Chapter 5 Uniform Circular Motion and Gravitation

5.1 Centripetal Acceleration

Summary

  • Establish the expression for centripetal acceleration.
  • Explain the centrifuge.

We know from kinematics that acceleration is a change in velocity, either in its magnitude or in its direction, or both. In uniform circular motion, the direction of the velocity changes constantly, so there is always an associated acceleration, even though the magnitude of the velocity might be constant. You experience this acceleration yourself when you turn a corner in your car. (If you hold the wheel steady during a turn and move at constant speed, you are in uniform circular motion.) What you notice is a sideways acceleration because you and the car are changing direction. The sharper the curve and the greater your speed, the more noticeable this acceleration will become. In this section we examine the direction and magnitude of that acceleration.

Figure 1 shows an object moving in a circular path at constant speed. The direction of the instantaneous velocity is shown at two points along the path. Acceleration is in the direction of the change in velocity, which points directly toward the center of rotation (the center of the circular path). This pointing is shown with the vector diagram in the figure. We call the acceleration of an object moving in uniform circular motion (resulting from a net external force) the centripetal acceleration (ac); centripetal means “toward the center” or “center seeking.”

The given figure shows a circle, with a triangle having vertices A B C made from the center to the boundry. A is at the center and B and C points are at the circle path. Lines A B and A C act as radii and B C is a chord. Delta theta is shown inside the triangle, and the arc length delta s and the chord length delta r are also given. At point B, velocity of object is shown as v one and at point C, velocity of object is shown as v two. Along the circle an equation is shown as delta v equals v sub 2 minus v sub 1.
Figure 1. The directions of the velocity of an object at two different points are shown, and the change in velocity Δv is seen to point directly toward the center of curvature. (See small inset.) Because ac = Δv/Δt, the acceleration is also toward the center; ac is called centripetal acceleration. (Because Δθ is very small, the arc length Δs is equal to the chord length Δr for small time differences.)

The direction of centripetal acceleration is toward the center of curvature, but what is its magnitude? Note that the triangle formed by the velocity vectors and the one formed by the radii r and Δs are similar. Both the triangles ABC and PQR are isosceles triangles (two equal sides). The two equal sides of the velocity vector triangle are the speeds v1 = v2 = v. Using the properties of two similar triangles, we obtain

[latex]\boldsymbol{\frac{\Delta{v}}{v}}[/latex][latex]\boldsymbol{=}[/latex][latex]\boldsymbol{\frac{\Delta{s}}{r}}.[/latex]

Acceleration is Δv/Δt, and so we first solve this expression for Δv:

[latex]\boldsymbol{\Delta{v}\:=}[/latex][latex]\boldsymbol{\frac{v}{r}}[/latex][latex]\boldsymbol{\Delta{s}}.[/latex]

Then we divide this by Δt, yielding

[latex]\boldsymbol{\frac{\Delta{v}}{\Delta{t}}}[/latex][latex]\boldsymbol{=}[/latex][latex]\boldsymbol{\frac{v}{r}\times\frac{\Delta{s}}{\Delta{t}}}.[/latex]

Finally, noting that Δv/Δt = ac and that Δs/Δt = v, the linear or tangential speed, we see that the magnitude of the centripetal acceleration is

[latex]\boldsymbol{a_{\textbf{c}}\:=}[/latex][latex]\boldsymbol{\frac{v^2}{r}},[/latex]

which is the acceleration of an object in a circle of radius r at a speed v. So, centripetal acceleration is greater at high speeds and in sharp curves (smaller radius), as you have noticed when driving a car. But it is a bit surprising that ac is proportional to speed squared, implying, for example, that it is four times as hard to take a curve at 100 km/h than at 50 km/h. A sharp corner has a small radius, so that ac is greater for tighter turns, as you have probably noticed.

A centrifuge (see Figure 2b) is a rotating device used to separate specimens of different densities. High centripetal acceleration significantly decreases the time it takes for separation to occur, and makes separation possible with small samples. Centrifuges are used in a variety of applications in science and medicine, including the separation of single cell suspensions such as bacteria, viruses, and blood cells from a liquid medium and the separation of macromolecules, such as DNA and protein, from a solution. Centrifuges are often rated in terms of their centripetal acceleration relative to acceleration due to gravity (g); maximum centripetal acceleration of several hundred thousand g is possible in a vacuum. Human centrifuges, extremely large centrifuges, have been used to test the tolerance of astronauts to the effects of accelerations larger than that of Earth’s gravity.

