Roational Motion Physics

 

Introduction to rotational motion

  1. Rotational Motion Physics Concepts
  2. Rotational Motion Physics Pdf
  3. Rotational Motion Physics

What is rotational motion

Pure Rotational motion is the motion of a body about a fixed axis. If a rigid body is moved in such a way such that all the particles constituting it undergoes circular motion about a common axis then that type of motion is rotational motion.

Rotational motion of a rigid body around a fixed axis is a special case of rotational motion. During rotational motion there is also a possibility of axis changing its orientation. We can not explain the concepts of wobbling or precession by using fixed axis hypothesis for rotational motion.(wikipedia)

Here in this article about rotational motion I am assuming that you are comfortable in describing motion in physics. In addition to this you are also aware of types of motion in physics.. After completing this lesson you would be able to explain Rotational motion and know the difference between rotational motion and circular motion. We will also look at rotational motion examples from activities around us in our daily life.

Till now in our study of force and laws of motion we have always analyzed motion of an object by considering it as a particle even when the size of the object is not negligible. In this process we represent object under consideration as a point mass and shape and size of the object remains irrelevant while discussing the particular problem under consideration. Here in the study of rotational mechanics, this point mass or point particle model is inadequate for problems involving rigid body motion i.e. rigid body undergoing both translational motion and rotational motion. One more reason for not considering the body as particle is that all the particles of the body do not undergo same linear displacement.

As an example consider the motion of a wheel, we cannot consider a wheel as a single particle because different parts of the wheel in motion has different velocities and acceleration.
Here in rotational mechanics, we will consider rigid bodies and their motion. Rigid bodies have definite shape and size and they are capable of having both rotational and translational motion.


What is a rigid body?
The rigid body is a body with a perfectly defined and unchanging shape that is no matter how the body moves, the distance between any two particles within the body remains constant.
Although the way we define rigid body gives us the definition of an idealized rigid body and real materials always deforms on the application of force and this idealized rigid body assumption can be used freely for the substances where deformation is negligibly small and can be neglected.

Kinematics is the description of motion. The kinematics of rotational motion describes the relationships among rotation angle, angular velocity, angular acceleration, and time. Let us start by finding an equation relating,. To determine this equation, we recall a familiar kinematic equation for translational, or straight-line, motion. Rotational Motion of a Rigid Body Rotational motion is more complicated than linear motion, and only the motion of rigid bodies will be considered here. A rigid body is an object with a mass that holds a rigid shape, such as a phonograph turntable, in contrast to the sun, which is a ball of gas.

Motion of a rigid body in general can be considered to consist of a translation of center of mass of the body plus rotation of the body about an axis through the center of mass as shown below in the figure:

Get rotational motion formulas list by following this link.

Rotational motion Examples

We see rotational motion examples in our daily life. Some of the examples of rotational motion are

  1. Rotation of earth about its own axis create the cycle of day and night.
  2. Motion of wheel, gears, motors, etc is rotational motion.
  3. Motion of the blades of the helicopter is also rotatory motion.
  4. A door, swiveling on its hinges as you open or close it.
  5. A spinning top, motion of a Ferris Wheal in an amusement park.

If you want to learn more about rotational dynamics and clear your concepts further then there is a very good book by MTG Interactive Physics: Rotational Dynamics where you can clear your concepts further and gain knowledge accordingly.
For those of you who are interested in video tutorials can look at this video course by udemy
Mastering Rotational Kinematics and Dynamics
In this course, you can learn concepts like Angular variables, Concept of Angular Velocity and Acceleration, Rolling, Rotation and Torque, Mechanics of a Rolling Mass on an Inclined Plane, Physics behind a YOYO and other such related concepts. What I like about this course is how the instructor builds concepts of rotational mechanics as he takes the course further. There are enough solved examples for students which help them improve their problem-solving skills.


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Learning Objectives

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

  • Observe the kinematics of rotational motion.
  • Derive rotational kinematic equations.
  • Evaluate problem solving strategies for rotational kinematics.
Just by using our intuition, we can begin to see how rotational quantities like θ, ω, and α are related to one another. For example, if a motorcycle wheel has a large angular acceleration for a fairly long time, it ends up spinning rapidly and rotates through many revolutions. In more technical terms, if the wheel’s angular acceleration α is large for a long period of time t, then the final angular velocity ω and angle of rotation θ are large. The wheel’s rotational motion is exactly analogous to the fact that the motorcycle’s large translational acceleration produces a large final velocity, and the distance traveled will also be large.

