Calculating the Angle of a Suspended Mass

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A 6.10 kg mass is suspended from a 1.51 m string and revolves horizontally at a speed of 3.24 m/s, prompting a calculation of the angle between the string and the vertical. The equations of motion involve components of tension and gravitational force, leading to a quadratic equation. The initial calculations yielded a quadratic with two potential solutions, but confusion arose over the validity of these results. It was clarified that the cosine of the angle must be between -1 and 1, and a negative cosine indicates an unphysical angle. The correct approach involves using the cosine value to find the angle, ensuring proper conversion between degrees and radians.
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Homework Statement



A mass of 6.10 kg is suspended from a 1.51 m long string. It revolves in a horizontal circle as shown in the figure
.http://capa2.cc.huji.ac.il/res/msu/physicslib/msuphysicslib/11_Force_Motion_Adv/graphics/prob03_pendulum.gif
The tangential speed of the mass is 3.24 m/s. Calculate the angle between the string and the vertical.

Homework Equations



F=mv^2/R

The Attempt at a Solution



The F component at y is:
Tcosa-mg=0
Tcosa=mg
T=mg/cosa
The F component at x is:
Tsina=ma
Tsina=mv^2/R
R=lsina
mg*sina/cosa=mv^2/lsina
lmgsin^2a=cosamv^2
lgsin^2a=v^2cosa

sin^2a=1-cos^2a
after putting all the numbers i get a final equation which is
10.798cos^2+10.4976cosa-14.798
the solution of the equation was
1-0.78
2-1.753

ofcourse i take the first one
BUT!its not correct!
pleasew tell me where i was wrong
 
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Your method is correct, but you made a mistake solving the quadratic. I recommend that you solve it symbolically and put in the numbers at the very end.
 
I am doing my calculations over and over again but am getting the same answer !
I can't see where i was wrong!
 
HardestPart said:
... after putting all the numbers i get a final equation which is
10.798cos^2+10.4976cosa-14.798
...

I should be able to help you if you show the equation (in symbols) before you put in the numbers and then what numbers you put in all laid out.
 
the equation is:
l*g(1-cos^2a)=v^2cosa
l*g*cos^2a+v^2cosa-l*g=0
that is befor i put in the numbers
after:
1.51*9.8*cos^2a+3.24^2cosa-1.51*9.8=0
14.798cos^2a+10.4976cosa-14.798=0
that is the final equation

did i make any mistake here?
 
HardestPart said:
the equation is:
l*g(1-cos^2a)=v^2cosa
l*g*cos^2a+v^2cosa-l*g=0
that is befor i put in the numbers
after:
1.51*9.8*cos^2a+3.24^2cosa-1.51*9.8=0
14.798cos^2a+10.4976cosa-14.798=0
that is the final equation

did i make any mistake here?

Not here, but you made a mistake in your first posting of the quadratic.


HardestPart said:
after putting all the numbers i get a final equation which is
10.798cos^2+10.4976cosa-14.798
 
I solved the equation with the correct one:
I get two answers
1-0.706
2--1.415

the first one is incorrect
could it be the second one?can it be negative?
 
HardestPart said:
I solved the equation with the correct one:
I get two answers
1-0.706
2--1.415

the first one is incorrect

How do you know it is incorrect? What is the correct answer?

[/quote]could it be the second one?can it be negative?[/QUOTE]

Don't forget you are solving for the cosine of an angle which cannot be less that -1. Also, a negative cosine means that the angle is greater than 90o which is unphysical.
 
I know its incorrect-thats complicated-
but I don't know what is the correct answer

Do you mean after i get my answers for the equation
I have to do SHIF COS A ->the solution i get
and than i get my angle?
 
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HardestPart said:
I know its incorrect-thats complicated-
but I don't know what is the correct answer

Do you mean after i get my answers for the equation
I have to do SHIF COS A ->the solution i get
and than i get my angel?

I don't know about SHIF COS A, but the 0.706 that you found is the cosine of the angle. You need to find the angle itself from it and be sure you have the degrees/radians switches set properly.
 
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