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

Δ v v = Δ r r or Δ v = v r Δ r .

Figure a shows a circle with center at point C. We are shown radius r of t and radius r of t, which are an angle Delta theta apart, and the chord length delta r connecting the ends of the two radii. Vectors r of t, r of t plus delta t, and delta r form a triangle. At the tip of vector r of t, the velocity is shown as v of t and points up and to the right, tangent to the circle. . At the tip of vector r of t plus delta t, the velocity is shown as v of t plus delta t and points up and to the left, tangent to the circle. Figure b shows the vectors v of t and v of t plus delta t with their tails together, and the vector delta v from the tip of v of t to the tip of v of t plus delta t. These three vectors form a triangle. The angle between the v of t and v of t plus delta t is theta.
(a) A particle is moving in a circle at a constant speed, with position and velocity vectors at times t and t + Δ t . (b) Velocity vectors forming a triangle. The two triangles in the figure are similar. The vector Δ v points toward the center of the circle in the limit Δ t 0 .

We can find the magnitude of the acceleration from

a = lim Δ t 0 ( Δ v Δ t ) = v r ( lim Δ t 0 Δ r Δ t ) = v 2 r .

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

a C = v 2 r .

The direction of the acceleration vector is toward the center of the circle ( [link] ). 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.”

A circle is shown with a purple arrow labeled as vector a sub c pointing radially inward and a green arrow tangent to the circle and labeled v. The arrows are shown with their tails at the same point on the circle.
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.

Questions & Answers

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
Aislinn Reply
cm
tijani
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John Reply
what is physics
Siyaka Reply
A mouse of mass 200 g falls 100 m down a vertical mine shaft and lands at the bottom with a speed of 8.0 m/s. During its fall, how much work is done on the mouse by air resistance
Jude Reply
Can you compute that for me. Ty
Jude
what is the dimension formula of energy?
David Reply
what is viscosity?
David
what is inorganic
emma Reply
what is chemistry
Youesf Reply
what is inorganic
emma
Chemistry is a branch of science that deals with the study of matter,it composition,it structure and the changes it undergoes
Adjei
please, I'm a physics student and I need help in physics
Adjanou
chemistry could also be understood like the sexual attraction/repulsion of the male and female elements. the reaction varies depending on the energy differences of each given gender. + masculine -female.
Pedro
A ball is thrown straight up.it passes a 2.0m high window 7.50 m off the ground on it path up and takes 1.30 s to go past the window.what was the ball initial velocity
Krampah Reply
2. A sled plus passenger with total mass 50 kg is pulled 20 m across the snow (0.20) at constant velocity by a force directed 25° above the horizontal. Calculate (a) the work of the applied force, (b) the work of friction, and (c) the total work.
Sahid Reply
you have been hired as an espert witness in a court case involving an automobile accident. the accident involved car A of mass 1500kg which crashed into stationary car B of mass 1100kg. the driver of car A applied his brakes 15 m before he skidded and crashed into car B. after the collision, car A s
Samuel Reply
can someone explain to me, an ignorant high school student, why the trend of the graph doesn't follow the fact that the higher frequency a sound wave is, the more power it is, hence, making me think the phons output would follow this general trend?
Joseph Reply
Nevermind i just realied that the graph is the phons output for a person with normal hearing and not just the phons output of the sound waves power, I should read the entire thing next time
Joseph
Follow up question, does anyone know where I can find a graph that accuretly depicts the actual relative "power" output of sound over its frequency instead of just humans hearing
Joseph
"Generation of electrical energy from sound energy | IEEE Conference Publication | IEEE Xplore" ***ieeexplore.ieee.org/document/7150687?reload=true
Ryan
what's motion
Maurice Reply
what are the types of wave
Maurice
answer
Magreth
progressive wave
Magreth
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Muhammad Reply
fine, how about you?
Mohammed
hi
Mujahid
A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the 3.00 m of the string?
yasuo Reply
Who can show me the full solution in this problem?
Reofrir Reply
Practice Key Terms 4

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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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