A schematic of the “London Eye” can be seen in the figure below. Represents the movement of a passenger represented by the point P, which follows the circular path, called c:
The passenger occupies position P at moment t and the angular position corresponding to that moment is ϕ.
From time t a period of time Δt elapses. In this period, the punctual passenger’s new position is P’ and the angular position has increased by an angle Δϕ.
How is angular velocity calculated?
For rotation magnitudes, the Greek letters are often used in order to differentiate them from linear magnitudes. So, initially, the average angular velocity ω m is defined as the angle traveled over a given period of time.
Then the ratio Δϕ / Δt will represent the average angular velocity ω m between the moments tet + Δt.
If you want to calculate the angular velocity just at moment t, you will need to calculate the ratio Δϕ / Δt when Δ0:
Relationship between linear and angular velocity
The linear velocity v , is the ratio between the distance covered and the time needed to cover it.
In the figure above, the traversed arc is Δs. But this arc is proportional to the traversed angle and the radius, fulfilling the following relationship, valid as long as Δϕ is measured in radians:
Δs = r · Δϕ
If we divide the previous expression between the time interval Δt and we take the limit when Δ
uniform rotation movement
The photo is the famous «London Eye», a rotating wheel 135 m high that turns slowly, so that people can board the booths at its base and enjoy the sights of London. Source: PixabayA rotational movement is uniform if, at any observed moment, the traversed angle is the same in the same period of time.
If the rotation is uniform, the angular velocity at any time coincides with the average angular velocity.
Also, when a complete turn is made, the angle traveled is 2π (equivalent to 360º). Therefore, in a uniform rotation, the angular velocity ω is related to the period T, by the following formula:
f = 1 / T
That is, in a uniform rotation, angular velocity is related to frequency by:
ω = 2π ・ f
Solved Angular Velocity Exercises
The cabins of the large rotating wheel known as the “ London Eye ” move slowly. The speed of the cabins is 26 cm / s and the wheel is 135 m in diameter.
With this data, calculate:
i) The angular speed of the wheel
ii) The rotation frequency
iii) The time it takes for a cabin to turn completely.
i) The velocity v in m / s is: v = 26 cm / s = 0.26 m / s.
The radius is half the diameter: r = (135 m) / 2 = 67.5 m
v = r ・ ω => ω = v / r = (0.26 m / s) / (67.5 m) = 0.00385 rad / s
ii) ω = 2π ・ f => f = ω / 2π = (0.00385 rad / s) / (2π rad) = 6.13 x 10 -4 turns / s
f = 6.13 x 10^-4 lap / s = 0.0368 lap / min = 2.21 lap / hour.
iii) T = 1 / f = 1 / 2.21 lap / hour = 0.45311 hour = 27 min 11 sec
A toy car moves on a circular track with a radius of 2m. At 0 s, its angular position is 0 rad, but after a time t, its angular position is given by:
φ (t) = 2 · t
i) The angular velocity
ii) Linear velocity at any time.
i) The angular velocity is that derived from the angular position: ω = φ ‘(t) = 2.
That is, the toy car always has a constant angular velocity equal to 2 rad / s.
ii) The linear speed of the car is: v = r · ω = 2 m · 2 rad / s = 4 m / s = 14.4 Km / h
The same car from the previous exercise starts to stop. Its angular position as a function of time is given by the following expression:
φ (t) = 2 · t – 0.5 · t 2
i) Angular velocity at any time
ii) linear velocity at any time
iii) The time it takes to stop from the moment it starts to decelerate
iv) The traveled angle
v) distance covered
i) The angular velocity is that derived from the angular position: ω = φ ‘(t)
ω (t) = φ ‘(t) = (2 · t – 0.5 · t 2 )’ = 2 – t
ii) The linear speed of the car at any time is given by:
v (t) = r · ω (t) = 2 · (2 - t) = 4-2 t
iii) The time it takes to stop from the moment it starts to decelerate is determined by knowing when the velocity v(t) becomes zero.
v (t) = 4-2 t = 0 => t = 2
In other words, it stops 2 s after starting to brake.
iv) In the interval of 2s from the moment it starts to brake until it stops, an angle given by φ (2) is traversed:
φ (2) = 2 · 2 – 0.5 · 2 ^ 2 = 4 – 2 = 2 rad = 2 x 180 / π = 114.6 degrees
v) In the interval of 2 s from the moment it starts to brake until it stops, the distance s is given by:
s = r · · = 2m · 2 rad = 4 m
The wheels of a car are 80 cm in diameter. If the car travels at 100 km/h. Find: i) the angular speed of rotation of the wheels, ii) the frequency of rotation of the wheels, iii) The number of turns the wheel makes in a 1 hour course.
i) First of all, let’s convert the car’s speed to Km / ham / s
v = 100 Km / h = (100 / 3.6) m / s = 27.78 m / s
The angular speed of rotation of the wheels is given by:
ω = v / r = (27.78 m / s) / (0.4 m) = 69.44 rad / s
ii) Wheel rotation frequency is given by:
f = ω / 2π = (69.44 rad / s) / (2π rad) = 11.05 revolutions / s
The rotation frequency is usually expressed in revolutions per minute rpm
f = 11.05 lap / s = 11.05 lap / (1/60) min = 663.15 rpm
iii) The number of turns the wheel makes in an hour is calculated knowing that 1 hour = 60 min and that the frequency is the number of turns N divided by the time these N turns occur.
f = N / t => N = f · t = 663.15 (rounds / min) x 60 min = 39788.7 rounds.