# In Uniform Circular Motion Which of the Following Are Constant

## In Uniform Circular Motion Which of the Following Are Constant

### Learning Objectives

By the end of this department, you volition be able to:

• Solve for the centripetal acceleration of an object moving on a circular path.
• Use the equations of round move to find the position, velocity, and dispatch of a particle executing circular motion.
• Explicate the differences between centripetal acceleration and tangential acceleration resulting from nonuniform circular motion.
• Evaluate centripetal and tangential acceleration in nonuniform circular motion, and discover the total dispatch vector.

Compatible circular motility is a specific type of motion in which an object travels in a circle with a constant speed. For example, any signal on a propeller spinning at a constant rate is executing uniform circular motion. Other examples are the 2d, minute, and hr hands of a sentry. It is remarkable that points on these rotating objects are really accelerating, although the rotation rate is a constant. To see this, nosotros must analyze the motility in terms of vectors.

### Centripetal Acceleration

In i-dimensional kinematics, objects with a constant speed have cipher acceleration. However, in two- and three-dimensional kinematics, fifty-fifty if the speed is a constant, a particle can have acceleration if it moves along a curved trajectory such as a circle. In this case the velocity vector is changing, or $d\overset{\to }{v}\text{/}dt\ne 0.$ This is shown in (Effigy). Every bit the particle moves counterclockwise in time $\text{Δ}t$ on the circular path, its position vector moves from $\overset{\to }{r}(t)$ to $\overset{\to }{r}(t+\text{Δ}t).$ The velocity vector has abiding magnitude and is tangent to the path equally it changes from $\overset{\to }{5}(t)$ to $\overset{\to }{five}(t+\text{Δ}t),$ changing its direction only. Since the velocity vector $\overset{\to }{v}(t)$ is perpendicular to the position vector $\overset{\to }{r}(t),$ the triangles formed by the position vectors and $\text{Δ}\overset{\to }{r},$ and the velocity vectors and $\text{Δ}\overset{\to }{five}$ are like. Furthermore, since $|\overset{\to }{r}(t)|=|\overset{\to }{r}(t+\text{Δ}t)|$ and $|\overset{\to }{v}(t)|=|\overset{\to }{five}(t+\text{Δ}t)|,$ the two triangles are isosceles. From these facts we can make the assertion

$\frac{\text{Δ}v}{v}=\frac{\text{Δ}r}{r}$ or $\text{Δ}v=\frac{5}{r}\text{Δ}r.$

Figure 4.eighteen(a) A particle is moving in a circle at a constant speed, with position and velocity vectors at times $t$ and $t+\text{Δ}t.$ (b) Velocity vectors forming a triangle. The two triangles in the effigy are similar. The vector $\text{Δ}\overset{\to }{5}$ points toward the center of the circumvolve in the limit $\text{Δ}t\to 0.$

We tin find the magnitude of the acceleration from

$a=\underset{\text{Δ}t\to 0}{\text{lim}}(\frac{\text{Δ}v}{\text{Δ}t})=\frac{v}{r}(\underset{\text{Δ}t\to 0}{\text{lim}}\frac{\text{Δ}r}{\text{Δ}t})=\frac{{five}^{2}}{r}.$

The direction of the dispatch can as well be establish by noting that every bit $\text{Δ}t$ and therefore $\text{Δ}\theta$ arroyo zero, the vector $\text{Δ}\overset{\to }{v}$ approaches a direction perpendicular to $\overset{\to }{5}.$ In the limit $\text{Δ}t\to 0,$$\text{Δ}\overset{\to }{5}$ is perpendicular to $\overset{\to }{v}.$ Since $\overset{\to }{5}$ is tangent to the circle, the acceleration $d\overset{\to }{v}\text{/}dt$ points toward the center of the circumvolve. Summarizing, a particle moving in a circle at a constant speed has an acceleration with magnitude

${a}_{\text{C}}=\frac{{5}^{2}}{r}.$

The direction of the acceleration vector is toward the center of the circle ((Effigy)). This is a radial acceleration and is chosen the
centripetal dispatch, which is why we requite information technology the subscript c. The give-and-take
centripetal
comes from the Latin words
centrum
(meaning “center”) and
petere
(meaning to seek”), and thus takes the meaning “heart seeking.”

