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α t 0 t d t = ω 0 ω f d ω .

Setting t 0 = 0 , we have

α t = ω f ω 0 .

We rearrange this to obtain

ω f = ω 0 + α t ,

where ω 0 is the initial angular velocity. [link] is the rotational counterpart to the linear kinematics equation v f = v 0 + a t . With [link] , we can find the angular velocity of an object at any specified time t given the initial angular velocity and the angular acceleration.

Let’s now do a similar treatment starting with the equation ω = d θ d t . We rearrange it to obtain ω d t = d θ and integrate both sides from initial to final values again, noting that the angular acceleration is constant and does not have a time dependence. However, this time, the angular velocity is not constant (in general), so we substitute in what we derived above:

t 0 t f ( ω 0 + α t ) d t = θ 0 θ f d θ ; t 0 t ω 0 d t + t 0 t α t d t = θ 0 θ f d θ = [ ω 0 t + α ( ( t ) 2 2 ) ] t 0 t = ω 0 t + α ( t 2 2 ) = θ f θ 0 ,

where we have set t 0 = 0 . Now we rearrange to obtain

θ f = θ 0 + ω 0 t + 1 2 α t 2 .

[link] is the rotational counterpart to the linear kinematics equation found in Motion Along a Straight Line for position as a function of time. This equation gives us the angular position of a rotating rigid body at any time t given the initial conditions (initial angular position and initial angular velocity) and the angular acceleration.

We can find an equation that is independent of time by solving for t in [link] and substituting into [link] . [link] becomes

θ f = θ 0 + ω 0 ( ω f ω 0 α ) + 1 2 α ( ω f ω 0 α ) 2 = θ 0 + ω 0 ω f α ω 0 2 α + 1 2 ω f 2 α ω 0 ω f α + 1 2 ω 0 2 α = θ 0 + 1 2 ω f 2 α 1 2 ω 0 2 α , θ f θ 0 = ω f 2 ω 0 2 2 α

or

ω f 2 = ω 0 2 + 2 α ( Δ θ ) .

[link] through [link] describe fixed-axis rotation for constant acceleration and are summarized in [link] .

Kinematic equations
Angular displacement from average angular velocity θ f = θ 0 + ω t
Angular velocity from angular acceleration ω f = ω 0 + α t
Angular displacement from angular velocity and angular acceleration θ f = θ 0 + ω 0 t + 1 2 α t 2
Angular velocity from angular displacement and angular acceleration ω f 2 = ω 0 2 + 2 α ( Δ θ )

Applying the equations for rotational motion

Now we can apply the key kinematic relations for rotational motion to some simple examples to get a feel for how the equations can be applied to everyday situations.

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/s 2 for 2.00 s ( [link] ).

(a) What is the final angular velocity of the reel after 2 s?

(b) How many revolutions does the reel make?

Figure is a drawing of a fishing line coming off a rotating reel. Rotation radius is 4.5 cm, rotation takes place in the counterclockwise direction.
Fishing line coming off a rotating reel moves linearly.

Strategy

Identify the knowns and compare with the kinematic equations for constant acceleration. Look for the appropriate equation that can be solved for the unknown, using the knowns given in the problem description.

Solution

  1. We are given α and t and want to determine ω . The most straightforward equation to use is ω f = ω 0 + α t , since all terms are known besides the unknown variable we are looking for. We are given that ω 0 = 0 (it starts from rest), so
    ω f = 0 + ( 110 rad/s 2 ) ( 2.00 s ) = 220 rad/s .
  2. We are asked to find the number of revolutions. Because 1 rev = 2 π rad , we can find the number of revolutions by finding θ in radians. We are given α and t , and we know ω 0 is zero, so we can obtain θ by using
    θ f = θ i + ω i t + 1 2 α t 2 = 0 + 0 + ( 0.500 ) ( 110 rad/s 2 ) ( 2.00 s ) 2 = 220 rad .

    Converting radians to revolutions gives
    Number of rev = ( 220 rad ) 1 rev 2 π rad = 35.0 rev .

Significance

This example illustrates that relationships among rotational quantities are highly analogous to those among linear quantities. 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.)

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Practice Key Terms 1

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Source:  OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5
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