Rotational Kinematics

Introduction

Rotational kinematics describes the motion of rigid bodies rotating about a fixed axis. It relates angular position, velocity, and acceleration without considering the forces or moments causing the motion. These relationships are directly analogous to linear kinematics.

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Angular Position, Velocity, and Acceleration

Angular position ( θ ) (\theta) is measured in r a d rad , and θ = s r \theta = \frac{s}{r} .

Angular velocity ( ω ) (\omega) is measured in r a d / s rad/s , and ω = d θ d t \omega = \frac{d\theta}{dt}

v t = ω r v_{t} = \omega r

Angular acceleration ( α ) (\alpha) is measured in r a d / s 2 rad/s^{2} , and α = d ω d t \alpha = \frac{d\omega}{dt}

a t = α r a_{t} = \alpha r and a n = ω 2 r a_{n} = \omega^{2}r

Constant Angular Acceleration

The equations are analogous to the constant-acceleration equations for linear motion:

θ = θ 0 + ω 0 t + 1 2 α t 2 \theta = \theta_{0} + \omega_{0}t + \frac{1}{2}\alpha t^{2}

ω = ω 0 + α t \omega = \omega_{0} + \alpha t

ω 2 = ω 0 2 + 2 α ( θ − θ 0 ) \omega^{2} = \omega_{0}^{2} + 2\alpha(\theta - \theta_{0})

Rolling Without Slipping

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For pure rolling,

s G = r θ s_{G} = r\theta

v G = r ω v_{G} = r\omega

a G = r α a_{G} = r\alpha

Varying Angular Acceleration

When α \alpha is changing over time, calculus is required:

α ( t ) = d ω d t \alpha(t) = \frac{d\omega}{dt} ⇒ \Rightarrow ω = ∫ α ( t ) d t \omega = \int_{}^{}\alpha(t)dt

ω ( t ) = d θ d t \omega(t) = \frac{d\theta}{dt}^{} ⇒ \Rightarrow θ = ∫ ω ( t ) d t \theta = \int_{}^{}\omega(t)dt

When α = f ( θ ) \alpha = f(\theta) , use α d θ = ω d ω \alpha d\theta = \omega d\omega : ∫ θ 0 θ α ( θ ) d θ = ∫ ω 0 ω ω d ω \int_{\theta_{0}}^{\theta}\alpha(\theta)d\theta = \int_{\omega_{0}}^{\omega}\omega d\omega

Example 1:
Find the magnitude of velocity and acceleration of point A on the disk when t = 0.5 s if ω = ( 5 t 2 + 2 ) r a d / s \omega = (5t^{2} + 2)\ rad/s .

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α = d ω d t = 10 t r a d / s 2 \alpha = \frac{d\omega}{dt} = 10t\ rad/s^{2}

At t = 0.5 s:

ω = 5 ( 0.5 ) 2 + 2 = 3.25 r a d / s \omega = 5(0.5)^{2} + 2 = 3.25\ rad/s

α = 10 ( 0.5 ) = 5 r a d / s 2 \alpha = 10(0.5) = 5\ rad/s^{2}

v t = ω r = ( 3.25 r a d / s ) ( 0.8 m ) = 2.6 m / s v_{t} = \omega r = (3.25\ rad/s)(0.8\ m) = 2.6\ m/s\ \ \ \ ← \leftarrow ans.

a t = α r = ( 5 r a d / s 2 ) ( 0.8 m ) = 4 m / s 2 a_{t} = \alpha r = (5\ rad/s^{2})(0.8\ m) = 4\ m/s^{2}

a n = ω 2 r = ( 3.25 r a d / s ) 2 ( 0.8 m ) = 8.45 m / s 2 a_{n} = \omega^{2}r = (3.25\ rad/s)^{2}(0.8\ m) = 8.45\ m/s^{2}

a = a t 2 + a n 2 = 4 2 + 8.45 2 = 9.35 m / s 2 a = \sqrt{a_t^2 + a_n^2} = \sqrt{4^2 + 8.45^2} = 9.35\,m/s^2 ← \leftarrow ans.

Example 2:
Find the number of revolutions a wheel must undergo to acquire a clockwise angular velocity of 15 rad/s with initial angular velocity of 10 rad/s and constant angular acceleration of 3 rad/s2.

ω 2 = ω 0 2 + 2 α ( θ − θ 0 ) \omega^{2} = \omega_{0}^{2} + 2\alpha(\theta - \theta_{0})

( 15 r a d / s ) 2 = ( 10 r a d / s ) 2 + 2 ( 3 r a d / s 2 ) ( θ ) ⇒ θ = 20.83 r a d (15\ rad/s)^{2} = (10\ rad/s)^{2} + 2(3\ rad/s^{2})(\theta)\ \ \Rightarrow \ \ \theta = 20.83\ rad

( 20.83 r a d ) ( 1 r e v o l u t i o n 2 π r a d ) = 3.32 r e v o l u t i o n s (20.83\ rad)(\frac{1\ revolution}{2\pi\ rad}) = 3.32\ revolutions

Example 3:
Find: Angular velocity of gear B when t = 2 s if gear A has an angular acceleration of (4t3) rad/s2 and the gear is initially turning at 20 rad/s.

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Find angular velocity of gear A:

ω = ∫ α ( t ) d t = ω O + ∫ 0 2 4 t 3 d t = 20 + 16 = 36 r a d / s \omega = \int_{}^{}\alpha(t)dt = \omega_{O} + \int_{0}^{2}4t^{3}dt = 20 + 16 = 36\ rad/s

Since the gears roll without slipping:

v A = v B v_{A} = v_{B}

r A ω A = r B ω B r_{A}\omega_{A} = r_{B}\omega_{B}

( 0.05 m ) ( 36 r a d / s ) = ( 0.15 m ) ( ω B ) (0.05\ m)(36\ rad/s) = (0.15\ m)(\omega_{B})

ω B = 12 r a d / s \omega_{B} = 12\ rad/s