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Physics
A device for training astronauts and jet fighter pilots is designed to rotate the trainee in a horizontal circle of radius 11.0 m .if the force felt by the trainee is 7.80 times her own weight, how fast is she rotating? express your answer in both (a)m/s and (b)rev/s.
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Best Answer

The velocity of the trainee is 29 m/s or 0.42 rev/s

Further explanation

Acceleration is rate of change of velocity.

[tex]\large {\boxed {a = \frac{v - u}{t} } }[/tex]

[tex]\large {\boxed {d = \frac{v + u}{2}~t } }[/tex]

a = acceleration (m / s²)v = final velocity (m / s)

u = initial velocity (m / s)

t = time taken (s)

d = distance (m)

Centripetal Acceleration of circular motion could be calculated using following formula:

[tex]\large {\boxed {a_s = v^2 / R} }[/tex]

a = centripetal acceleration ( m/s² )

v = velocity ( m/s )

R = radius of circle ( m )

Let us now tackle the problem!

Given:

Radius of horizontal circle = R = 11.0 m

Force Felt by the Trainee = F = 7.80w

Unknown:

Velocity of Rotation = v = ?

Solution:

[tex]F = ma[/tex]

[tex]F = m\frac{v^2}{R}[/tex]

[tex]7.80w = m\frac{v^2}{R}[/tex]

[tex]7.80mg = m\frac{v^2}{R}[/tex]

[tex]7.80g = \frac{v^2}{R}[/tex]

[tex]7.80 \times 9.8 = \frac{v^2}{11.0}[/tex]

[tex]v^2 = 840.84[/tex]

[tex]v \approx 29 ~m/s[/tex]

[tex]\omega = \frac{v}{R}[/tex]  → in rad/s

[tex]\omega = \frac{v}{2 \pi R}[/tex]  → in rev/s

[tex]\omega = \frac{29}{2 \pi \times 11.0}[/tex]

[tex]\omega \approx 0.42 ~ rev/s[/tex]

Answer details

Grade: High School

Subject: Physics

Chapter: Circular Motion

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle , Speed , Time , Rate , Circular , Ball , Centripetal

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