Kinematics is the description of motion. The kinematics of rotational motion describes the relationships among rotation angle, angular velocity, angular acceleration, and time. Let us start by finding an equation relating ω, α, and t. To determine this equation, we recall a familiar kinematic equation for translational, or straight-line, motion:

[latex]v={v}_{0}+{at}[/latex] (constant a)

Rotational Motion Physics Concepts

Note that in rotational motion a = at, and we shall use the symbol a for tangential or linear acceleration from now on. As in linear kinematics, we assume a is constant, which means that angular acceleration α is also a constant, because a = . Now, let us substitute v = and a = into the linear equation above:

=0 + rat.

The radius r cancels in the equation, yielding

ω =ω0 + at. (constant a)

where ω0 is the initial angular velocity. This last equation is a kinematic relationship among ω, α, and t —that is, it describes their relationship without reference to forces or masses that may affect rotation. It is also precisely analogous in form to its translational counterpart.

Making Connections

Kinematics for rotational motion is completely analogous to translational kinematics, first presented in One-Dimensional Kinematics. Kinematics is concerned with the description of motion without regard to force or mass. We will find that translational kinematic quantities, such as displacement, velocity, and acceleration have direct analogs in rotational motion.

Starting with the four kinematic equations we developed in One-Dimensional Kinematics, we can derive the following four rotational kinematic equations (presented together with their translational counterparts):

Table 1. Rotational Kinematic Equations
RotationalTranslational
[latex]theta =bar{omega }t[/latex][latex]x=bar{v}t[/latex]
ω = ω0 + αtv = vo + at(constant α, a)
[latex]theta ={omega }_{0}t+frac{1}{2}{alpha t}^{2}[/latex][latex]x={v}_{0}t+frac{1}{2}{text{at}}^{2}[/latex](constant α, a)
ω2 = ω02+ 2αθv2 = vo2 + 2ax(constant α, a)

In these equations, the subscript 0 denotes initial values (θ0, x0, and t0 are initial values), and the average angular velocity [latex]bar{omega}[/latex] and average velocity [latex]bar{v}[/latex] are defined as follows:

Motion

[latex]bar{omega}=frac{{omega }_{0}+omega }{2}text{ and }overline{v}=frac{{v}_{0}+v}{2}[/latex].

The equations given above in Table 1 can be used to solve any rotational or translational kinematics problem in which a and α are constant.

Problem-Solving Strategy for Rotational Kinematics

  1. Examine the situation to determine that rotational kinematics (rotational motion) is involved. Rotation must be involved, but without the need to consider forces or masses that affect the motion.
  2. Identify exactly what needs to be determined in the problem (identify the unknowns). A sketch of the situation is useful.
  3. Make a list of what is given or can be inferred from the problem as stated (identify the knowns).
  4. Solve the appropriate equation or equations for the quantity to be determined (the unknown). It can be useful to think in terms of a translational analog because by now you are familiar with such motion.
  5. Substitute the known values along with their units into the appropriate equation, and obtain numerical solutions complete with units. Be sure to use units of radians for angles.
  6. Check your answer to see if it is reasonable: Does your answer make sense?

Example 1. Calculating the Acceleration of a Fishing Reel

A deep-sea fisherman hooks a big fish that swims away from the boat pulling the fishing line from his fishing reel. The whole system is initially at rest and the fishing line unwinds from the reel at a radius of 4.50 cm from its axis of rotation. The reel is given an angular acceleration of 110 rad/s2 for 2.00 s as seen in Figure 1. (a) What is the final angular velocity of the reel? (b) At what speed is fishing line leaving the reel after 2.00 s elapses? (c) How many revolutions does the reel make? (d) How many meters of fishing line come off the reel in this time?

Strategy

In each part of this example, the strategy is the same as it was for solving problems in linear kinematics. In particular, known values are identified and a relationship is then sought that can be used to solve for the unknown.

Solution for (a)

Here α and t are given and ω needs to be determined. The most straightforward equation to use is ω = ω0+αt because the unknown is already on one side and all other terms are known. That equation states that

We are also given that ω0 = 0 (it starts from rest), so that

ω = 0 + (110 rad/s2)(2.00 s) = 220 rad/s

Solution for (b)

Now that ω is known, the speed v can most easily be found using the relationship

where the radius r of the reel is given to be 4.50 cm; thus,

Note again that radians must always be used in any calculation relating linear and angular quantities. Also, because radians are dimensionless, we have m × rad = m .

Solution for (c)

Here, we are asked to find the number of revolutions. Because 1 rev=2π rad, we can find the number of revolutions by finding θ in radians. We are given α and t, and we know ω0 is zero, so that θ can be obtained using [latex]theta = omega_{0}t+frac{1}{2}{{alpha t}}^{2}[/latex].

