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Kinematics of Rotational Motion

Module by: OpenStax College. E-mail the author

Summary:

  • 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 θθ size 12{θ} {}, ωω size 12{ω} {}, and αα size 12{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 αα size 12{α} {} is large for a long period of time tt size 12{α} {}, then the final angular velocity ωω size 12{ω} {} and angle of rotation θθ size 12{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 ωω size 12{ω} {}, αα size 12{α} {}, and tt size 12{t} {}. To determine this equation, we recall a familiar kinematic equation for translational, or straight-line, motion:

v = v 0 + at       ( constant  a ) v = v 0 + at       ( constant  a ) size 12{v=v rSub { size 8{0} } + ital "at"" " \[ "constant "a \] } {}
(1)

Note that in rotational motion a=ata=at size 12{a=a rSub { size 8{t} } } {}, and we shall use the symbol aa size 12{a} {} for tangential or linear acceleration from now on. As in linear kinematics, we assume aa size 12{a} {} is constant, which means that angular acceleration αα size 12{α} {} is also a constant, because a=a= size 12{a=rα} {}. Now, let us substitute v=v= size 12{v=rω} {} and a=a= size 12{a=rα} {} into the linear equation above:

= 0 + rαt . = 0 + rαt . size 12{rω=rω rSub { size 8{0} } +rαt} {}
(2)

The radius rr size 12{r} {} cancels in the equation, yielding

ω = ω 0 + at       ( constant  a ) , ω = ω 0 + at       ( constant  a ) , size 12{ω=ω rSub { size 8{0} } + ital "at"" " \[ "constant "a \] ,} {}
(3)

where ω0ω0 size 12{ω rSub { size 8{0} } } {} is the initial angular velocity. This last equation is a kinematic relationship among ωω size 12{ω} {}, αα size 12{α} {}, and tt size 12{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
Rotational Translational
θ = ω ¯ t θ = ω ¯ t size 12{θ= {overline {ωt}} } {} x = v - t x = v - t size 12{x= { bar {v}}t} {}  
ω = ω 0 + αt ω = ω 0 + αt size 12{ω=ω rSub { size 8{0} } +αt} {} v = v 0 + at v = v 0 + at size 12{v=v rSub { size 8{0} } + ital "at"} {} (constant αα size 12{α} {}, aa size 12{a} {})
θ = ω 0 t + 1 2 αt 2 θ = ω 0 t + 1 2 αt 2 size 12{θ=ω rSub { size 8{0} } t+ { {1} over {2} } αt rSup { size 8{2} } } {} x = v 0 t + 1 2 at 2 x = v 0 t + 1 2 at 2 size 12{x=v rSub { size 8{0} } t+ { {1} over {2} } ital "at" rSup { size 8{2} } } {} (constant αα size 12{α} {}, aa size 12{a} {})
ω 2 = ω 0 2 + 2 αθ ω 2 = ω 0 2 + 2 αθ size 12{ω rSup { size 8{2} } =ω rSub { size 8{0} rSup { size 8{2} } } +2 ital "αθ"} {} v 2 = v 0 2 + 2 ax v 2 = v 0 2 + 2 ax (constant αα, aa)

In these equations, the subscript 0 denotes initial values (θ0θ0 size 12{θ rSub { size 8{0} } } {}, x0x0 size 12{x rSub { size 8{0} } } {}, and t0t0 size 12{t rSub { size 8{0} } } {} are initial values), and the average angular velocity ω-ω- size 12{ { bar {ω}}} {} and average velocity v-v- size 12{ { bar {v}}} {} are defined as follows:

ω ¯ = ω 0 + ω 2  and  v ¯ = v 0 + v 2 . ω ¯ = ω 0 + ω 2  and  v ¯ = v 0 + v 2 . size 12{ {overline {ω}} = { {ω rSub { size 8{0} } +ω} over {2} } " and " {overline {v}} = { {v rSub { size 8{0} } +v} over {2} } " " \( "constant "α, a \) } {}
(4)

