We are given that a mass is attached to a string and is spun is in a horizontal circle, this means that the free body diagram of the problem is the following:
We will add the forces in the vertical direction:
[tex]T_y-mg=0[/tex]
The forces add up to zero since there is no acceleration in the vertical direction. Therefore, we can add "mg" to both sides:
[tex]T_y=mg[/tex]
Now, The vertical component of the tension can be put in terms of the total tension using the following right triangle:
Therefore, we use the trigonometric function sine:
[tex]\cos x=\frac{T_y}{T}[/tex]
Multiplying both sides by "T"
[tex]T\cos x=T_y[/tex]
Substituting in the sum of vertical forces:
[tex]T\cos x=mg[/tex]
Now, Since the horizontal component of the tension is equivalent to the centripetal force, we have that:
[tex]T_x=ma_r[/tex]
Where:
[tex]\begin{gathered} r=\text{ radius of the circle} \\ a_r=\text{ radial acceleration} \end{gathered}[/tex]
We can use the trigonometric function cosine to determine the horizontal component of the tension:
[tex]\sin x=\frac{T_x}{T}[/tex]
Multiplying both sides by "T":
[tex]T\sin x=T_x[/tex]
Substituting we get:
[tex]T\sin x=ma_r[/tex]
Now, we divide both equations:
[tex]\frac{T\sin x}{T\cos x}=\frac{ma_r}{mg}[/tex]
Simplifying we get;
[tex]\tan x=\frac{a_r}{g}[/tex]
Now, we multiply both sides by "g":
[tex]g\tan x=a_r[/tex]
Now, in any circular motion, the period "P" is given by:
[tex]P=\frac{2\pi}{\omega}[/tex]
Where:
[tex]\begin{gathered} r=\text{ radius} \\ v=\text{ tangential velocity} \end{gathered}[/tex]
Also, the acceleration is given by:
[tex]a_r=\frac{4\pi^2r}{P^2}[/tex]
Substituting the expression for the acceleration we determined before we get:
[tex]g\tan x=\frac{4\pi^2r}{P^2}[/tex]
Substituting the expression for the period:
[tex]g\tan x=\frac{4\pi^2r}{(\frac{2\pi}{\omega})^2}[/tex]
Solving the square:
[tex]g\tan x=\frac{4\pi^2r}{\frac{4\pi^2}{\omega^2}}[/tex]
Solving the fraction:
[tex]g\tan x=\frac{4\pi^2r\omega^2}{4\pi^2}[/tex]
Simplifying:
[tex]g\tan x=r\omega^2[/tex]
Now, we can put the radius in terms of the length of the spring using the following triangle:
using the function sine we get:
[tex]L\sin x=r[/tex]
Substituting in the previous equation we get:
[tex]g\tan x=L\sin x\omega^2[/tex]
Now, we decompose the tangent:
[tex]\frac{g\sin x}{\cos x}=L\sin x\omega^2[/tex]
Simplifying:
[tex]\frac{g}{\cos x}=L\omega^2[/tex]
Now, we invert both sides:
[tex]\frac{\cos x}{g}=\frac{1}{L\omega^2}[/tex]
Multiplying both sides by "g" we get:
[tex]\cos x=\frac{g}{L\omega^2}[/tex]
The angular velocity is the angle divided by time, since it takes 0.644 s to complete one revolution this means that :
[tex]\omega=\frac{2\pi}{0.644}[/tex]
substituting the values:
[tex]\cos x=\frac{9.8}{1.44(\frac{2\pi}{0.644})^2}[/tex]
Solving the operations:
[tex]\cos x=0.071[/tex]
Now, going back to the sum of forces in the vertical direction:
[tex]T\cos x=mg[/tex]
Dividing both sides by "g":
[tex]\frac{T\cos x}{g}=m[/tex]
Substituting the values:
[tex]\frac{(2.91)(0.071)}{9.8}=m[/tex]
Solving the operations:
[tex]0.021=m[/tex]
Therefore, the mass of the ball is 0.021 kg.