Figure 4.xix The centripetal acceleration vector points toward the middle of the round path of motion and is an acceleration in the radial direction. The velocity vector is too shown and is tangent to the circumvolve.

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

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### Example

#### Creating an Acceleration of 1g

A jet is flying at 134.i m/s along a straight line and makes a turn along a circular path level with the ground. What does the radius of the circumvolve take to be to produce a centripetal dispatch of 1
one thousand
on the pilot and jet toward the heart of the round 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 dispatch equal to the acceleration of gravity: $9.8{\,\text{yard/s}}^{2}={five}^{two}\text{/}r.$

Solving for the radius, nosotros observe

$r=\frac{(134.ane\,\text{m}\text{/}{\text{south})}^{two}}{9.8\,\text{m}\text{/}{\text{southward}}^{2}}=1835\,\text{yard}=one.835\,\text{km}\text{.}$

#### Significance

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

A flywheel has a radius of xx.0 cm. What is the speed of a point on the border of the flywheel if it experiences a centripetal acceleration of $900.0\,{\text{cm}\text{/}\text{s}}^{ii}?$

Centripetal acceleration can take a wide range of values, depending on the speed and radius of curvature of the round path. Typical centripetal accelerations are given in the following tabular array.

Typical Centripetal Accelerations
Object Centripetal Dispatch (one thousand/stwo
or factors of
g)
Earth around the Lord’s day $5.93\,×\,{10}^{-3}$
Moon around the Earth $ii.73\,×\,{10}^{-3}$
Satellite in geosynchronous orbit 0.233
Outer edge of a CD when playing $5.78$
Jet in a butt roll (2–3
g)
Roller coaster (5
grand)
Electron orbiting a proton in a simple Bohr model of the cantlet $ix.0\,×\,{10}^{22}$

### Equations of Motion for Compatible Circular Motion

A particle executing circular movement can exist described past its position vector $\overset{\to }{r}(t).$ (Figure) shows a particle executing circular move in a counterclockwise management. As the particle moves on the circle, its position vector sweeps out the angle $\theta$ with the
x-centrality. Vector $\overset{\to }{r}(t)$ making an bending $\theta$ with the
ten-axis is shown with its components along the
x– and
y-axes. The magnitude of the position vector is $A=|\overset{\to }{r}(t)|$ and is as well the radius of the circumvolve, and then that in terms of its components,

$\overset{\to }{r}(t)=A\,\text{cos}\,\omega t\lid{i}+A\,\text{sin}\,\omega t\chapeau{j}.$

Here, $\omega$ is a abiding called the
angular frequency

of the particle. The angular frequency has units of radians (rad) per 2nd and is simply the number of radians of angular mensurate through which the particle passes per 2d. The angle $\theta$ that the position vector has at any particular fourth dimension is $\omega t$.

If
T
is the period of motion, or the time to complete one revolution ($two\pi$ rad), then

$\omega =\frac{2\pi }{T}.$

Figure four.twenty
The position vector for a particle in round motion with its components along the x- and y-axes. The particle moves counterclockwise. Angle $\theta$ is the angular frequency $\omega$ in radians per second multiplied by t.

