • #20
Josh, the answer in your scan is wrong. When i did it and got 12.91 m, it said that I was within 10% of the correct answer.
 
  • #21
wwshr87 said:
Sorry for the confusion, just making sure you know what's going on.

its okay.
wasnt paying attn to the diagram .. as you can see i ripped yours in mine.
 
  • #22
mandy9008 said:
Josh, the answer in your scan is wrong. When i did it and got 12.91 m, it said that I was within 10% of the correct answer.

read the second page =]
 
  • #23
Are you guys doing mastering physics?
 
  • #24
wwshr87 said:
Are you guys doing mastering physics?

if you mean uni for phys.. ima go to uft or waterloo for aerospace engineering/mechanical engineering.. I am currently gr 11.. writing gr 12 phys exam on friday.
 
  • #25
joshmdmd, alright good luck. You seem to be on the right track by taking physics in high school.
 
  • #26
Correct! :)
 
  • #27
wwshr87 said:
joshmdmd, alright good luck. You seem to be on the right track by taking physics in high school.

thanks man, you mastering atm?

all the people in my class think ima idiot bc I am doing 11/12 phys this year and 11/12 chem next year.. lol.. they say ill forget phys before uni
 
  • #28
I am about to begin a master's in electrical engineering. I think you are doing great you'll never totally forget this things.
 
  • #29
wwshr87 said:
I am about to begin a master's in electrical engineering. I think you are doing great you'll never totally forget this things.

that means a lot bro.. have fun and enjoy yourself... schools always funner than work =]
 

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