The equations given above in Table 1 can be used to solve any rotational or translational kinematics problem in which aa size 12{a} {} and αα size 12{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 110rad/s2110rad/s2 size 12{"110""rad/s" rSup { size 8{2} } } {} 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 αα size 12{α} {} and tt size 12{α} {} are given and ωω size 12{ω} {} needs to be determined. The most straightforward equation to use is ω=ω0+αtω=ω0+αt size 12{ω=ω rSub { size 8{0} } +αt} {} because the unknown is already on one side and all other terms are known. That equation states that

ω=ω0+αt .ω=ω0+αt . size 12{ω=ω rSub { size 8{0} } +αt"."} {}
(5)

We are also given that ω0=0ω0=0 size 12{ω rSub { size 8{0} } =0} {} (it starts from rest), so that

ω=0+110 rad/s22.00 s=220rad/s .ω=0+110 rad/s22.00 s=220rad/s . size 12{ω=0+ left ("110"" rad/s" rSup { size 8{2} } right ) left (2 "." "00"" s" right )="220 rad/s."} {}
(6)

Solution for (b)

Now that ωω size 12{ω} {} is known, the speed vv size 12{v} {} can most easily be found using the relationship

v= ,v= , size 12{v=rω","} {}
(7)

where the radius rr size 12{α} {} of the reel is given to be 4.50 cm; thus,

v=0.0450 m220 rad/s=9.90 m/s.v=0.0450 m220 rad/s=9.90 m/s. size 12{v= left (0 "." "0450"" m" right ) left ("220"" rad/s" right )=9 "." "90"" m/s."} {}
(8)

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 m × rad = m size 12{m times "rad"=m} {} .

Solution for (c)

Here, we are asked to find the number of revolutions. Because 1 rev=2π rad1 rev=2π rad size 12{1" rev"=2π" rad"} {}, we can find the number of revolutions by finding θθ size 12{θ} {} in radians. We are given αα size 12{α} {} and tt size 12{t} {}, and we know ω0ω0 size 12{ω rSub { size 8{ {} rSub { size 6{0} } } } } {} is zero, so that θθ size 12{θ} {} can be obtained using θ=ω0t+12αt2θ=ω0t+12αt2 size 12{θ=ω rSub { size 8{0} } t+ { {1} over {2} } αt rSup { size 8{2} } } {}.

θ = ω 0 t + 1 2 αt 2 = 0 + 0.500 110 rad/s 2 2.00 s 2 = 220 rad . θ = ω 0 t + 1 2 αt 2 = 0 + 0.500 110 rad/s 2 2.00 s 2 = 220 rad . alignl { stack { size 12{θ=ω rSub { size 8{0} } t+ { {1} over {2} } αt rSup { size 8{2} } } {} # " "=0+ left (0 "." "500" right ) left ("110"" rad/s" rSup { size 8{2} } right ) left (2 "." "00"" s" right ) rSup { size 8{2} } ="220"" rad" {} } } {}
(9)

Converting radians to revolutions gives

θ=220 rad1 rev2π rad=35.0 rev.θ=220 rad1 rev2π rad=35.0 rev. size 12{θ= left ("220"" rad" right ) { {1" rev"} over {2π" rad"} } ="35" "." 0" rev."} {}
(10)

Solution for (d)

The number of meters of fishing line is xx size 12{x} {}, which can be obtained through its relationship with θθ size 12{θ} {}:

x = = 0.0450 m 220 rad = 9.90 m . x = = 0.0450 m 220 rad = 9.90 m . size 12{x=rθ= left (0 "." "0450"" m" right ) left ("220"" rad" right )=9 "." "90"" m"} {}
(11)

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 1 and Example 2 consider relationships between rotational and linear quantities associated with a fishing reel.
The figure shows a fishing reel, with radius equal to 4.5 centimeters. The direction of rotation of the reel is counterclockwise. The rotational quantities are theta, omega and alpha, and x, v, a are linear or translational quantities. The reel, fishing line, and the direction of motion have been separately indicated by curved arrows pointing toward those parts.