Velocity and dispatch tin exist obtained from the position part by differentiation:

$\overset{\to }{v}(t)=\frac{d\overset{\to }{r}(t)}{dt}=\text{−}A\omega \,\text{sin}\,\omega t\hat{i}+A\omega \,\text{cos}\,\omega t\hat{j}.$

It can be shown from (Figure) that the velocity vector is tangential to the circumvolve at the location of the particle, with magnitude $A\omega .$ Similarly, the acceleration vector is found by differentiating the velocity:

$\overset{\to }{a}(t)=\frac{d\overset{\to }{v}(t)}{dt}=\text{−}A{\omega }^{2}\,\text{cos}\,\omega t\hat{i}-A{\omega }^{2}\,\text{sin}\,\omega t\hat{j}.$

From this equation we see that the acceleration vector has magnitude $A{\omega }^{2}$ and is directed opposite the position vector, toward the origin, because $\overset{\to }{a}(t)=\text{−}{\omega }^{two}\overset{\to }{r}(t).$

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### Example

#### Round Move of a Proton

A proton has speed $five\,×\,{10}^{6}\,\text{m/southward}$ and is moving in a circle in the
xy
r
= 0.175 thou. What is its position in the
xy
aeroplane at time $t=2.0\,×\,{10}^{-seven}\,\text{s}=200\,\text{ns?}$ At
t
= 0, the position of the proton is $0.175\,\text{m}\lid{i}$ and it circles counterclockwise. Sketch the trajectory.

#### Solution

From the given data, the proton has flow and athwart frequency:

$T=\frac{two\pi r}{five}=\frac{2\pi (0.175\,\text{m})}{v.0\,×\,{ten}^{6}\,\text{k}\text{/}\text{s}}=2.20\,×\,{10}^{-seven}\,\text{s}$

$\omega =\frac{two\pi }{T}=\frac{2\pi }{2.20\,×\,{x}^{-7}\,\text{s}}=2.856\,×\,{10}^{7}\,\text{rad}\text{/}\text{s}.$

The position of the particle at $t=2.0\,×\,{ten}^{-seven}\,\text{due south}$ with
A
= 0.175 m is

$\begin{array}{cc}\hfill \overset{\to }{r}(two.0\,×\,{10}^{-seven}\text{s})& =A\,\text{cos}\,\omega (two.0\,×\,{10}^{-7}\,\text{s})\lid{i}+A\,\text{sin}\,\omega (two.0\,×\,{ten}^{-7}\,\text{southward})\chapeau{j}\,\text{k}\hfill \\ & =0.175\text{cos}[(2.856\,×\,{ten}^{7}\,\text{rad}\text{/}\text{s})(2.0\,×\,{10}^{-vii}\,\text{s})]\lid{i}\hfill \\ & \hfill +0.175\text{sin}[(2.856\,×\,{x}^{vii}\,\text{rad}\text{/}\text{s})(two.0\,×\,{10}^{-7}\,\text{s})]\chapeau{j}\,\text{chiliad}\\ & =0.175\text{cos}(5.712\,\text{rad})\hat{i}+0.175\text{sin}(5.712\,\text{rad})\hat{j}=0.147\chapeau{i}-0.095\hat{j}\,\text{m}.\hfill \finish{assortment}$

From this upshot nosotros encounter that the proton is located slightly below the
x-centrality. This is shown in (Figure).

Effigy four.21
Position vector of the proton at $t=ii.0\,×\,{x}^{-7}\text{s}=200\,\text{ns}\text{.}$ The trajectory of the proton is shown. The angle through which the proton travels along the circumvolve is 5.712 rad, which a little less than i complete revolution.

#### Significance

We picked the initial position of the particle to exist on the
x-axis. This was completely arbitrary. If a different starting position were given, we would have a dissimilar final position at
t
= 200 ns.

### Nonuniform Circular Move

Circular motion does not take to be at a abiding speed. A particle can travel in a circle and speed upwardly or slow down, showing an acceleration in the direction of the motility.

