Introduction
Motion in one dimension are concepts such as displacement, speed and velocity, and acceleration, We will be following these concepts, but in this case, for a circular trajectory, We will end the lesson by studying the centripetal force, acceleration, and the origins of both.
Uniform Circular Motion
First, let us look at the new type of trajectory, a circle, and define some fundamental quantities. We will define uniform circular motion as the motion of an object traveling at a constant speed on a circular path. One rotation around the trajectory is called a revolution.
Uniform Circular Motion
Uniform circular motion is the motion of an object traveling at a constant speed on a circular path.
Angular Displacement
What quantity can describe motion on this circular trajectory? In between the two positions shown in Figure 8.1, there is a distance, arc length, and an angle (revealed by the two radii). The angle described by an object moving around the circular trajectory will be called angular displacement. Also similar to the linear displacement, angular displacement has a positive and negative direction: The positive direction is the counterclockwise rotation, and the negative direction is the clockwise rotation.
When the angular displacement is small, we can determine the angle in radians as:
where s is the arc length and r is the radius.
The conversion factor between radians and degrees can be found if we think of 2π radians that are equivalent to 360°. Then:
2π radians = 360°
1 radian = 57.3°
Example
A rock at the end of a string is spun around in a circle of radius 0.50 m. Between two close instances, the arc length is 10.0 cm. Find the angular displacement in both radians and degrees.
Solution
First, convert the quantities to SI. Next, list the data and solve the problem.
r = 0.5 m
s = 10 cm = 10 cm · 1 m/100 cm = 0.1 m
θ = ?
Tangential Velocity
Although the uniform circular motion is characterized by constant speed, the direction of motion changes all the time, and therefore, the velocity will have different directions, as is shown in Figures 8.3 and 8.4. At two different times, the object occupies different positions on the circle. This defines the linear displacement, and by dividing it by time, we can determine the linear velocity. In Figure 8.3, we are shown two distant positions on the circle and the corresponding displacement, d, whereas in Figure 8.4, the displacement is very small, d → 0.
As the two dotted lines show, the change in position modifies the direction of the displacement, and for a very small interval of time, Δt → 0, displacement becomes tangent to the radius and so does the velocity. So, at every point on the trajectory, the velocity will be tangent to the circle pointing in the direction of motion and defining tangential velocity.
The magnitude of the average velocity (for uniform circular motion, this is also the value of the instantaneous speed) is given by:
And the direction is tangent to the trajectory at every point. For one revolution, the time is called a period, and its symbol is T. T is measured in seconds (s).
The inverse of the period is called the frequency and defines the number of rotations per second. Frequency is measured in s^{–1} lor hertz (Hz), named for Heinrich Rudolf Hertz (1857–1894), and its symbol is f.
Then the expression of the speed becomes:
v = 2 · π · r · f
Example
As in the previous example, a rock at the end of a string is spun around in a circle of radius 0.50 m. Between two close instances, the arc length is 10.0 cm, and the time is 0.2 s. Find the average speed and the period.
Solution
First, convert the quantities to SI, list the data, and solve the problem.
r = 0.5 m
s = 10 cm = 10 cm · 1 m/100 cm = 0.1 m
v = ?
T = ?
First, we find the average speed:
And the period is:
T = 6.3 s
Angular Speed
The average angular speed is given by the angular displacement and the time interval.
The unit of measurement for the angular velocity is radian per second.
If we consider an object moving in a uniform circular motion with a speed v, then during one period, the object moves around a circle once and the average angular speed is:
So if we consider Δt = T:
And because the object describes equal angles in equal times, then the average acceleration and the instantaneous acceleration are equal:
Considering the expressions in the previous section:
w = 2 · π · f
Example
Consider a couple of pieces of paper on a spinning 45.0 rpm record, one at a distance 10.0 cm from the center, and the other at 5.00 cm. Find the angular speed and tangential speed for each of the pieces.
Solution
The data in the problem requires conversion to SI.
1 rpm = 1 rotation/60 s
45 rpm = 45 rotations/60 s = .75 rotations/s
T = 1/.75 s = 1.33 s
r_{1} = 5.00 cm 0.05 m
r_{2} = 10.0 cm = 0.1 m
w = ?
v = ?
For the piece that is closest to the center:
v_{1} = 2 · π · r_{1}/T
v_{1} = 0.075π · m/s = 0.24 m/s
v_{2} = 2 · π · r_{2}/T
v_{2} = 0.15π · m/s = 0.47 m/s
w = 2 · π/T = 4.7 radians/s
Note
The centripetal force does not have a source of interaction (in the same way charge is the source for the electrical force or gravity for weight). Rather, it is due to a nonequalized force or system of forces that act on the object.
Centripetal Force and Acceleration
Let us review the concepts learned: For uniform circular motion, although the speed and angular speed are constant, the velocity is not because the direction of the object constantly changes. In this case, an acceleration associated with the motion can be defined.
Similar to the determination in the previous paragraph regarding the direction of the tangential velocity, we can argue that the change in direction for a finite interval of time re1;ults in an acceleration, as shown in Figure 8.6. The acceleration is directed toward the final velocity vector that, although equal in absolute value with the initial vector, is definitely in a different direction.
For small intervals of time, the displacement is very small and the two vectors—initial and final velocity—are also in proximity. The difference between the two vectors once again determines the vector acceleration, which approaches a direction perpendicular to the initial velocity vector. The acceleration vector points toward the center of the circle, as can be seen in the inset of Figure 8.7.
The two acceleration vectors look different in these two figures. Their direction is different, as is the size. This last difference is due to the time interval, which is different in the two figures: Larger time has passed between the two positions for the first figure. The acceleration defined by these two figures due to the change in the direction and directed toward the center of the ciruclar motion is called centripetal acceleration and is due to the curving of the trajectory and umcompensated forces acting on the object. The corresponding force is called the centripetal force.
As a consequence of the change in direction and the variation of the velocity of the object, the centripetal acceleration can be expressed as:
a_{c} = v^{2}/r
And then the centripetal force is:
F_{c} = m · a_{c}
F_{c} = m · v^{2}/r
And using the relationship established between angular and tangential speeds:
F_{c} = m · a_{c} = m · (w · r)^{2}/r = m · w^{2} · r
F_{c} = m · w^{2} · r
Centripetal Force
The centripetal force is the net force required to keep an object moving with speed v on a circular path of radius r. The direction of F_{c} is along the radius and toward the center of the circle.
Example
Find the centripetal acceleration and force exerted on a 1,500 kg car taking a curve at a speed of 60 mi/h if the curvature of the road has a radius of 50 m.
Solution
First, list the data and convert to SI because we have different units for the same quantities. Then, set the equations and solve for the unknown.
m = 1,500 kg
v = 60 mi/h = 60 · 1,609 m/3,600 s = 27 m/s
r = 50m
a_{c} = ?
F_{c} = ?
The curve is part of a circle, so at the change of direction, a centripetal acceleration and force will be developed by the change in velocity even if the car keeps running at the same speed.
To determine the first unknown, we have the following expression:
a_{c} = v^{2}/r
and replacing the tangential speed and the radius, we get:
a_{c} = (27 m/s)^{2}/(50 m)
a_{c} = 15 m/s^{2}
And for the second unknown:
F_{c} = m · v^{2}/r
Or:
F_{c} = m · a_{c}
F_{c} = 1,500 kg · 15 m/s^{2} = 22,500 N
Practice problems of this concept can be found at: Uniform Rotational Motion Practice Questions