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 300rad/s2300rad/s2 size 12{"300"`"rad/s" rSup { size 8{2} } } {}. How long does it take the reel to come to a stop?

Strategy

We are asked to find the time tt size 12{α} {} 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/sω0=220 rad/s size 12{ω rSub { size 8{0} } ="220"" rad/s"} {} and the final angular velocity ωω size 12{ω} {} is zero. The angular acceleration is given to be α=300rad/s2α=300rad/s2 size 12{α= - "300" "rad/s" rSup { size 8{2} } } {}. Examining the available equations, we see all quantities but t are known in ω=ω0+αt,ω=ω0+αt, size 12{ω=ω rSub { size 8{0} } +αt} {} making it easiest to use this equation.

Solution

The equation states

ω=ω0+αt .ω=ω0+αt . size 12{ω=ω rSub { size 8{0} } +αt"."} {}
(12)

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

t=ωω0α=0220 rad/s300rad/s2=0.733 s.t=ωω0α=0220 rad/s300rad/s2=0.733 s. size 12{t= { {ω - ω rSub { size 8{0} } } over {α} } = { {0 - "220"" rad/s"} over { - "300""rad/s" rSup { size 8{2} } } } =0 "." "733"" s."} {}
(13)

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 0.250rad/s20.250rad/s2 size 12{0 "." "250"`"rad/s" rSup { size 8{2} } } {}. 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 xx size 12{x} {}, and in (b) we are asked to find ωω size 12{ω} {} and vv size 12{v} {}. We are given the number of revolutions θθ size 12{θ} {}, the radius of the wheels rr size 12{r} {}, and the angular acceleration αα size 12{α} {}.

Solution for (a)

The distance xx size 12{x} {} is very easily found from the relationship between distance and rotation angle:

θ = x r . θ = x r . size 12{θ= { {x} over {r} } } {}
(14)

Solving this equation for xx size 12{x} {} yields

x=rθ.x=rθ. size 12{x=rθ.} {}
(15)

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:

θ = 200 rev rad 1 rev = 1257 rad . θ = 200 rev rad 1 rev = 1257 rad . size 12{θ= left ("200"" rev" right ) { {2π" rad"} over {"1 rev"} } ="1257"" rad"} {}
(16)

Now we can substitute the known values into x=x= size 12{x=rθ} {} to find the distance the train moved down the track:

x = = 0.350 m 1257 rad = 440 m . x = = 0.350 m 1257 rad = 440 m . size 12{x=rθ= left (0 "." "350"`m right ) left ("1257"" rad" right )="440"" m"} {}
(17)

Solution for (b)

We cannot use any equation that incorporates tt to find ωω, because the equation would have at least two unknown values. The equation ω2= ω02+2αθω2= ω02+2αθ will work, because we know the values for all variables except ωω:

ω2= ω02+2αθω2= ω02+2αθ
(18)

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

ω = 0 + 2 ( 0 . 250  rad/s 2 ) ( 1257  rad ) 1 / 2 = 25.1 rad/s. ω = 0 + 2 ( 0 . 250  rad/s 2 ) ( 1257  rad ) 1 / 2 = 25.1 rad/s. alignl { stack { size 12{ω= left [0+2 \( 0 "." "250"" rad/s" rSup { size 8{2} } \) \( "1257"" rad" \) right ] rSup { size 8{1/2} } "." } {} # ="25" "." 1" rad/s" {} } } {}
(19)

We can find the linear velocity of the train, vv size 12{v} {}, through its relationship to ωω size 12{ω} {}:

v = = 0.350 m 25.1 rad/s = 8.77 m/s . v = = 0.350 m 25.1 rad/s = 8.77 m/s . size 12{v=rω= left (0 "." "350"" m" right ) left ("25" "." 1" rad/s" right )=8 "." "77"" m/s"} {}
(20)

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).
The figure shows a fly that has landed on the rotating plate of the microwave. The direction of rotation of the plate, omega, is counterclockwise and is shown with an arrow.