[latex]begin{array}{lll}theta &=& omega_{0}t+frac{1}{2}{{alpha t}}^{2} &=& 0+left(text{0.500}right)left(text{110}{text{rad/s}}^{2}right){left(text{2.00 s}right)}^{2}=text{220 rad}.end{array}[/latex]

Converting radians to revolutions gives

[latex]theta =(220 text{ rad})frac{1 text{ rev}}{2pi text{rad}}=35.0 text{ rev}[/latex]

Solution for (d)

The number of meters of fishing line is x, which can be obtained through its relationship with θ:

Discussion

This example illustrates that relationships among rotational quantities are highly analogous to those among linear quantities. We also see in this example how linear and rotational quantities are connected. The answers to the questions are realistic. After unwinding for two seconds, the reel is found to spin at 220 rad/s, which is 2100 rpm. (No wonder reels sometimes make high-pitched sounds.) The amount of fishing line played out is 9.90 m, about right for when the big fish bites.

Figure 1. Fishing line coming off a rotating reel moves linearly. Example 10.3 and Example 10.4 consider relationships between rotational and linear quantities associated with a fishing reel.

Example 2. Calculating the Duration When the Fishing Reel Slows Down and Stops

Now let us consider what happens if the fisherman applies a brake to the spinning reel, achieving an angular acceleration of -300 rad/s2. How long does it take the reel to come to a stop?

Strategy

We are asked to find the time t for the reel to come to a stop. The initial and final conditions are different from those in the previous problem, which involved the same fishing reel. Now we see that the initial angular velocity is ω0 = 220 rad/s and the final angular velocity ω is zero. The angular acceleration is given to be α = -300 rad/s2. Examining the available equations, we see all quantities but t are known in ω = ω0+αt, making it easiest to use this equation.

Solution

The equation states

ω = ω0 + αt.

We solve the equation algebraically for t, and then substitute the known values as usual, yielding

[latex]t=frac{omega -{omega }_{0}}{alpha }=frac{0-text{220 rad/s}}{-text{300}{text{rad/s}}^{2}}=0text{.}text{733 s}[/latex].

Discussion

Note that care must be taken with the signs that indicate the directions of various quantities. Also, note that the time to stop the reel is fairly small because the acceleration is rather large. Fishing lines sometimes snap because of the accelerations involved, and fishermen often let the fish swim for a while before applying brakes on the reel. A tired fish will be slower, requiring a smaller acceleration.

Example 3. Calculating the Slow Acceleration of Trains and Their Wheels

Large freight trains accelerate very slowly. Suppose one such train accelerates from rest, giving its 0.350-m-radius wheels an angular acceleration of 0.250 rad/s2. After the wheels have made 200 revolutions (assume no slippage): (a) How far has the train moved down the track? (b) What are the final angular velocity of the wheels and the linear velocity of the train?

Strategy

In part (a), we are asked to find x, and in (b) we are asked to find ω and v. We are given the number of revolutions θ, the radius of the wheels r, and the angular acceleration α.

Roational Motion Physics

Solution for (a)

The distance x is very easily found from the relationship between distance and rotation angle:

Solving this equation for x yields

x = rθ.

Before using this equation, we must convert the number of revolutions into radians, because we are dealing with a relationship between linear and rotational quantities:

[latex]theta =left(text{200}text{rev}right)frac{2pi text{rad}}{text{1 rev}}=text{1257}text{rad}[/latex].

Now we can substitute the known values into x= to find the distance the train moved down the track:

x = rθ = (0.350 m)(1257 rad) = 440 m

Solution for (b)

We cannot use any equation that incorporates t to find ω, because the equation would have at least two unknown values. The equation [latex]{omega }^{2}={{omega }_{0}}^{2}+2alphatheta [/latex] will work, because we know the values for all variables except ω:

[latex]{omega }^{2}={{omega }_{0}}^{2}+2alphatheta[/latex]

Roational Motion Physics

Taking the square root of this equation and entering the known values gives

[latex]begin{array}{lll}omega & =& {left[0+2left(0text{.}text{250}{text{ rad/s}}^{2}right)left(text{1257}text{ rad}right)right]}^{1/2} & =& text{25.1 rad/s.}end{array}[/latex]

We can find the linear velocity of the train, v, through its relationship to ω:

Discussion

The distance traveled is fairly large and the final velocity is fairly slow (just under 32 km/h).

There is translational motion even for something spinning in place, as the following example illustrates. Figure 2 shows a fly on the edge of a rotating microwave oven plate. The example below calculates the total distance it travels.

Figure 2. The image shows a microwave plate. The fly makes revolutions while the food is heated (along with the fly).

