Younes Sina's Lecture at Pellissippi State Community College

1. Chapter 8
Rotational Kinematics
Younes Sina

2. 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]

3. 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 / Δt
α = (ωf - ωi) /Δt
α = (8.0 rd/s - 5.0 rd/s) / 6.0s = 0.50 rd/s2

4. 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
vf
2 - 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
ωf
2 - 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 )

5. s = Rθ
v = Rω
at = Rα (tangential acceleration)
Displacement
Velocity
Acceleration

6. 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

7. DEarth=12,742 km
v = Rω
ω=2πf=2π/T=6.28/T
f=1/T
T= 24 hr= (24).(3600)s= 86400s

8. Example : As a car starts accelerating (from rest) along a straight road at a rate of
2.4 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.

9. Solution:
(a) at = Rα
R = at /α
R = [2.4 m/s2] / [6.86 rd/s2]
R = 0.35m = 14 in
(b) α = (ωf - ωi)/Δt
α Δt = ωf - ωi
ωf = ωi + α Δt
ωf = 0 +(6.86rd/s2)(3.0s) = 21 rd/s
(c) θ = (1/2)α t2 + ωi t
θ = (1/2)( 6.86 rd/s2)(3.0s)2 + (0) (3.0s) = 31 rd

10. 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.

11. Solution:
ω = 3600 rev/min = 3600 (6.28 rd / 60s) = 377 rd/s
(a) α = (ωf - ωi) /Δt
α = (377- 0 ) / 4.00s = 94.3 rd/s2
(b) θ = (1/2) α t2 + ωi t
θ = (1/2)(94.3 rd/s2)(4.00s)2 + 0 = 754 rd
(c) at = Rα
at = (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/s
At 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 = v2
2 / 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)

12. Homework :
problems 1 through 5 chapter 8