Answer :
Answer:
The angle is 55.5°
Explanation:
Using the second Newton's law:
[tex]Tcos\theta =mg[/tex] (eq. 1)
[tex]tsin\theta =\frac{mv^{2} }{r}[/tex] (eq. 2)
Dividing both equations:
[tex]\frac{sin^{2}\theta }{cos\theta } =\frac{v^{2} }{gL} \\\frac{1-cos^{2}\theta }{cos\theta } =\frac{v^{2} }{gL}[/tex]
Where
v = 4.97 m/s
g = 9.8 m/s²
L = 2.11 m
Replacing:
[tex]\frac{1-cos^{2}\theta }{cos\theta } =\frac{4.97^{2} }{9.8*2.11}=1.2\\cos^{2} \theta +1.2cos\theta -1=0\\\theta =55.5[/tex]
The value of the angle the string makes with the pole is 55.4⁰.
The given parameters;
- mass of the ball, m = 5.59 kg
- length of the string, L = 2.11 m
- speed of the ball, v = 4.97 m/s
The vertical and horizontal component of tension on the string is calculated as follows;
[tex]Tcos (\theta) = mg \ \ ----(1) \\\\Tsin(\theta) = \frac{mv^2}{r} = \frac{mv^2}{lsin(\theta)} \\\\Tsin^2 \theta = \frac{mv^2}{l} \\\\T(1- cos^2 \theta) = \frac{mv^2}{l} \ ---(2) \\\\from \ (1), \ \ \ cos \theta = \frac{mg}{T}[/tex]
substitute the value of cosθ into (2)
[tex]T(1 - \frac{m^2g^2}{T^2} )= \frac{mv^2}{l} \\\\T - \frac{m^2g^2}{T} = \frac{mv^2}{l}\\\\T - \frac{(5.59)^2 (9.8)^2 }{T} = \frac{5.59\times 4.97^2}{2.11} \\\\T^2 - 3001.1 = 65.44T\\\\T^2 -65.44\ T - 3001.1 = 0\\\\Solve \ the \ quadratic \ equation \ using \ formula \ method;\\\\a = 1, \ b = -65.44 , \ c = -3001.1 \\\\T = \frac{-b \ \ +/- \ \ \sqrt{b^2 - 4ac} }{2a} \\\\T = \frac{-(-65.44) \ \ +/- \ \ \sqrt{(-65.44)^2 - 4(1\times -3001.1)} }{2(1)}\\\\T = 96.52[/tex]
The value of the angle is calculated as follows;
[tex]cos (\theta ) = \frac{mg}{T} \\\\cos(\theta) = \frac{5.59\times 9.8}{96.52} \\\\cos(\theta) = 0.5676\\\\\theta = cos^{-1} (0.5676)\\\\\theta = 55.4 ^0[/tex]
Thus, the value of the angle the string makes with the pole is 55.4⁰.
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