In compatible circular motion, the particle executing circular motion has a constant speed and the circle is at a fixed radius. If the speed of the particle is changing as well, and then we introduce an boosted acceleration in the management tangential to the circle. Such accelerations occur at a point on a meridian that is irresolute its spin rate, or whatsoever accelerating rotor. In Displacement and Velocity Vectors we showed that centripetal acceleration is the time rate of change of the management of the velocity vector. If the speed of the particle is changing, and so it has a
tangential acceleration
that is the time rate of change of the magnitude of the velocity:

${a}_{\text{T}}=\frac{d|\overset{\to }{5}|}{dt}.$

The direction of tangential acceleration is tangent to the circle whereas the direction of centripetal acceleration is radially inwards toward the center of the circle. Thus, a particle in round motion with a tangential acceleration has a
total acceleration
that is the vector sum of the centripetal and tangential accelerations:

$\overset{\to }{a}={\overset{\to }{a}}_{\text{C}}+{\overset{\to }{a}}_{\text{T}}.$

The acceleration vectors are shown in (Figure). Note that the ii dispatch vectors ${\overset{\to }{a}}_{\text{C}}$ and ${\overset{\to }{a}}_{\text{T}}$ are perpendicular to each other, with ${\overset{\to }{a}}_{\text{C}}$ in the radial direction and ${\overset{\to }{a}}_{\text{T}}$ in the tangential direction. The total dispatch $\overset{\to }{a}$ points at an angle between ${\overset{\to }{a}}_{\text{C}}$ and ${\overset{\to }{a}}_{\text{T}}.$

Effigy 4.22
The centripetal acceleration points toward the center of the circumvolve. The tangential acceleration is tangential to the circumvolve at the particle’s position. The total acceleration is the vector sum of the tangential and centripetal accelerations, which are perpendicular.

### Example

#### Total Acceleration during Circular Motion

A particle moves in a circumvolve of radius
r
= 2.0 m. During the time interval from
t
= ane.5 s to
t
= 4.0 s its speed varies with time co-ordinate to

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$v(t)={c}_{1}-\frac{{c}_{2}}{{t}^{2}},\enspace{c}_{1}=4.0\,\text{m}\text{/}\text{s,}\enspace{c}_{2}=6.0\,\text{one thousand}·\text{southward}\text{.}$

What is the total acceleration of the particle at
t
= 2.0 s?

#### Strategy

Nosotros are given the speed of the particle and the radius of the circle, so we tin summate centripetal acceleration hands. The direction of the centripetal acceleration is toward the heart of the circumvolve. We find the magnitude of the tangential acceleration past taking the derivative with respect to fourth dimension of $|v(t)|$ using (Figure) and evaluating it at
t
= 2.0 due south. We utilize this and the magnitude of the centripetal dispatch to notice the total acceleration.

#### Solution

Centripetal dispatch is

$5(2.0\text{s})=(4.0-\frac{six.0}{{(2.0)}^{2}})\text{chiliad}\text{/}\text{south}=two.5\,\text{m}\text{/}\text{s}$

${a}_{\text{C}}=\frac{{v}^{2}}{r}=\frac{(two.5\,\text{yard}\text{/}{\text{s})}^{2}}{2.0\,\text{m}}=3.1\,\text{g}\text{/}{\text{due south}}^{ii}$

directed toward the center of the circle. Tangential acceleration is

${a}_{\text{T}}=|\frac{d\overset{\to }{five}}{dt}|=\frac{2{c}_{2}}{{t}^{3}}=\frac{12.0}{{(2.0)}^{3}}\text{grand}\text{/}{\text{s}}^{two}=ane.5\,\text{m}\text{/}{\text{due south}}^{ii}.$

Full acceleration is

$|\overset{\to }{a}|=\sqrt{{iii.i}^{2}+{1.5}^{2}}\text{m}\text{/}{\text{due south}}^{2}=iii.44\,\text{one thousand}\text{/}{\text{s}}^{two}$

and $\theta ={\text{tan}}^{-1}\frac{3.1}{1.5}=64\text{°}$ from the tangent to the circle. See (Figure).

Figure 4.23
The tangential and centripetal acceleration vectors. The net acceleration $\overset{\to }{a}$ is the vector sum of the two accelerations.

