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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

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Practice Key Terms 1

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