Example 4. Calculating the Distance Traveled by a Fly on the Edge of a Microwave Oven Plate

A person decides to use a microwave oven to reheat some lunch. In the process, a fly accidentally flies into the microwave and lands on the outer edge of the rotating plate and remains there. If the plate has a radius of 0.15 m and rotates at 6.0 rpm, calculate the total distance traveled by the fly during a 2.0-min cooking period. (Ignore the start-up and slow-down times.)

Strategy

First, find the total number of revolutions θ, and then the linear distance x traveled.[latex]theta =bar{omega}t[/latex] can be used to find θ because [latex]bar{omega}[/latex] is given to be 6.0 rpm.

Solution

Entering known values into [latex]theta =bar{omega} t[/latex] gives

[latex]theta =bar{omega }t=left(text{6.0 rpm}right)left(text{2.0 min}right)=text{12 rev}[/latex].

As always, it is necessary to convert revolutions to radians before calculating a linear quantity like x from an angular quantity like θ:

[latex]theta =left(text{12 rev}right)left(frac{2pi text{rad}}{text{1 rev}}right)=75.4 text{ rad}[/latex].

Now, using the relationship between x and θ, we can determine the distance traveled:

Discussion

Rotational motion physics wallah

Quite a trip (if it survives)! Note that this distance is the total distance traveled by the fly. Displacement is actually zero for complete revolutions because they bring the fly back to its original position. The distinction between total distance traveled and displacement was first noted in One-Dimensional Kinematics.

Check Your Understanding

Rotational kinematics has many useful relationships, often expressed in equation form. Are these relationships laws of physics or are they simply descriptive? (Hint: the same question applies to linear kinematics.)

Solution

Rotational kinematics (just like linear kinematics) is descriptive and does not represent laws of nature. With kinematics, we can describe many things to great precision but kinematics does not consider causes. For example, a large angular acceleration describes a very rapid change in angular velocity without any consideration of its cause.

Section Summary

  • Kinematics is the description of motion.
  • The kinematics of rotational motion describes the relationships among rotation angle, angular velocity, angular acceleration, and time.
  • Starting with the four kinematic equations we developed in the One-Dimensional Kinematics, we can derive the four rotational kinematic equations (presented together with their translational counterparts) seen in Table 1.
  • In these equations, the subscript 0 denotes initial value ([latex]{x}_{0}[/latex] and [latex]{t}_{0}[/latex] are initial values), and the average angular velocity [latex]bar{omega}[/latex] and average velocity [latex]bar{v}[/latex] are defined as follows:
    [latex]bar{omega }=frac{{omega }_{0}+omega }{2}text{ and }bar{v}=frac{{v}_{0}+v}{2}[/latex].

Problems & Exercises

1. With the aid of a string, a gyroscope is accelerated from rest to 32 rad/s in 0.40 s. (a) What is its angular acceleration in rad/s2? (b) How many revolutions does it go through in the process?

2. Suppose a piece of dust finds itself on a CD. If the spin rate of the CD is 500 rpm, and the piece of dust is 4.3 cm from the center, what is the total distance traveled by the dust in 3 minutes? (Ignore accelerations due to getting the CD rotating.)

3. A gyroscope slows from an initial rate of 32.0 rad/s at a rate of 0.700 rad/s2. (a) How long does it take to come to rest? (b) How many revolutions does it make before stopping?

4. During a very quick stop, a car decelerates at 700 m/s2.

(a) What is the angular acceleration of its 0.280-m-radius tires, assuming they do not slip on the pavement?
(b) How many revolutions do the tires make before coming to rest, given their initial angular velocity is 95.0 rad/s?
(c) How long does the car take to stop completely?
(d) What distance does the car travel in this time?
(e) What was the car’s initial velocity?
(f) Do the values obtained seem reasonable, considering that this stop happens very quickly?

Rotational Motion Physics Pdf

Figure 3. Yo-yos are amusing toys that display significant physics and are engineered to enhance performance based on physical laws. (credit: Beyond Neon, Flickr)

5. Everyday application: Suppose a yo-yo has a center shaft that has a 0.250 cm radius and that its string is being pulled.

(a) If the string is stationary and the yo-yo accelerates away from it at a rate of 1.50 m/s2, what is the angular acceleration of the yo-yo?
(b) What is the angular velocity after 0.750 s if it starts from rest?
(c) The outside radius of the yo-yo is 3.50 cm. What is the tangential acceleration of a point on its edge?

Glossary

kinematics of rotational motion:
describes the relationships among rotation angle, angular velocity, angular acceleration, and time

Selected Solutions to Problems & Exercises

1. (a) [latex]80{text{rad/s}}^{2}[/latex] (b) 1.0 rev

Rotational motion physics examples

3. (a) 45.7 s (b) 116 rev

Rotational Motion Physics

5. (a) 600 rad/s2 (b) 450 rad/s (c) 21.0 m/s