#### Significance

The directions of centripetal and tangential accelerations can exist described more than conveniently in terms of a polar coordinate system, with unit vectors in the radial and tangential directions. This coordinate system, which is used for motion along curved paths, is discussed in particular later in the book.

### Summary

• Uniform circular motion is move in a circle at constant speed.
• Centripetal acceleration ${\overset{\to }{a}}_{\text{C}}$ is the acceleration a particle must have to follow a circular path. Centripetal acceleration always points toward the center of rotation and has magnitude ${a}_{\text{C}}={v}^{2}\text{/}r.$
• Nonuniform circular motion occurs when there is tangential acceleration of an object executing circular movement such that the speed of the object is changing. This acceleration is called tangential acceleration ${\overset{\to }{a}}_{\text{T}}.$ The magnitude of tangential dispatch is the time rate of change of the magnitude of the velocity. The tangential acceleration vector is tangential to the circumvolve, whereas the centripetal acceleration vector points radially inward toward the middle of the circumvolve. The total acceleration is the vector sum of tangential and centripetal accelerations.
• An object executing uniform circular motion can be described with equations of motion. The position vector of the object is $\overset{\to }{r}(t)=A\,\text{cos}\,\omega t\hat{i}+A\,\text{sin}\,\omega t\chapeau{j},$ where
A
is the magnitude $|\overset{\to }{r}(t)|,$ which is also the radius of the circumvolve, and $\omega$ is the angular frequency.

### Conceptual Questions

Can centripetal dispatch modify the speed of a particle undergoing circular motion?

Tin tangential acceleration change the speed of a particle undergoing circular move?

### Issues

A flywheel is rotating at thirty rev/south. What is the total angle, in radians, through which a point on the flywheel rotates in 40 due south?

A particle travels in a circle of radius ten 1000 at a constant speed of twenty m/south. What is the magnitude of the acceleration?

${a}_{\text{C}}=forty\,\text{m}\text{/}{\text{s}}^{two}$

Cam Newton of the Carolina Panthers throws a perfect football spiral at viii.0 rev/s. The radius of a pro football is eight.five cm at the heart of the short side. What is the centripetal acceleration of the laces on the football?

A fairground ride spins its occupants inside a flying saucer-shaped container. If the horizontal round path the riders follow has an 8.00-yard radius, at how many revolutions per minute are the riders subjected to a centripetal dispatch equal to that of gravity?

A runner taking office in the 200-thou dash must run around the finish of a runway that has a round arc with a radius of curvature of 30.0 m. The runner starts the race at a constant speed. If she completes the 200-m dash in 23.2 s and runs at abiding speed throughout the race, what is her centripetal dispatch as she runs the curved portion of the rails?

What is the acceleration of Venus toward the Sunday, assuming a circular orbit?

An experimental jet rocket travels around Earth along its equator only above its surface. At what speed must the jet travel if the magnitude of its dispatch is
g?

A fan is rotating at a constant 360.0 rev/min. What is the magnitude of the acceleration of a point on ane of its blades 10.0 cm from the axis of rotation?

A point located on the second hand of a big clock has a radial dispatch of $0.1\text{cm}\text{/}{\text{s}}^{2}.$ How far is the point from the axis of rotation of the 2nd hand?

### Glossary

angular frequency
$\omega ,$ rate of modify of an angle with which an object that is moving on a circular path
centripetal acceleration
component of acceleration of an object moving in a circle that is directed radially in toward the center of the circumvolve
tangential acceleration
magnitude of which is the time rate of change of speed. Its direction is tangent to the circle.
total acceleration
vector sum of centripetal and tangential accelerations

## In Uniform Circular Motion Which of the Following Are Constant

Source: https://courses.lumenlearning.com/suny-osuniversityphysics/chapter/4-4-uniform-circular-motion/

Originally posted 2022-08-01 20:42:52.

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