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A toy plane is flying in a horizontal circle by being attached to a 0.75 meter string. The plane has a mass of 101.7 grams and makes five circular revolutions in 5.1 seconds. Find the tension force acting upon the plane. X Newtons Enter a number Determine the angle at which the string is pulled below the horizontal. degrees

User LLF
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1 Answer

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Answer:

F = 2,894 N

Step-by-step explanation:

For this exercise let's use Newton's second law

F = m a

The acceleration is centripetal

a = v² / r

Angular and linear variables are related.

v = w r

Let's replace

F = m w² r

The radius r and the length of the rope is related

cos is = r / L

r = L cos tea

Let's replace

F = m w² L cos θ

Let's reduce the magnitudes to the SI system

m = 101.7 g (1 kg / 1000g) = 0.1017 kg

θ = 5 rev (2π rad / rev) = 31,416 rad

w = θ / t

w = 31.416 / 5.1

w = 6.16 rad / s

F = 0.1017 6.16² 0.75 cos θ

F = 2,894 cos θ

The maximum value of F is for θ equal to zero

F = 2,894 N

User Sawako
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