Example 1: How Does the Centripetal Acceleration of a Car Around a Curve Compare with That Due to Gravity?

What is the magnitude of the centripetal acceleration of a car following a curve of radius 500 m at a speed of 25.0 m/s (about 90 km/h)? Compare the acceleration with that due to gravity for this fairly gentle curve taken at highway speed. See Figure 2(a).

Strategy

Because v and r are given, we use the expression [latex]\boldsymbol{a_{\textbf{c}}=\dfrac{v^2}{r}}[/latex].

Solution

Entering the given values of v = 25.0 m/s and r = 500 m into the first expression for ac gives

[latex]\boldsymbol{a_\textbf{c}} = \boldsymbol{\dfrac{v^2}{r}} = \boldsymbol{\dfrac{25.0^2}{500} = 1.25}~\textbf{m/s}^2[/latex]

Discussion

To compare this with the acceleration due to gravity (g = 9.80 m/s2), we take the ratio of ac /g = (1.25 m/s2)/(9.80m/s2) = 0.128. Thus, ac = 0.128 g and is noticeable especially if you were not wearing a seat belt.

In figure a, a car shown from top is running on a circular road around a circular path. The center of the park is termed as the center of this circle and the distance from this point to the car is taken as radius r. The linear velocity is shown in perpendicular direction toward the front of the car, shown as v the centripetal acceleration is shown with an arrow pointed towards the center of rotation. In figure b, a centrifuge is shown an object of mass m is rotating in it at a constant speed. The object is at the distance equal to the radius, r, of the centrifuge. The centripetal acceleration is shown towards the center of rotation, and the velocity, v is shown perpendicular to the object in the clockwise direction.
Figure 2. (a) The car following a circular path at constant speed is accelerated perpendicular to its velocity, as shown. The magnitude of this centripetal acceleration is found in Example 1. (b) A particle of mass in a centrifuge is rotating at constant angular velocity . It must be accelerated perpendicular to its velocity or it would continue in a straight line. The magnitude of the necessary acceleration is found in Example 2.

Example 2: What is the speed of an Ultracentrifuge?

Calculate the speed of a point 7.50 cm from the axis of a spinning ultracentrifuge if the centripetal acceleration at that point is [latex]\boldsymbol{4.63\times10^6}~\textbf{m/s}^2[/latex]

Strategy

Both r and ac are given, so we can use the equation [latex]\boldsymbol{a_{\textbf{c}} = \dfrac{v^2}{r}}[/latex] to determine the speed of the centrifuge.

Solution

First we solve for the speed of the centrifugre

[latex]\boldsymbol{a_{\textbf{c}} = \dfrac{v^2}{r} \rightarrow v =\sqrt{a_{\textbf{c}}r}}[/latex]

Converting 7.50 cm to meters and substituting known values gives

[latex]\boldsymbol{v =\sqrt{a_{\textbf{c}}r}=\sqrt{(4.63\times10^6)(0.0750)} = 590~\textbf{m/s}}.[/latex]

Discussion

The given centripetal acceleration is 472,000 times as strong as g. It is no wonder that such high velocity centrifuges are called ultracentrifuges. The extremely large accelerations involved greatly decrease the time needed to cause the sedimentation of blood cells or other materials.

Of course, a net external force is needed to cause any acceleration, just as Newton proposed in his second law of motion. So a net external force is needed to cause a centripetal acceleration. In the next section on centripetal force, we will consider the forces involved in circular motion.

