Question

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.

Answer 1

The speed of trainee in ${{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}$  is $\boxed{29{\text{ }}{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}}$  and in ${{{\text{rev}}}\mathord{\left/{\vphantom{{{\text{rev}}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}$  is $\boxed{0.42{\text{ }}{{{\text{rev}}}\mathord{\left/{\vphantom{{{\text{rev}}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}}$ .

Explanation:

The radius of horizontal circle is $11{\text{ m}}$ .and the force is equal to $7.8$  times the weight of trainee.

Our aim is to obtain the velocity or speed of trainee in both ${{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}$  and ${{{\text{rev}}}\mathord{\left/{\vphantom{{{\text{rev}}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}$ .

The weight of the trainee is calculated as,

$W=mg$

The force is equal to 7.8 times the weight of trainee and is shown below.

$F=7.8mg$

The expression for centripetal force is shown below.

${F_{{\text{centripetal}}}}=\frac{{m{v^2}}}{r}$                                  ......(1)

The radius of circle is $11{\text{ m}}$ .

The centripetal force is equal to the force exerted by trainee.

So, substitute $7.8mg$  for ${F_{{\text{centripetal}}}}$  and $11$  for $r$  in equation (1) to obtain the value of velocity in ${{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}$ .

$\begin{aligned}7.8mg&=\frac{{m{v^2}}}{{11}}\\7.8g&=\frac{{{v^2}}}{{11}}\\{v^2}&=85.8g\\\end{aligned}$

The acceleration due to gravity is $9.8{{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}^{\text{2}}}$ .

Now, the velocity is calculated as,

$\begin{gathered}{v^2}=85.8\left({9.8}\right)\\=840.84\\v=\sqrt{840.84}\\=28.99\\\approx29{\text{ }}{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}\\\end{gathered}$

Therefore, the velocity of trainee in ${{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}$  is approximately $29{\text{ }}{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}$ .

The expression for angular velocity in ${{{\text{rev}}}\mathord{\left/{\vphantom{{{\text{rev}}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}$  is shown below.

$\begin{aligned}\omega&=\frac{v}{2\pi r}\end{aligned}$         ... (2)

The obtained velocity is $29{\text{ }}{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}$ , so substitute $29$  for $v$  and $11$  for $r$  in equation (2) to obtain the angular velocity.

$\begin{aligned}\omega&=\frac{29}{2\pi(11)}\\&=0.419\\&\approx0.42\text{ rev/s}\end{aligned}$

Therefore, the angular velocity in ${{{\text{rev}}}\mathord{\left/{\vphantom{{{\text{rev}}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}$  is $0.42{\text{ }}{{{\text{rev}}}\mathord{\left/{\vphantom{{{\text{rev}}}{\text{s}}}}\right.\kern-\nulldelimiterspace} {\text{s}}}$ .

Thus, the speed of trainee in ${{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}$  is $\boxed{29{\text{ }}{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}}$  and in ${{{\text{rev}}}\mathord{\left/{\vphantom{{{\text{rev}}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}$  is $\boxed{0.42{\text{ }}{{{\text{rev}}}\mathord{\left/{\vphantom{{{\text{rev}}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}}$ .

Learn More:

1. Linear momentum brainly.com/question/11947870

2. Motion and velocity brainly.com/question/6955558

3. Centripetal Force brainly.com/question/7420923

Answer Details:

Grade: High School

Subject: Physics

Chapter: Circular Motion

Keywords:

Device, astronauts, jet, pilots, rotation, trainee, horizontal, force, weight, fast, m/s, rev/s, tangential, velocity, speed, angular, centripetal.

Answer 2

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

Further explanation

Acceleration is rate of change of velocity.

$\large {\boxed {a = \frac{v - u}{t} } }$

$\large {\boxed {d = \frac{v + u}{2}~t } }$

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:

$\large {\boxed {a_s = v^2 / R} }$

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:

$F = ma$

$F = m\frac{v^2}{R}$

$7.80w = m\frac{v^2}{R}$

$7.80mg = m\frac{v^2}{R}$

$7.80g = \frac{v^2}{R}$

$7.80 \times 9.8 = \frac{v^2}{11.0}$

$v^2 = 840.84$

$v \approx 29 ~m/s$

$\omega = \frac{v}{R}$  → in rad/s

$\omega = \frac{v}{2 \pi R}$  → in rev/s

$\omega = \frac{29}{2 \pi \times 11.0}$

$\omega \approx 0.42 ~ rev/s$

Learn more

• Velocity of Runner : brainly.com/question/3813437

• Kinetic Energy : brainly.com/question/692781

• Acceleration : brainly.com/question/2283922

• The Speed of Car : brainly.com/question/568302
• Uniform Circular Motion : brainly.com/question/2562955
• Trajectory Motion : brainly.com/question/8656387

Answer details

Grade: High School

Subject: Physics

Chapter: Circular Motion

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