# Circular motion

#### HardestPart

1. The problem statement, all variables and given/known data

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
The tangential speed of the mass is 3.24 m/s. Calculate the angle between the string and the vertical.
2. Relevant equations

F=mv^2/R

3. 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

Last edited by a moderator:

#### kuruman

Homework Helper
Gold Member
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.

#### HardestPart

I am doing my calculations over and over again but am getting the same answer !
I cant see where i was wrong!

#### kuruman

Homework Helper
Gold Member
... 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.

#### HardestPart

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?

#### kuruman

Homework Helper
Gold Member
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.

after putting all the numbers i get a final equation which is
10.798cos^2+10.4976cosa-14.798

#### HardestPart

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?

#### kuruman

Homework Helper
Gold Member
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.

#### HardestPart

I know its incorrect-thats complicated-
but I dont 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?

#### kuruman

Homework Helper
Gold Member
I know its incorrect-thats complicated-
but I dont 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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