PHET EXPLORATIONS: LADYBUG MOTION 2D

Learn about position, velocity and acceleration vectors. Move the ladybug by setting the position, velocity or acceleration, and see how the vectors change. Choose linear, circular or elliptical motion, and record and playback the motion to analyze the behavior.

image
Figure 3. Ladybug Motion in 2D

Section Summary

  • Centripetal acceleration ac is the acceleration experienced while in uniform circular motion. It always points toward the center of rotation. It is perpendicular to the linear velocity[latex]\boldsymbol{v}[/latex]and has the magnitude
[latex]\boldsymbol{a_{\textbf{c}}\:=}[/latex][latex]\boldsymbol{\frac{v^2}{r}}[/latex]
  • The unit of centripetal acceleration is m/s2.

Conceptual Questions

1: Can centripetal acceleration change the speed of circular motion? Explain.

Problems & Exercises

1: A fairground ride spins its occupants inside a flying saucer-shaped container. If the horizontal circular path the riders follow has an 8.00 m radius, at how many revolutions per minute will the riders be subjected to a centripetal acceleration whose magnitude is 1.50 times that due to gravity?

2: A runner taking part in the 200 m dash must run around the end of a track that has a circular arc with a radius of curvature of 30 m. If he completes the 200 m dash in 23.2 s and runs at constant speed throughout the race, what is the magnitude of his centripetal acceleration as he runs the curved portion of the track?

3: Taking the age of Earth to be about 4 × 109 years and assuming its orbital radius of 1.5 × 1011 has not changed and is circular, calculate the approximate total distance Earth has traveled since its birth (in a frame of reference stationary with respect to the Sun). Recall that the circumference of a circle is [latex]2\pi r[/latex].

4: The propeller of a World War II fighter plane is 2.30 m in diameter and the tip of the blade spins at 145 m/s. What is the centripetal acceleration of the propeller tip under these conditions? Calculate it in meters per second squared and convert to multiples of g.

5: Integrated Concepts

Riders in an amusement park ride shaped like a Viking ship hung from a large pivot are rotated back and forth like a rigid pendulum. Sometime near the middle of the ride, the ship is momentarily motionless at the top of its circular arc. The ship then swings down under the influence of gravity.

(a) Assuming negligible friction, find the speed of the riders at the bottom of its arc, given the system’s center of mass travels in an arc having a radius of 14.0 m and the riders are near the center of mass.

(b) What is the centripetal acceleration at the bottom of the arc?

(c) Draw a free body diagram of the forces acting on a rider at the bottom of the arc.

(d) Find the force exerted by the ride on a 60.0 kg rider and compare it to her weight.

(e) Discuss whether the answer seems reasonable.

6: Unreasonable Results

A mother pushes her child on a swing so that his speed is 9.00 m/s at the lowest point of his path. The swing is suspended 2.00 m above the child’s center of mass.

(a) What is the magnitude of the centripetal acceleration of the child at the low point?

(b) What is the magnitude of the force the child exerts on the seat if his mass is 18.0 kg?

(c) What is unreasonable about these results?

(d) Which premises are unreasonable or inconsistent?

Glossary

centripetal acceleration
the acceleration of an object moving in a circle, directed toward the center
ultracentrifuge
a centrifuge optimized for spinning a rotor at very high speeds

Solutions

Problems & Exercises

1: [latex]\boldsymbol{10.8~\textbf{m/s}}[/latex]

3: [latex]\boldsymbol{4\times10^{21}\textbf{ m}}[/latex]

5: (a) [latex]\boldsymbol{23.4\textbf{ m/s}}[/latex] (b) [latex]\boldsymbol{39.1\textbf{ m/s}^2}[/latex] (c)

A rectangle with a base longer than the height. A vertical line with arrowheads on both ends passes through the rectangle, bisecting the horizontal sides. The top of the arrow is labeled N, and the bottom is labeled w.
Figure 4.

(d) [latex]\boldsymbol{1.76\times10^3\textbf{ N}\text{ or }3.00\:\omega},[/latex] that is, the normal force (upward) is three times her weight. (e) This answer seems reasonable, since she feels like she’s being forced into the chair MUCH stronger than just by gravity.

6: (a) [latex]\boldsymbol{40.5\textbf{ m/s}^2}[/latex] (b) 905 N (c) The force in part (b) is very large. The acceleration in part (a) is too much, about 4 g. (d) The speed of the swing is too large. At the given velocity at the bottom of the swing, there is enough kinetic energy to send the child all the way over the top, ignoring friction.

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