Rotational kinematics

For motion with constant angular acceleration α = (ωf - ωi)/(tf - ti) = Δω/Δt we have
Δω = ωΔt,  ωf = ωi + αΔt.
This is a vector equation.  It yields an equation for each Cartesian component.
For the z-component we have
ωzf = ωzi + αzΔt.

If an object is rotating about the z-axis, and there are no other components of ω and α, we often drop the index z and just write
ωf = ωi + αΔt.

The angular displacement θ about the z-axis is then given by
∆θ = ∫titf (dθ/dt)dt =  ∫titf ωdt = ωi(tf - ti) + ½ α (tf - ti)2.


Kinematic equations for motion with constant angular acceleration

If an object is rotating about a fixed z-axis with constant angular acceleration α, we have Δω = αΔt,

ωf = ωi + α(tf - ti).

The angular displacement θ about the z-axis is then given by

θf = θi + ωi(tf - ti) +  ½α(tf - ti)2.

These equations are of the same form as the equations for linear motion with constant acceleration a. 
For motion along the x-axis we have

vf = vi + a(tf - ti),
xf - xi = vxi∆t + ½ax(tf - ti)2.

If we replace x by θ and a by α, then the kinematic equations for linear motion along the x-axis transform into the kinematic equations for rotational motion about the z-axis.

Link:  Rotational Kinematics equation solver

Problem:

An airliner arrives at the terminal and the engines are shut off.  The rotor of one of the engines has an initial clockwise angular speed of 2000 rad/s.  The engines rotation slows with an angular acceleration of magnitude 80 rad/s2.
(a)  Determine the angular speed after 10 s.
(b)  How long does it take the rotor to come to rest?

Solution:
(a)  In this problem the initial angular velocity ωi and the angular acceleration α are given.  If we choose the direction of the initial angular acceleration to be the z-direction, then
ωf = ωi - α(tf - ti),
since α is in the negative z-direction. 
At t = 0, ωi = 2000/s.
At t = 10 s we have ωf  = 2000/s - (80/s2)(10 s) = 1200/s.
(b) Setting ωf = ωi - α(tf - ti) = 0 we can find the time it takes the rotor to come to rest.
2000/s - (80/s2)t = 0,  t = (2000/80)s = 25s is the time it takes the rotor to come to rest.

Problem:

A rotating wheel requires 3 s to rotate 37 revolutions.  Its angular velocity at the end of the 3 s interval is 98 rad/s.  What is the constant angular acceleration of the wheel?

Solution:
Using θf = θi + ωi(tf - ti) + ½α(tf - ti)2 with θi = 0, we have 37*2π = ωi*(3 s) + ½α(3 s)2.
Using ωf = ωi + α(tf - ti) we have 98/s = ωi + α(3 s).
We solve this equation for ωi  ωi = 98 s - α(3 s), and insert it into the first equation.
37*2π = (98/s)(3 s) - α(3 s)2 + ½α(3 s)2, 74π = 294 - α*(9 s2) + α*(4.5 s2),
(4.5 s2)*α = 294 - 74π, α = 13.67/s2 is the constant acceleration of the wheel.


imageWhen a wheel rotates about the z-axis, each point on the wheel has the same angular speed.  The linear speed v of a point P, however, depends on its distance from the axis of rotation. 
When a point P goes through an angular displacement of 2π, then its distance traveled is s = 2πr.
When a point P goes through an angular displacement of π, then its distance traveled is s = πr.
When a point P goes through an angular displacement of θ, then its distance traveled is s = θr.
In terms of the angular speed ω, the speed v of the point P therefore is

v = ds/dt = rdθ/dt = rω,

if r is constant; v is the tangential velocity of the point P.

Link:  Tangential and angular speed  (Youtube)

The tangential acceleration a of a point P moving along a circular path is given in terms of its angular acceleration by at = dv/dt = rdω/dt = rα.
The radial or centripetal acceleration is ar = v2/r = rω2.
The total acceleration is given by
a = (at2 + ar2)½ = (r2α2 + r2ω4)½ = r(α2 + ω4)½.

Problem:

If a car's wheels are replaced with wheels of a larger diameter, will the reading of the speedometer change?  Explain!

Solution:
The sensor for the speedometer senses the angular speed of the wheel.  Using vnominal = rnominalω, the speedometer displays the correct speed if the tires have the nominal radius.  If you put larger tires on your car, then your actual speed vactual = ractualω is greater than the displayed speed vnominal = rnominalω.

Problem:

A car accelerates uniformly from rest and reaches a speed of 22 m/s in 9 s  If the diameter of a tire is 58 cm, find
(a)  the number of revolutions the tire makes during this motion, assuming no slipping, and
(b)  the final rotational speed of the tire in revolutions per second.

Solution:
(a)  The acceleration of the car is a = v/t = (22 m/s)/(9 s) = 2.44 m/s2.
The distance traveled in 9 s is d = ½at2 = (½*2.44*81) m = 99 m.
The circumference of the tire is π*0.58 m = 1.82 m.
The number of revolution made by the wheel is 99/1.82 = 54.3.
(b)  The final linear speed of the tire is v = 22 m/s.  Using v = ωr, ω = v/r the final angular speed is ω = 75.9/s.  The number of revolutions per second is ω/2π = 12/s.