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Wize University Physics Textbook (Master) > Angular / Rotational Motion

Rotation Kinematics

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Kinematics Equations for Rotation


The parameters describing the translational motion have an analogue in rotational motion.

Angular velocity is the equivalent of linear velocity, angular acceleration is analogue to linear acceleration, and etc.



Kinematic equations for rotational motions are very similar to kinematic equations for linear motion. For a rotation with a constant angular acceleration:

θf=12αt2+ωit+θi\theta_f=\frac12\alpha t^2+\omega_it+\theta_i
ωf=αt+ωi\omega_f=\alpha t+\omega_i
ωf2−ωi2=2αΔθ\omega_f^2-\omega_i^2=2\alpha\Delta\theta


Watch Out!
Note that for a uniform circular motion, α=0\alpha=0.

Exam Tip
We can combine second and third equations above to find another useful equation which does not depend on angular acceleration:
12(ωf+ωi)t=Δθ\frac{1}{2}(\omega_f+\omega_i)t=\Delta\theta



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Example: Rotating Flywheel


A flywheel turns through 60 revolutions as it slows from an angular speed of 2 rad/s to a stop.

(a) Assuming a constant angular acceleration, find the time for it to come to rest.
(b) What is its angular acceleration?
(c) How much time is required for it to complete the first 20 of the 60 revolutions?


Solution:

Part a)

As angular acceleration is not given, we use the below formula to find the time for the flywheel to stop:


θ−θ0=12(ω+ω0)⋅t\theta-\theta_0=\frac{1}{2}\left(\omega+\omega_0\right)\cdot t

(60)⋅2π=12(2+0)⋅t\left(60\right)\cdot2\pi=\frac{1}{2}\left(2+0\right)\cdot t
Note that here the flywheel turns through 60 revolutions before it stops, since each revolution covers 2π  rad2\pi \ \ rad. The net change in the angle for this rotation is 60(2π)  rad60(2\pi)\ \ rad. From the above equation:
t=377 st=377\ s

Part b)

Now we can use another kinematic equation for rotation to solve for the angular acceleration:

ω=α⋅t+ω0\omega=\alpha\cdot t+\omega_0

0=α⋅377+20=\alpha\cdot377+2


α=−0.0053  rad/s2\alpha =-0.0053 \ \ rad/s^2


Note that here the angular acceleration is negative which means that the angular speed is decreasing!

Part c)

We first need to find the angular speed when the flywheel makes 20 revs:

ω2−ω02=2⋅α⋅(θ−θ0)\omega^2-\omega_0^2=2\cdot\alpha\cdot\left(\theta-\theta_0\right)

ω2−22=2⋅(−0.005)⋅(20⋅2π−0)\omega^2-2^2=2\cdot\left(-0.005\right)\cdot\left(20\cdot2\pi-0\right)

ω=1.7 rads\omega=1.7\ \frac{rad}{s}
Now we use the formula below to find time:
θ−θ0=12(ω+ω0)⋅t\theta-\theta_0=\frac{1}{2}\left(\omega+\omega_0\right)\cdot t

20⋅2π−0=12(1.7+2)⋅t20\cdot2\pi-0=\frac{1}{2}\left(1.7+2\right)\cdot t

t=68 st=68\ s


A 10 kg mass is hanging from a rope wrapped around a cylindrical pulley. The falling mass causes the pulley to rotate. The mass of the pulley is 3 kg. (The moment of inertia of a solid cylinder with radius R and mass of M about its axis of symmetry is I=12MR2I=\frac{1}{2}MR^2)

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What is the acceleration of the falling mass? (choices are in m/s2m/s^2)

Extra Practice
The revolutions of your truck engine climb steadily from 1200 rpm to 5500 rpm in 2.7s. What is
a) the engine's angular acceleration? (in $rad/s^2$)
b) the tangential acceleration of a point on the edge of the engine's 3.5cm diameter crankshaft? (in $m/s^2$)
Practice: Quick rotational dynamics
The engine of a turbine has a torque of 100 Nm. The engine rotates a 20 m long and 100 kg blade. How long does it take the blade to reach to 400 rpm? Model the blade as a thin rod.
A merry-go-round starts from rest and is accelerating to 100 revolutions per second in 25 seconds. Calculate the angular acceleration α\alpha and the number of revolutions the merry-go-round completes in 1 minute.
The revolutions of your truck engine climb steadily from 1200 rpm to 5500 rpm in 2.7s. What is a) the engine's angular acceleration? b) the tangential acceleration of a point on the edge of the engine's 3.5cm diameter crankshaft? c) the number of revolutions the engine makes during this time?
A disk with moment of inertia of 10 kg m2 and a radius of 1 m rotates on a frictionless axis perpendicular to the disk and through its center. A force of 10 N is applied tangentially to the rim. If the disk starts at rest, what is its angular velocity after 2 turns?
If a wheel of radius 0.1m is rotating at 100rpm, find the velocity at the outer rim of the wheel.
A stone of mass 9.5kg is hung vertically from the end of a rope that is wound around a pulley on a boat. The moment of inertia of the pulley is 0.47kg m2 and the radius is 0.25m. The stone is released from rest and falls toward the water, causing the pulley to rotate. After the stone has fallen 2.5 m, what is the angular velocity of the pulley?
rotational motion
A disc rotates through 10 radians in 4 seconds. The disc experienced uniform acceleration. If the disc starts from rest, what is the angular velocity after four seconds?