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  • 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 size 12{θ} {} , ω size 12{ω} {} , and α 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 α size 12{α} {} is large for a long period of time t size 12{α} {} , then the final angular velocity ω size 12{ω} {} and angle of rotation θ 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 ω size 12{ω} {} , α size 12{α} {} , and t 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 ) size 12{v=v rSub { size 8{0} } + ital "at"" " \[ "constant "a \] } {}

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

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

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

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

where ω 0 size 12{ω rSub { size 8{0} } } {} is the initial angular velocity. This last equation is a kinematic relationship among ω size 12{ω} {} , α size 12{α} {} , and t 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):

Rotational kinematic equations
Rotational Translational
θ = ω ¯ t size 12{θ= {overline {ωt}} } {} x = v - t size 12{x= { bar {v}}t} {}
ω = ω 0 + αt size 12{ω=ω rSub { size 8{0} } +αt} {} v = v 0 + at size 12{v=v rSub { size 8{0} } + ital "at"} {} (constant α size 12{α} {} , a size 12{a} {} )
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 size 12{x=v rSub { size 8{0} } t+ { {1} over {2} } ital "at" rSup { size 8{2} } } {} (constant α size 12{α} {} , a size 12{a} {} )
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 (constant α , a )

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

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} } " " \( "constant "α, a \) } {}

The equations given above in [link] can be used to solve any rotational or translational kinematics problem in which a size 12{a} {} and α 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 ?

Questions & Answers

how did you get 1640
Noor Reply
If auger is pair are the roots of equation x2+5x-3=0
Peter Reply
Wayne and Dennis like to ride the bike path from Riverside Park to the beach. Dennis’s speed is seven miles per hour faster than Wayne’s speed, so it takes Wayne 2 hours to ride to the beach while it takes Dennis 1.5 hours for the ride. Find the speed of both bikers.
MATTHEW Reply
420
Sharon
from theory: distance [miles] = speed [mph] × time [hours] info #1 speed_Dennis × 1.5 = speed_Wayne × 2 => speed_Wayne = 0.75 × speed_Dennis (i) info #2 speed_Dennis = speed_Wayne + 7 [mph] (ii) use (i) in (ii) => [...] speed_Dennis = 28 mph speed_Wayne = 21 mph
George
Let W be Wayne's speed in miles per hour and D be Dennis's speed in miles per hour. We know that W + 7 = D and W * 2 = D * 1.5. Substituting the first equation into the second: W * 2 = (W + 7) * 1.5 W * 2 = W * 1.5 + 7 * 1.5 0.5 * W = 7 * 1.5 W = 7 * 3 or 21 W is 21 D = W + 7 D = 21 + 7 D = 28
Salma
Devon is 32 32​​ years older than his son, Milan. The sum of both their ages is 54 54​. Using the variables d d​ and m m​ to represent the ages of Devon and Milan, respectively, write a system of equations to describe this situation. Enter the equations below, separated by a comma.
Aaron Reply
find product (-6m+6) ( 3m²+4m-3)
SIMRAN Reply
-42m²+60m-18
Salma
what is the solution
bill
how did you arrive at this answer?
bill
-24m+3+3mÁ^2
Susan
i really want to learn
Amira
I only got 42 the rest i don't know how to solve it. Please i need help from anyone to help me improve my solving mathematics please
Amira
Hw did u arrive to this answer.
Aphelele
hi
Bajemah
-6m(3mA²+4m-3)+6(3mA²+4m-3) =-18m²A²-24m²+18m+18mA²+24m-18 Rearrange like items -18m²A²-24m²+42m+18A²-18
Salma
complete the table of valuesfor each given equatio then graph. 1.x+2y=3
Jovelyn Reply
x=3-2y
Salma
y=x+3/2
Salma
Hi
Enock
given that (7x-5):(2+4x)=8:7find the value of x
Nandala
3x-12y=18
Kelvin
please why isn't that the 0is in ten thousand place
Grace Reply
please why is it that the 0is in the place of ten thousand
Grace
Send the example to me here and let me see
Stephen
A meditation garden is in the shape of a right triangle, with one leg 7 feet. The length of the hypotenuse is one more than the length of one of the other legs. Find the lengths of the hypotenuse and the other leg
Marry Reply
how far
Abubakar
cool u
Enock
state in which quadrant or on which axis each of the following angles given measure. in standard position would lie 89°
Abegail Reply
hello
BenJay
hi
Method
I am eliacin, I need your help in maths
Rood
how can I help
Sir
hmm can we speak here?
Amoon
however, may I ask you some questions about Algarba?
Amoon
hi
Enock
what the last part of the problem mean?
Roger
The Jones family took a 15 mile canoe ride down the Indian River in three hours. After lunch, the return trip back up the river took five hours. Find the rate, in mph, of the canoe in still water and the rate of the current.
cameron Reply
Shakir works at a computer store. His weekly pay will be either a fixed amount, $925, or $500 plus 12% of his total sales. How much should his total sales be for his variable pay option to exceed the fixed amount of $925.
mahnoor Reply
I'm guessing, but it's somewhere around $4335.00 I think
Lewis
12% of sales will need to exceed 925 - 500, or 425 to exceed fixed amount option. What amount of sales does that equal? 425 ÷ (12÷100) = 3541.67. So the answer is sales greater than 3541.67. Check: Sales = 3542 Commission 12%=425.04 Pay = 500 + 425.04 = 925.04. 925.04 > 925.00
Munster
difference between rational and irrational numbers
Arundhati Reply
When traveling to Great Britain, Bethany exchanged $602 US dollars into £515 British pounds. How many pounds did she receive for each US dollar?
Jakoiya Reply
how to reduced echelon form
Solomon Reply
Jazmine trained for 3 hours on Saturday. She ran 8 miles and then biked 24 miles. Her biking speed is 4 mph faster than her running speed. What is her running speed?
Zack Reply
d=r×t the equation would be 8/r+24/r+4=3 worked out
Sheirtina
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Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
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