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 θθ size 12{θ} {}, and then the linear distance xx size 12{x} {} traveled. θ=ω¯tθ=ω¯t size 12{θ= {overline {ωt}} } {} can be used to find θθ size 12{θ} {} because ω - ω - size 12{ { bar {ω}}} {} is given to be 6.0 rpm.

Solution

Entering known values into θ=ω¯tθ=ω¯t size 12{θ= {overline {ωt}} } {} gives

θ=ω-t=6.0 rpm2.0 min=12 rev.θ=ω-t=6.0 rpm2.0 min=12 rev.
(21)

As always, it is necessary to convert revolutions to radians before calculating a linear quantity like xx size 12{x} {} from an angular quantity like θθ size 12{θ} {}:

θ=12 rev2π rad1 rev=75.4 rad.θ=12 rev2π rad1 rev=75.4 rad. size 12{θ= left ("12"" rev" right ) left ( { {2π" rad"} over {"1 rev"} } right )="75" "." 4" rad"} {}
(22)

Now, using the relationship between xx size 12{x} {} and θθ size 12{θ} {}, we can determine the distance traveled:

x = = 0 . 15  m 75 . 4  rad = 11  m . x = = 0 . 15  m 75 . 4  rad = 11  m . size 12{x=rθ= left (0 "." "15"" m" right ) left ("75" "." 4" rad" right )="11" "." 3" m"} {}
(23)

Discussion

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 values (x0x0 size 12{x rSub { size 8{0} } } {} and t0t0 size 12{t rSub { size 8{0} } } {} are initial values), and the average angular velocity ω-ω- size 12{ { bar {ω}}} {} and average velocity v-v- size 12{ { bar {v}}} {} are defined as follows:
    ω ¯ = ω 0 + ω 2  and  v ¯ = v 0 + v 2 . ω ¯ = ω 0 + ω 2  and  v ¯ = v 0 + v 2 . size 12{ {overline {ω}} = { {ω rSub { size 8{0} } +ω} over {2} } " and " {overline {v}} = { {v rSub { size 8{0} } +v} over {2} } " " \( "constant "α, a \) } {}
    (24)

Problems & Exercises

Exercise 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?

Solution

(a) 80 rad/s280 rad/s2 size 12{80 rad/s" rSup { size 8{2} } } {}

(b) 1.0 rev

Exercise 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.)

Exercise 3

A gyroscope slows from an initial rate of 32.0 rad/s at a rate of 0.700 rad/s20.700 rad/s2 size 12{0 "." "700"`"rad/s" rSup { size 8{2} } } {}.

(a) How long does it take to come to rest?

(b) How many revolutions does it make before stopping?

Solution

(a) 45.7 s

(b) 116 rev

Exercise 4

During a very quick stop, a car decelerates at 7.00 m/s27.00 m/s2 size 12{7 "." "00"`"m/s" rSup { size 8{2} } } {}.

(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/s95.0 rad/s size 12{"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?

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)
The figure shows the left arm of a man with tattoo imprints and wearing a glove. He is circulating a yo-yo toy, which is in mid air and connected by the string to his hand. Some people are standing in the background watching the yo-yo trick.

Exercise 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/s21.50 m/s2 size 12{1 "." "50"`"m/s" rSup { size 8{2} } } {}, 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?

Solution

a) 600 rad/s2600 rad/s2 size 12{ {underline {6"00 rad/s" rSup { size 8{2} } }} } {}

b) 450 rad/s

c) 21.0 m/s

Glossary

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

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