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Wize AP Physics C: Mechanics Textbook > Unit 5: Rotation (14-20%)

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)