For motion with constant angular acceleration** α** = (**ω**_{f}
- **ω**_{i})/(t_{f} - t_{i})
= Δ**ω**/Δt we have

Δ**ω **=
**ω**Δt, ** ω**_{f
}= **ω**_{i }+ **α**Δt.

This is a vector equation. It yields an equation for each Cartesian component.

For the z-component we have

ω_{zf }= ω_{zi} + α_{z}Δt.

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

ω_{f }= ω_{i }+
αΔt.

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

∆θ = ∫_{ti}^{tf
}(dθ/dt)dt = ∫_{ti}^{tf
}ωdt = ω_{i}(t_{f} - t_{i})
+ ½ α (t_{f} - t_{i})^{2}.

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

ω_{f} = ω_{i }+ α(t_{f} - t_{i}).

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

θ_{f} = θ_{i} + ω_{i}(t_{f}
- t_{i}) + ½α(t_{f} - t_{i})^{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

v_{f} = v_{i }+ a(t_{f} - t_{i}),

x_{f }- x_{i }= v_{xi}∆t + ½a_{x}(t_{f} -
t_{i})^{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

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/s^{2}.

(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
**ω**_{i} and the
angular acceleration **α** are given. If
we choose the direction of the initial angular acceleration
to be the z-direction, then

ω_{f }= ω_{i} - α(t_{f} - t_{i}),

since α is in the negative z-direction.

At t = 0, ω_{i
}= 2000/s.

At t = 10 s we have ω_{f } = 2000/s - (80/s^{2})(10
s) = 1200/s.

(b) Setting ω_{f }= ω_{i} - α(t_{f}
- t_{i}) = 0 we can find the time it takes the rotor to come
to rest.

2000/s - (80/s^{2})t = 0, t =
(2000/80)s = 25s is the time it takes the rotor to come to rest.

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
θ_{f }= θ_{i} + ω_{i}(t_{f} - t_{i})
+ ½α(t_{f} - t_{i})^{2 }with θ_{i }= 0, we have 37*2π = ω_{i}*(3 s)
+ ½α(3 s)^{2}.

Using ω_{f }= ω_{i }+
α(t_{f }- t_{i}) we have 98/s = ω_{i }+ α(3 s).

We solve this equation
for ω_{i} ω_{i
}= 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 s^{2})^{ }+ α*(4.5 s^{2}),

(4.5 s^{2})*α
= 294 - 74π, α = 13.67/s^{2} is the constant
acceleration of the wheel.

When 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 a_{t} = dv/dt = rdω/dt = rα.

The **
radial or centripetal acceleration** is a_{r} = v^{2}/r =
rω^{2}.

The **total acceleration** is given by

a
= (a_{t}^{2} + a_{r}^{2})^{½} =
(r^{2}α^{2} + r^{2}ω^{4})^{½}
= r(α^{2} + ω^{4})^{½}.

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 v_{nominal }= r_{nominal}ω, 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 v_{actual }= r_{actual}ω
is greater than the displayed speed v_{nominal }= r_{nominal}ω.

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/s^{2}.

The distance
traveled in 9 s is d = ½at^{2 }= (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.