Energy question - How much will this rope swinger rise

[phys1618]

Homework Statement

JayJay is running topspeed of 8 meters/seconds, he grabs a string handing vetical fr. a tall-tree. how high will he swings upwards,b.does the length of the string affect the total height?c. if it tkaes him 2s to reach that height, how long is the string.d.explain why the lengh of the string does or doesn't affect the heght.e. find the horizontal distance of the swing.

Homework Equations

PE=mgh KE=1/2(mv2) v=(2piA)/T
KEmax=1/2(mv2)max=PEmax=1/2(kA2)
T=s[o(sq.root of l/g)

The Attempt at a Solution

what i did was using PE and equals it to KE, because this will cancel out the m(unknown). so i have 3.2m as the height he will swing upward. I said yes to b. because if the length of the string is longer then, the total height will increase also. for c, i used the v eq. and have 2.55m for the lengthof the string. I don't know how to find the horizontal distant of the swing. thank you for the help.

 

[Doc Al]

what i did was using PE and equals it to KE, because this will cancel out the m(unknown). so i have 3.2m as the height he will swing upward.

OK.

I said yes to b. because if the length of the string is longer then, the total height will increase also.

When you found the height (part a) did you need to use the length of the string?

for c, i used the v eq. and have 2.55m for the lengthof the string.

How did you solve for the length? I don't understand.


Also: "please help" is not a good title for a thread. (All threads in Homework Help are requests for help. ) Try to be more descriptive.

 

[phys1618]

eheheh sorry for the title, i'll be more descriptive next time. sorry.

for part b. no, i didint use te length of the string to find part a. so i guess the answer is no then. when it said total height, i thought it meant the height he jumped and the vine length.

for c. i used the v=(2piA)/T
he 2 seconds and his velocity 8m/s was given. so
8m/s= 2pi(A)/2s, so A=2.55m , A=length
i hope i made it clear enough. thank you

 

[phys1618]

you know i just thought of, for e. can i use v=d/t ? because v and t is given. so vt=d. 8*2=16m?

 

[Doc Al]

for c. i used the v=(2piA)/T

That's an equation for the speed for uniform circular motion--not relevant here.

Hint: Think pendulum.

 

[phys1618]

so does it make sense to us eht T=2pi(sq.root(l/g)? i appreciate your help and coping with my ignorance =x

 

[Doc Al]

so does it make sense to us eht T=2pi(sq.root(l/g)?

Yes, that's what you need. Use it wisely.

 

[phys1618]

when i used the pendulum eq.
2s = 2pi(sq.root l/9.8 m/s2)
i got l= .9929 m

so now i got 3.2m as the height he swing upwards
b. no, the length of the string doesn't affect the total height
d. because i didnt need the height of the string to find the height jayjay swing upwards
c. length of the vine is .9929m which is approx. 1meters
e. to find the horizontal distance of the swing use v=d/t
which is 8m/s=d/2s = horizontal distance is 16 m
is that right/good?

 

[Doc Al]

when i used the pendulum eq.
2s = 2pi(sq.root l/9.8 m/s2)

Careful. 2s is the time it takes to go from the bottom position to the highest point, not the period of the pendulum motion. What is the full period?

 

[phys1618]

can you give me a hint on how to find the full period of the pendulum motion?

 

[Doc Al]

can you give me a hint on how to find the full period of the pendulum motion?

For a full period, the pendulum must complete a full cycle. That means if you start counting when the pendulum is at the bottom and moving to the right (say) the cycle won't be complete until its gone all around and once again is at the bottom and moving to the right.

In your problem, the "pendulum" only swings for a portion of the cycle. What fraction of the period does it swing when it goes from the bottom to the top?

 

[phys1618]

1/4?

 

[Doc Al]

1/4?

Right!

 

[phys1618]

yay! i finally got that right...gosh..i thought i'll never get it...thank you thank you.

ok...so let's try this again.

so now i got 3.2m as the height he swing upwards
b. no, the length of the string doesn't affect the total height
d. because i didnt need the height of the string to find the height jayjay swing upwards
c. length of the vine is (1/4)2s = 2pi(sq.root l/9.8 m/s2) which is .062059225m...is that an odd answer for length of a string?...i think i did something wrong??
e. to find the horizontal distance of the swing use v=d/t
which is 8m/s=d/2s = horizontal distance is 16 m

is that right/good?

 

[Doc Al]

c. length of the vine is (1/4)2s = 2pi(sq.root l/9.8 m/s2) which is .062059225m...is that an odd answer for length of a string?...i think i did something wrong??

Yes, you made a mistake. The time given (2 s) is 1/4 of the period. So what's the full period?

e. to find the horizontal distance of the swing use v=d/t
which is 8m/s=d/2s = horizontal distance is 16 m

No. To find the horizontal distance you'll use the length of the string and the height of the swing and a bit of trig/geometry. (Note that 8 m/s is only the speed at the bottom of the swing; as the swing proceeds, the speed and direction changes.)

