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Learning Objectives

By the end of this section, you will be able to:

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

Uniform circular motion is a specific type of motion in which an object travels in a circle with a constant speed. For example, any point on a propeller spinning at a constant rate is executing uniform circular motion. Other examples are the second, minute, and hour hands of a watch. It is remarkable that points on these rotating objects are actually accelerating, although the rotation rate is a constant. To see this, we must analyze the motion in terms of vectors.

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 [latex] doverset{to }{v}text{/}dtne 0. [/latex] This is shown in (Figure). As the particle moves counterclockwise in time [latex] text{Δ}t [/latex] on the circular path, its position vector moves from [latex] overset{to }{r}(t) [/latex] to [latex] overset{to }{r}(t+text{Δ}t). [/latex] The velocity vector has constant magnitude and is tangent to the path as it changes from [latex] overset{to }{v}(t) [/latex] to [latex] overset{to }{v}(t+text{Δ}t), [/latex] changing its direction only. Since the velocity vector [latex] overset{to }{v}(t) [/latex] is perpendicular to the position vector [latex] overset{to }{r}(t), [/latex] the triangles formed by the position vectors and [latex] text{Δ}overset{to }{r}, [/latex] and the velocity vectors and [latex] text{Δ}overset{to }{v} [/latex] are similar. Furthermore, since [latex] |overset{to }{r}(t)|=|overset{to }{r}(t+text{Δ}t)| [/latex] and [latex] |overset{to }{v}(t)|=|overset{to }{v}(t+text{Δ}t)|, [/latex] the two triangles are isosceles. From these facts we can make the assertion

[latex] frac{text{Δ}v}{v}=frac{text{Δ}r}{r} [/latex] or [latex] text{Δ}v=frac{v}{r}text{Δ}r. [/latex]

Figure 4.18 (a) A particle is moving in a circle at a constant speed, with position and velocity vectors at times [latex] t [/latex] and [latex] t+text{Δ}t. [/latex] (b) Velocity vectors forming a triangle. The two triangles in the figure are similar. The vector [latex] text{Δ}overset{to }{v} [/latex] points toward the center of the circle in the limit [latex] text{Δ}tto 0. [/latex]

We can find the magnitude of the acceleration from

[latex] a=underset{text{Δ}tto 0}{text{lim}}(frac{text{Δ}v}{text{Δ}t})=frac{v}{r}(underset{text{Δ}tto 0}{text{lim}}frac{text{Δ}r}{text{Δ}t})=frac{{v}^{2}}{r}. [/latex]

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

[latex] {a}_{text{C}}=frac{{v}^{2}}{r}. [/latex]

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.

Example

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: [latex] 9.8{,text{m/s}}^{2}={v}^{2}text{/}r. [/latex]

Solving for the radius, we find

[latex] r=frac{(134.1,text{m}text{/}{text{s})}^{2}}{9.8,text{m}text{/}{text{s}}^{2}}=1835,text{m}=1.835,text{km}text{.} [/latex]

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.

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 [latex] 900.0,{text{cm}text{/}text{s}}^{2}? [/latex]

Show Solution

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/s² or factors of g)
Earth around the Sun	[latex] 5.93,×,{10}^{-3} [/latex]
Moon around the Earth	[latex] 2.73,×,{10}^{-3} [/latex]
Satellite in geosynchronous orbit	0.233
Outer edge of a CD when playing	[latex] 5.78 [/latex]
Jet in a barrel roll	(2–3 g)
Roller coaster	(5 g)
Electron orbiting a proton in a simple Bohr model of the atom	[latex] 9.0,×,{10}^{22} [/latex]

Equations of Motion for Uniform Circular Motion

A particle executing circular motion can be described by its position vector [latex] overset{to }{r}(t). [/latex] (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 [latex] theta [/latex] with the x-axis. Vector [latex] overset{to }{r}(t) [/latex] making a

Chapter 3uniform Circular Motionmr.'s Learning Website