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http://i20.tinypic.com/f9mb1i.jpg

The answer seems kinda weird to so i want some people to check over my work and see if i have made any mistakes.

Thanx

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- Thread starter AFG34
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- #1

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http://i20.tinypic.com/f9mb1i.jpg

The answer seems kinda weird to so i want some people to check over my work and see if i have made any mistakes.

Thanx

- #2

Shooting Star

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- #3

Shooting Star

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You have done a lot of messy calculations. If you use symbols and plug in the values in the last step, then the chances of error are much less, and it’s easier to revise.

If l=length of string, H=height where the hand is holding the string, and b is the angle of the string with the vertical (here b=60 deg), then the required distance is given by:

2H*l*(sin b)*(tan b)*(1 – cos b).

The intriguing part is that the result is

- #4

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So what you are saying is the speed of the rock while being swung is the same as x-component of speed when the vine breaks?

When the vine breaks, doesn't it become a projectile?

- #5

Shooting Star

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The rock was moving with constant speed in a horizontal circle. The direction of the velo at each point was tangential to the circle. If let go, it flew off horizontally, which gave the direction of the initial velo. That velo was the x component of the initial velo. The y comp was zero.

Yes, it does become a projectile, and that's how I got the derived relation. If you can't get it, tell me.

- #6

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So my first two steps are correct right? (where i find acceleration and velocity).

Once the vine breaks and the rock becomes a projectile, then why do we ignore y component of acceleration?

So we ignore gravity and assume that the rock just travels in a straight line (60 from the vertical) and hits the ground?

- #7

Shooting Star

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So we ignore gravity and assume that the rock just travels in a straight line (60 from the vertical) and hits the ground?

No, no, we don't ignore the y comp of accn. There is, in fact,

Tell me once more which portion of my earlier post you did not understand.

- #8

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can you explain this part pleaseThe intriguing part is that the result is independent of g. The reason is that it’s a given condition that the vine breaks if the angle with the vertical is 60 deg. That may be true if the g is as on earth, but otherwise such a given condition is unphysical.

also...

5.83m/s = Vx?

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- #9

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this is as soon as the vine breaks, there is initial velocity in y-direction

- #10

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But try to understand what I have said before, that you should plug in the numerical values only at the end to get best results. The value of Vx explicitly was not at all necessary to find. Still,

Why is it interesting that the result is independent of g? Well, suppose you have a very heavy mass and you are trying to whirl it about with a string. Wouldn't you expect the string to break very soon if the g is more, which means the weight of the object is more?

But in your problem, even if you go to Jupiter, the string will only break when you are whirling it around quite fast to make its angle with the vertical 60 deg. Physically, we would expect the string to break much earlier in such a high gravity.

What has happened is that the angle 60 deg may be realistic for a vine to break on earth, so it has been given in the problem, but it is a bad way of giving a constraint in a problem. When does a string break? When the tension exceeds a certain value. Giving that critical tension would have been better Physics.

But for beginners, all this is good practice.

Now, derive the expression that I had obtained by considering the mass as just a projectile after the vine snaps. Should be easy for you.

- #11

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this is as soon as the vine breaks, there is initial velocity in y-direction

No. NO.

I presume y-axis is the vertical axis.

- #12

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we have only worked with formulas such as D=V1t +.5at^2 where

d = displacement

v1=initial velocity

t=time

a=accleration

and other basic kinemtaic equations

and some circular motion formulas such as:

Centripital force = mv^2/r

where m=mass, v=velocity, r=radius

there are other formulas with frequency and period but that is irrelevant to this question.

...so i have no idea of how to derive that expression.

this is the only way i can think of doing it:

my second try:

acceleration-y= 9.8m/s/s down

initial velocity-y= 0m/s

displacement-y= 1 meter (this is the height of the rock when the vine snaps)

time= ?

using the equation i described above (D=V1t +.5at^2)

i used the quadratic equation and got a time of 0.45seconds.

velocity-x= 5.83m/s (60degrees from vertical)

acceleration-x=0

t=0.45s

displacement-x=?

using the same formula i get 2.6m

i called a friend and he got the same answer.

d = displacement

v1=initial velocity

t=time

a=accleration

and other basic kinemtaic equations

and some circular motion formulas such as:

Centripital force = mv^2/r

where m=mass, v=velocity, r=radius

there are other formulas with frequency and period but that is irrelevant to this question.

...so i have no idea of how to derive that expression.

this is the only way i can think of doing it:

my second try:

acceleration-y= 9.8m/s/s down

initial velocity-y= 0m/s

displacement-y= 1 meter (this is the height of the rock when the vine snaps)

time= ?

using the equation i described above (D=V1t +.5at^2)

i used the quadratic equation and got a time of 0.45seconds.

velocity-x= 5.83m/s (60degrees from vertical)

acceleration-x=0

t=0.45s

displacement-x=?

using the same formula i get 2.6m

i called a friend and he got the same answer.

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- #13

Shooting Star

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(I made a careless mistake. I got x^2 as 6 and in a hurry forgot to take the sqrt.)

The ans is sqrt 6 = 2.4 m. Your ans of 2.6 m is absolutely all right because you have multiplied decimals in between.

When I get a bit of time, I'll give you all the steps of the calculation, and tell you how not to plug in numerical values at every step and waste time.

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Thanx, that would be great :D

- #15

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i checked with my teacher, and she said the corrent answer is 3m(2.9...)

- #16

Shooting Star

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I'll check once more. Give me some time.

- #17

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Now draw the diagram. The man is at the centre of the circle of radius ‘r’ = sqrt 3 m. (r=2*sin 60). The initial position of the rock is on the circumference. The direction of ‘d’ is perp to ‘r’. It’s a right angled triangle. So, the dist where the rock hits the ground is the hypotenuse of the triangle made by d and r and so is equal to exactly 3 m.

I hope it’s absolutely clear now…

- #18

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Thanks for the help, i got 8/10. damn these small mistakes.

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