Chapter 8

Rotational Kinematics

Younes Sina

In Chapter 2, the kinematics of straight line motion was studied where the variables were x, v, a, and t.

In this chapter, the kinematics of rotation will be studied where the variables are θ, ω, α, and t.

θ: Angular Displacement

ω: Angular Velocity

ω = Δθ/Δt ; [rd/s]

Example :A car tire is turning at a rate of 5.0 rd/s as the car travels on a straight road. The driver accelerates the car uniformly for 6.0 seconds.As a result, the angular speed of each tire increases to 8.0 rd/s.Find the angular acceleration of each tire during this period.

Solution:α = Δω / Δtα = (ωf - ωi) /Δtα = (8.0 rd/s - 5.0 rd/s) / 6.0s = 0.50 rd/s2

Linear Motion (Chapter 2) Relations Angular Motion (Chapter 8)

Variables: x, t, v, and a

v = Δx/Δt

a = Δv/Δt ; a = ( vf - vi ) / Δt

x = (1/2) a t2 + vi t

vf2 - vi

2 = 2ax

x = Rθ

v = Rω

a = Rα

or

at = Rα*

Variables: θ, t, ω, and α

ω = Δθ / Δt

α = Δω / Δt ; α = ( ωf - ωi ) / Δt

θ = (1/2) α t2 + ωi t

ωf2 - ωi

2 = 2αθ

Similarity between the linear motion and the angular motion formulas

(*) Note that linear acceleration (a) is the same quantity as tangential acceleration (at )

s = Rθ

v = Rω

at = Rα (tangential acceleration)

Displacement

Velocity

Acceleration

Problem: What is the linear speed of a person who lives on (a) the equator(b) the North Pole(c) between the equator and the North Pole

DEarth=12,742 km

v = Rω

DEarth=12,742 km

v = Rω

ω=2πf=2π/T=6.28/T

f=1/T

T= 24 hr= (24).(3600)s= 86400s

Example : As a car starts accelerating (from rest) along a straight road at a rate of2.4 m/s2, each of its tires gains an angular acceleration of 6.86 rd/s2. Calculate (a) the radius of its tires(b) the angular speed of every particle of the tires at t = 3.0s(c) the angle every particle of its tires sweeps during this 3.0-second period.

Solution:

(a) at = RαR = at /αR = [2.4 m/s2] / [6.86 rd/s2]R = 0.35m = 14 in(b) α = (ωf - ωi)/Δtα Δt = ωf - ωi

ωf = ωi + α Δtωf = 0 +(6.86rd/s2)(3.0s) = 21 rd/s(c) θ = (1/2)α t2 + ωi tθ = (1/2)( 6.86 rd/s2)(3.0s)2 + (0) (3.0s) = 31 rd

Example : The canister of a juicer has 333 grams of pulp distributed over its inside wall at an average radius of 8.00cm. It starts from rest and reaches its maximum angular speed of 3600 rpm in 4.00 seconds. For the pulp, determine(a) the angular acceleration(b) the angle (radians) it sweeps during this period(c) the tangential acceleration(d) the linear velocity at t =2.00s and t = 4.00s(e) the centripetal acceleration at t = 2.00s and t = 4.00s(f) the tangential and centripetal force on it at t = 2.00s and t = 4.00s.

Solution:ω = 3600 rev/min = 3600 (6.28 rd / 60s) = 377 rd/s(a) α = (ωf - ωi) /Δtα = (377- 0 ) / 4.00s = 94.3 rd/s2

(b) θ = (1/2) α t2 + ωi tθ = (1/2)(94.3 rd/s2)(4.00s)2 + 0 = 754 rd(c) at = Rαat = (0.0800m)(94.3 rd/s2) = 7.54 m/s2

(d) At t = 2.00s: α = (ωf - ωi) /Δt ; α Δt = ωf - ωi ; ωf = ωi + α Δt ; (ωf)1 = 0 +(94.3rd/s2)(2.0s) = 189 rd/sAt t = 4.00s:

(ωf)2 = 0 +(94.3rd/s2)(4.0s) = 377 rd/s(d) v1 = R(ωf)1 ; v1 = (0.0800m) (189 rd/s) = 15.1 m/s

v2 = R(ωf)2 ; v2 = (0.0800m) (377 rd/s) = 30.2 m/s(e) (ac)1 = v1

2 / R ; (ac)1 = (15.1 m/s)2 / 0.0800m = 2850 m/s2

(ac)2 = v22 / R ; (ac)2 = (30.2 m/s)2 / 0.0800m = 11400 m/s2

(f) Ft = Mat

(Ft)1 = (0.333 kg )(7.54 m/s2) = 2.51 N

(Ft )2 = 2.51 N (Constant tangential force):

Fc= Mac

(Fc)1 = (0.333 kg )( 2850 m/s2) = 949 N(Fc)2 = (0.333 kg )( 11400 m/s2) = 3800 N (Variable centripetal force)

Homework : problems 1 through 5 chapter 8