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notka56 [123]
3 years ago
6

A model airplane with a mass of 0.741kg is tethered by a wire so that it flies in a circle 30.9 m in radius. The airplane engine

provides anet thrust of 0.795 N perpendicular tothe tethering wire.(a) Find the torque the net thrust producesabout the center of the circle.N·m(b) Find the angular acceleration of the airplane when it is inlevel flight.rad/s2(c) Find the linear acceleration of the airplane tangent to itsflight path.m/s2
Physics
1 answer:
Tju [1.3M]3 years ago
5 0

Answer:

_T}=24.57Nm

ω = 0.0347 rad/s²

a ≅ 1.07 m/s²

Explanation:

Given that:

mass of the model airplane = 0.741 kg

radius of the wire = 30.9 m

Force = 0.795 N

The torque produced by the net thrust about the center of the circle can be calculated as:

_T } = Fr

where;

F represent the magnitude of the thrust

r represent the radius of the wire

Since we have our parameters in set, the next thing to do is to replace it into the above formula;

So;

_T}=(0.795)*(30.9)

_T}=24.57Nm

(b)

Find the angular acceleration of the airplane when it is in level flight rad/s²

_T}=I \omega

where;

I = moment of inertia

ω = angular acceleration

The moment of inertia (I) can also be illustrated as:

I = mr^2

I = ( 0.741) × (30.9)²

I = 0.741 × 954.81

I = 707.51 Kg.m²

_T}=I \omega

Making angular acceleration the subject of the formula; we have;

\omega = \frac{_T}{I}

ω = \frac{24.57}{707.51}

ω = 0.0347 rad/s²

(c)

Find the linear acceleration of the airplane tangent to its flight path.m/s²

the linear acceleration (a) can be given as:

a =  ωr

a = 0.0347 × 30.9

a = 1.07223 m/s²

a ≅ 1.07 m/s²

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