Answer :
Centripetal acceleration (Ac)
= (1/2)v^2/r^2
Gravity = 9.8
Angle with vertical tan = tan ^-1 (Ac/g)
You will get that the answer will be 76.1
Hope this helps
= (1/2)v^2/r^2
Gravity = 9.8
Angle with vertical tan = tan ^-1 (Ac/g)
You will get that the answer will be 76.1
Hope this helps
Answer: The correct answer is 75.6 degree.
Explanation:
Calculate the angular velocity.
[tex]\omega =\frac{2\pi }{T}[/tex]
Here, T is the time.
Put T=0.50 s.
[tex]\omega =\frac{2\pi }{0.50}[/tex]
[tex]\omega =12.6\ rad\ per\ second[/tex]
If T is the tension acting on the sinker then tension equals to the centripetal force due to the circular motion as the sinker is moving in the circular motion.
[tex]T=mL\omega ^{2}[/tex]
Here, m is the mass of the sinker and L is the length of the sinker.
By resolving the components of the tension, the vertical component of the tension equals to the weight of the sinker.
[tex]Tcos\theta =mg[/tex]
Calculate the angle that the fishing line makes with the vertical.
[tex]cos\theta =\frac{mg}{T}[/tex]
[tex]\theta =cos^{-1}(\frac{mg}{T})[/tex]
Put [tex]T=mL\omega ^{2}[/tex].
[tex]\theta =cos^{-1}(\frac{g}{L\omega ^{2}})[/tex]
Put L= 0.25 , g= 9.8 meter per second and [tex]\omega =12.6\ rad\ per\ second[/tex].
[tex]\theta =cos^{-1}(\frac{9.8}{(0.25)\(12.6) ^{2}})[/tex]
[tex]\theta =cos^{-1}(0.25)[/tex]
[tex]\theta =1.32[/tex]
Convert radian into degree by multiplying 57.3 degree.
[tex]\theta =75.6[/tex]
Therefore, the angle that the fishing line makes with the vertical is 75.6 degree.