Centripetal Acceleration

In one-dimensional kinematics, objects with a constant speed have zero acceleration. However, in two- and three-dimensional kinematics, even 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 (Figure). As 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 constant magnitude and is tangent to the path as it changes from $$ \overset{\to }{v}(t) $$ to $$ \overset{\to }{v}(t+\text{Δ}t), $$ 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 $$ \text{Δ}\overset{\to }{v} $$ are similar. Furthermore, since $$ |\overset{\to }{r}(t)|=|\overset{\to }{r}(t+\text{Δ}t)| $$ and $$ |\overset{\to }{v}(t)|=|\overset{\to }{v}(t+\text{Δ}t)|, $$ the two triangles are isosceles. From these facts we can make the assertion

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

Figure 4.18 (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 figure are similar. The vector $$ \text{Δ}\overset{\to }{v} $$ points toward the center of the circle in the limit $$ \text{Δ}t\to 0.$$

We can find the magnitude of the acceleration from

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

The direction of the acceleration vector is toward the center of the circle ((Figure)). This is a radial acceleration and is called the centripetal acceleration, which is why we give it the subscript c. The word centripetal comes from the Latin words centrum (meaning “center”) and petere (meaning to seek”), and thus takes the meaning “center seeking.”

Figure 4.19 The centripetal acceleration vector points toward the center of the circular path of motion and is an acceleration in the radial direction. The velocity vector is also shown and is tangent to the circle.

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

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.

Equations of Motion for Uniform Circular Motion

A particle executing circular motion can be described by its position vector $$ \overset{\to }{r}(t). $$ (Figure) shows a particle executing circular motion in a counterclockwise direction. As the particle moves on the circle, its position vector sweeps out the angle $$ \theta $$ with the x-axis. Vector $$ \overset{\to }{r}(t) $$ making an angle $$ \theta $$ with the x-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 also the radius of the circle, so that in terms of its components,

Here, $$ \omega $$ 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 $$ \theta $$ that the position vector has at any particular time is $$ \omega t$$.

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

Figure 4.20 The position vector for a particle in circular 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 acceleration can be obtained from the position function by differentiation:

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

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 $$ \overset{\to }{a}(t)=\text{−}{\omega }^{2}\overset{\to }{r}(t).$$

Example

Circular Motion of a Proton

A proton has speed $$ 5\,×\,{10}^{6}\,\text{m/s} $$ and is moving in a circle in the xy plane of radius r = 0.175 m. What is its position in the xy plane at time $$ t=2.0\,×\,{10}^{-7}\,\text{s}=200\,\text{ns?} $$ At t = 0, the position of the proton is $$ 0.175\,\text{m}\hat{i} $$ and it circles counterclockwise. Sketch the trajectory.

Solution

From the given data, the proton has period and angular frequency:

$$T=\frac{2\pi r}{v}=\frac{2\pi (0.175\,\text{m})}{5.0\,×\,{10}^{6}\,\text{m}\text{/}\text{s}}=2.20\,×\,{10}^{-7}\,\text{s}$$

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

The position of the particle at $$ t=2.0\,×\,{10}^{-7}\,\text{s} $$ with A = 0.175 m is

$$\begin{array}{cc}\hfill \overset{\to }{r}(2.0\,×\,{10}^{-7}\text{s})& =A\,\text{cos}\,\omega (2.0\,×\,{10}^{-7}\,\text{s})\hat{i}+A\,\text{sin}\,\omega (2.0\,×\,{10}^{-7}\,\text{s})\hat{j}\,\text{m}\hfill \\ & =0.175\text{cos}[(2.856\,×\,{10}^{7}\,\text{rad}\text{/}\text{s})(2.0\,×\,{10}^{-7}\,\text{s})]\hat{i}\hfill \\ & \hfill +0.175\text{sin}[(2.856\,×\,{10}^{7}\,\text{rad}\text{/}\text{s})(2.0\,×\,{10}^{-7}\,\text{s})]\hat{j}\,\text{m}\\ & =0.175\text{cos}(5.712\,\text{rad})\hat{i}+0.175\text{sin}(5.712\,\text{rad})\hat{j}=0.147\hat{i}-0.095\hat{j}\,\text{m}.\hfill \end{array}$$

From this result we see that the proton is located slightly below the x-axis. This is shown in (Figure).

Figure 4.21 Position vector of the proton at $$ t=2.0\,×\,{10}^{-7}\text{s}=200\,\text{ns}\text{.} $$ The trajectory of the proton is shown. The angle through which the proton travels along the circle is 5.712 rad, which a little less than one complete revolution.

Significance

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

Nonuniform Circular Motion

Circular motion does not have to be at a constant speed. A particle can travel in a circle and speed up or slow down, showing an acceleration in the direction of the motion.

In uniform 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, then we introduce an additional acceleration in the direction tangential to the circle. Such accelerations occur at a point on a top that is changing its spin rate, or any accelerating rotor. In Displacement and Velocity Vectors we showed that centripetal acceleration is the time rate of change of the direction of the velocity vector. If the speed of the particle is changing, then it has a tangential acceleration that is the time rate of change of the magnitude of the velocity:

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

The acceleration vectors are shown in (Figure). Note that the two acceleration 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 acceleration $$ \overset{\to }{a} $$ points at an angle between $$ {\overset{\to }{a}}_{\text{C}} $$ and $$ {\overset{\to }{a}}_{\text{T}}.$$

Figure 4.22 The centripetal acceleration points toward the center of the circle. The tangential acceleration is tangential to the circle 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 circle of radius r = 2.0 m. During the time interval from t = 1.5 s to t = 4.0 s its speed varies with time according to

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

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

Strategy

We are given the speed of the particle and the radius of the circle, so we can calculate centripetal acceleration easily. The direction of the centripetal acceleration is toward the center of the circle. We find the magnitude of the tangential acceleration by taking the derivative with respect to time of $$ |v(t)| $$ using (Figure) and evaluating it at t = 2.0 s. We use this and the magnitude of the centripetal acceleration to find the total acceleration.

Solution

Centripetal acceleration is

$$v(2.0\text{s})=(4.0-\frac{6.0}{{(2.0)}^{2}})\text{m}\text{/}\text{s}=2.5\,\text{m}\text{/}\text{s}$$

$${a}_{\text{C}}=\frac{{v}^{2}}{r}=\frac{(2.5\,\text{m}\text{/}{\text{s})}^{2}}{2.0\,\text{m}}=3.1\,\text{m}\text{/}{\text{s}}^{2}$$

directed toward the center of the circle. Tangential acceleration is

$${a}_{\text{T}}=|\frac{d\overset{\to }{v}}{dt}|=\frac{2{c}_{2}}{{t}^{3}}=\frac{12.0}{{(2.0)}^{3}}\text{m}\text{/}{\text{s}}^{2}=1.5\,\text{m}\text{/}{\text{s}}^{2}.$$

Total acceleration is

$$|\overset{\to }{a}|=\sqrt{{3.1}^{2}+{1.5}^{2}}\text{m}\text{/}{\text{s}}^{2}=3.44\,\text{m}\text{/}{\text{s}}^{2}$$

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 be described more 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 detail later in the book.