 

[phys1618]

eheh i thought that too...it should of been 2s*4=2pi(sq.root l/9.8m/s2) which is 15.88m for the length of the string.

so would i use a2 + b2 = c2?
if so would it be (3.2)2 + b2 = (15.88)2?
I really appreciate your help

 

[Doc Al]

eheh i thought that too...it should of been 2s*4=2pi(sq.root l/9.8m/s2) which is 15.88m for the length of the string.

Good.

so would i use a2 + b2 = c2?
if so would it be (3.2)2 + b2 = (15.88)2?

That's the right equation, but the height of the swing is not one of the sides of the triangle. Draw a picture of the initial and final positions of the string.

Also: Recalculate the height a bit more accurately.

 

[phys1618]

yea that makes sense, because the height of the swing is not involved in the triangle...
the position of the string from initial and final had changed but the length of the string didnt change. so a2 + b2=c2 a and b is actually 15.88
so (15.88)2 + (15.88)2 = distance of the horizontal swing squared
which is 22.34m

the more accurate height is 3.265m

is this all correct??

 

[Doc Al]

yea that makes sense, because the height of the swing is not involved in the triangle...

While the height of the swing is not a side of the triangle, it is needed to calculate one of the sides.

the position of the string from initial and final had changed but the length of the string didnt change. so a2 + b2=c2 a and b is actually 15.88
so (15.88)2 + (15.88)2 = distance of the horizontal swing squared

No. The triangle you want is a right triangle whose hypotenuse is the length of the string.

the more accurate height is 3.265m

Good. Use it to find the vertical side of the triangle above.

 

[phys1618]

hypotenuse is the length of the string, which is 15.88.
using a2 + b2 =c2
i don't know how to use 3.265m ..the only thing i can think of is ... cos 90 degrees=adj/hyp = 3.265m/15.88?

 

[Doc Al]

Please draw a diagram showing the string at the height of its swing. Then find the triangle you need on that diagram.

 

[phys1618]

does the 3.265m cut the length of the vine in half?
\
| \
| \
a| \ c
| \
| _____\

(7.94)2 + (3.265)2 = vetical side
63.0436 + 10.660225 = c2
73.703825=c2
8.585093185m = vertical

now take the vertical and length of the vine to find b(horizontal distance) for the bigger triangle.
(8.59m)2 + b2 = (15.88)2
73.703825 + b2= 252.1744
b2=178.470575
b=13.359m

 

[phys1618]

i splitted the diagrams into different angles in a different way from the one above and have the horizontal distance as 15.89m using a2 + b2 =c2

 

[Doc Al]

Picture the following. Imagine that the string hangs from ceiling to floor. Thus the distance between ceiling and floor equals the length of the string.

Now imagine that the string is at the point of maximum swing. So now the bottom of the string is at the height "h" above the floor. (You have determined this swing height already.) What's the distance between the bottom of the string and the ceiling? That's one side of the triangle that you want. (Call that vertical side "a".) The hypotenuse of the triangle is just the string itself, with length "c". The other side of the triangle, "b", is the horizontal distance of the swing.

Based on this, draw yourself another diagram and see if it makes more sense.

 

[phys1618]

ok i drew the diagram already...can i use
the original position of the string and substract 3.265 from that to get a?
15.88-3.265m= 12.615m is a

then use a2 + b2 = c2
(12.615)2 + b2= (15.88)2
159.138225 + b2 = 252.1744
b= 9.645526165m which is approximately 9.65m (horizontal distance)

b
___________
| \ <-c |
| \ length |
initial | \ vine | a
position| \ |
length | \ |
vine | / |
| / |
| / |
| / |
| / | height of swing
| / |
| /_________|

b

i tried my best to explain how i drew the diagram...

 

[phys1618]

erghh the diagram didnt show up =(

 

[phys1618]

well it as a big rectangle box
with 3 traingles in it
___
|>|
---

 

[phys1618]

you know...i read wut u said over and over to make sure i understand it...it totally makes a lot of sense to me! i think i have a clearer idea..so i hope wut i did was right...

 

[Doc Al]

ok i drew the diagram already...can i use
the original position of the string and substract 3.265 from that to get a?
15.88-3.265m= 12.615m is a

then use a2 + b2 = c2
(12.615)2 + b2= (15.88)2
159.138225 + b2 = 252.1744
b= 9.645526165m which is approximately 9.65m (horizontal distance)

Now you've got it.

 

[phys1618]

Yay! thank you so much for your help! i know that i was pretty slow at figuring it out, but you''ve patiently hellped me a lot..i actually understand it too...thank you so much! i greatly appreciate all the help and time... :)