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Let’s investigate some examples that illustrate the relative magnitudes of the velocity, radius, and centripetal acceleration.

Creating an acceleration of 1 g

A jet is flying at 134.1 m/s along a straight line and makes a turn along a circular path level with the ground. What does the radius of the circle have to be to produce a centripetal acceleration of 1 g on the pilot and jet toward the center of the circular trajectory?

Strategy

Given the speed of the jet, we can solve for the radius of the circle in the expression for the centripetal acceleration.

Solution

Set the centripetal acceleration equal to the acceleration of gravity: 9.8 m/s 2 = v 2 / r .

Solving for the radius, we find

r = ( 134.1 m / s ) 2 9.8 m / s 2 = 1835 m = 1.835 km .

Significance

To create a greater acceleration than g on the pilot, the jet would either have to decrease the radius of its circular trajectory or increase its speed on its existing trajectory or both.

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Check Your Understanding A flywheel has a radius of 20.0 cm. What is the speed of a point on the edge of the flywheel if it experiences a centripetal acceleration of 900.0 cm / s 2 ?

134.0 cm/s

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Centripetal acceleration can have a wide range of values, depending on the speed and radius of curvature of the circular path. Typical centripetal accelerations are given in the following table.

Typical centripetal accelerations
Object Centripetal Acceleration (m/s 2 or factors of g )
Earth around the Sun 5.93 × 10 −3
Moon around the Earth 2.73 × 10 −3
Satellite in geosynchronous orbit 0.233
Outer edge of a CD when playing 5.78
Jet in a barrel roll (2–3 g )
Roller coaster (5 g )
Electron orbiting a proton in a simple Bohr model of the atom 9.0 × 10 22

Equations of motion for uniform circular motion

A particle executing circular motion can be described by its position vector r ( t ) . [link] shows a particle executing circular motion in a counterclockwise direction. As the particle moves on the circle, its position vector sweeps out the angle θ with the x- axis. Vector r ( t ) making an angle θ with the x- axis is shown with its components along the x - and y -axes. The magnitude of the position vector is A = | r ( t ) | and is also the radius of the circle, so that in terms of its components,

r ( t ) = A cos ω t i ^ + A sin ω t j ^ .

Here, ω is a constant called the angular frequency    of the particle. The angular frequency has units of radians (rad) per second and is simply the number of radians of angular measure through which the particle passes per second. The angle θ that the position vector has at any particular time is ω t .

If T is the period of motion, or the time to complete one revolution ( 2 π rad), then

ω = 2 π T .
A circle radius r, centered on the origin of an x y coordinate system is shown. Radius r of t is a vector from the origin to a point on the circle and is at an angle of theta equal to omega t to the horizontal. The x component of vector r is the magnitude of r of t times cosine of omega t. The y component of vector r is the magnitude of r of t times sine of omega t. The circulation is counterclockwise around the circle.
The position vector for a particle in circular motion with its components along the x - and y -axes. The particle moves counterclockwise. Angle θ is the angular frequency ω in radians per second multiplied by t .

Velocity and acceleration can be obtained from the position function by differentiation:

v ( t ) = d r ( t ) d t = A ω sin ω t i ^ + A ω cos ω t j ^ .

It can be shown from [link] that the velocity vector is tangential to the circle at the location of the particle, with magnitude A ω . Similarly, the acceleration vector is found by differentiating the velocity:

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Source:  OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5
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