Respuesta :
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]
Learn more
- Velocity of Runner : https://brainly.com/question/3813437
- Kinetic Energy : https://brainly.com/question/692781
- Acceleration : https://brainly.com/question/2283922
- The Speed of Car : https://brainly.com/question/568302
- Uniform Circular Motion : https://brainly.com/question/2562955
- Trajectory Motion : https://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
The speed of trainee in [tex]{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}[/tex] is [tex]\boxed{29{\text{ }}{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}}[/tex] and in [tex]{{{\text{rev}}}\mathord{\left/{\vphantom{{{\text{rev}}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}[/tex] is [tex]\boxed{0.42{\text{ }}{{{\text{rev}}}\mathord{\left/{\vphantom{{{\text{rev}}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}}[/tex] .
Explanation:
The radius of horizontal circle is [tex]11{\text{ m}}[/tex] .and the force is equal to [tex]7.8[/tex] times the weight of trainee.
Our aim is to obtain the velocity or speed of trainee in both [tex]{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}[/tex] and [tex]{{{\text{rev}}}\mathord{\left/{\vphantom{{{\text{rev}}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}[/tex] .
The weight of the trainee is calculated as,
[tex]W=mg[/tex]
The force is equal to 7.8 times the weight of trainee and is shown below.
[tex]F=7.8mg[/tex]
The expression for centripetal force is shown below.
[tex]{F_{{\text{centripetal}}}}=\frac{{m{v^2}}}{r}[/tex] ......(1)
The radius of circle is [tex]11{\text{ m}}[/tex] .
The centripetal force is equal to the force exerted by trainee.
So, substitute [tex]7.8mg[/tex] for [tex]{F_{{\text{centripetal}}}}[/tex] and [tex]11[/tex] for [tex]r[/tex] in equation (1) to obtain the value of velocity in [tex]{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}[/tex] .
[tex]\begin{aligned}7.8mg&=\frac{{m{v^2}}}{{11}}\\7.8g&=\frac{{{v^2}}}{{11}}\\{v^2}&=85.8g\\\end{aligned}[/tex]
The acceleration due to gravity is [tex]9.8{{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}^{\text{2}}}[/tex] .
Now, the velocity is calculated as,
[tex]\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}[/tex]
Therefore, the velocity of trainee in [tex]{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}[/tex] is approximately [tex]29{\text{ }}{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}[/tex] .
The expression for angular velocity in [tex]{{{\text{rev}}}\mathord{\left/{\vphantom{{{\text{rev}}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}[/tex] is shown below.
[tex]\begin{aligned}\omega&=\frac{v}{2\pi r}\end{aligned}[/tex] ... (2)
The obtained velocity is [tex]29{\text{ }}{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}[/tex] , so substitute [tex]29[/tex] for [tex]v[/tex] and [tex]11[/tex] for [tex]r[/tex] in equation (2) to obtain the angular velocity.
[tex]\begin{aligned}\omega&=\frac{29}{2\pi(11)}\\&=0.419\\&\approx0.42\text{ rev/s}\end{aligned}[/tex]
Therefore, the angular velocity in [tex]{{{\text{rev}}}\mathord{\left/{\vphantom{{{\text{rev}}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}[/tex] is [tex]0.42{\text{ }}{{{\text{rev}}}\mathord{\left/{\vphantom{{{\text{rev}}}{\text{s}}}}\right.\kern-\nulldelimiterspace} {\text{s}}}[/tex] .
Thus, the speed of trainee in [tex]{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}[/tex] is [tex]\boxed{29{\text{ }}{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}}[/tex] and in [tex]{{{\text{rev}}}\mathord{\left/{\vphantom{{{\text{rev}}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}[/tex] is [tex]\boxed{0.42{\text{ }}{{{\text{rev}}}\mathord{\left/{\vphantom{{{\text{rev}}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}}[/tex] .
Learn More:
1. Linear momentum https://brainly.com/question/11947870
2. Motion and velocity https://brainly.com/question/6955558
3. Centripetal Force https://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.
