Chapter 8

Rotational Kinematics:

In Chapter 2, the kinematics of straight line motion was studied where the variables were x, v, a, and t.  In Chapter 2, the cause of motion (force) was not important and force was not used to calculate the acceleration (a) of motion.

In this chapter, the kinematics of rotation will be studied where the variables are θ, ω, α,  and t.   Although these quantities were discussed in Chapter 5, we will review them here again.

Angular Displacement (θ):  

In Fig. 1, angular displacement, θ, pronounced "theta" is the angle swept by radius R of the circle that points to a rotating particle or mass like M.

Angular Velocity (ω):  

Angular velocity, ω, is defined as the change in angular displacement, θ, per unit of time, t.  

ω = Δθ/Δt

Fig. 1

Example 1:  An object travels around a circle10.0 full turns in 2.5 seconds.  Calculate (a) its angular displacement, θ, during this period and (b) its average angular speed, ω.

Solution:  (a)    θ = 10.0 turns (6.28 rd/turn) = 62.8 radians.

                 (b)   ω  = Δθ/Δt   = 62.8rd / 2.5s  = 25 rd/s.


Angular Acceleration (α):  Angular acceleration α is the change in the angular velocity, ω, per unit of time, t.

α  = Δω/Δt

with the SI unit of rd/s2.  The symbol α is pronounced " Alpha."

Example 2:  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 =f - ωi) /Δt ;   α = (8.0rd/s -5.0rd/s)/6.0s  = 0.50 rd/s2.

There is a one to one correspondence between the linear motion formulas and the angular motion formulas that can be similarly derived here.  Table 1 shows this one to one correspondence.  If you know Chapter 2 formulas, it is like you already know Chapter 8 formulas.  The formulas in the middle column relate each angular variable to its counterpart linear variable.

Linear Motion (Chapter 2)

Relations Angular Motion (Chapter 8)
  Variables: x, t, v, and a

  v = Δx /Δt.

  a = Δv /Δ;  a = ( vf - vi )/Δt.

  x = (1/2) at2  + vi t.

  vf2 - vi2 = 2ax.

x = Rθ 

v = Rω

a = Rα


at = Rα*

   Variables:  θ, t, ω, and α

 ω = Δθ /Δt.

 α  = Δω /Δt  ; α  = f - ωi ) /Δt.

 θ = (1/2) α t2 + ωi t.

   ωf2 - ωi2  = 2αθ.

(*)  Linear acceleration a is the same quantity as tangential acceleration at .

Table 1

The Relations between Linear and Angular Variables:

Each of the angular variables θ, ω, and α is related to its corresponding linear variable x, v, and a or at by factor R, the radius of rotation.

 x = Rθ      ;      v = Rω     ;     at = Rα. 

This can be easily verified by the following simple mathematics:

Starting with  s = Rθ,  or x = Rθ and writing it as  Δx = RΔθ, and then dividing both sides by Δt, yields:

Δx/Δt = RΔθ/Δt  ;  the left side is v and the right side is  ; therefore,  v = Rω.


we may write v = Rω  as   Δv = RΔω   and then divided through by Δt to get:   

Δv/Δt  = R Δω/Δt  ;   the left side is  a and the right side is  ; therefore,  at = Rα.

Note that the acceleration vector a in circular motion is always tangent to the circle and is shown as at that means tangential acceleration.





s = Rθ


v = Rω




at = Rα   (tangential acceleration)



Note in the 3rd figure that there are two types of acceleration in rotational motion.  One type is at , the tangential acceleration that is responsible for the change in the magnitude of the linear velocity, v.  Pushing a merry-go-round gives it tangential acceleration because the pushing force is tangent to a circular path.

The other type is ac, the centripetal acceleration that is responsible for the change in the direction of v.  This acceleration is always directed toward the center of rotation (not shown here to keep the diagram more clear).  It was discussed in Chapter 5.   ac is also called the "normal acceleration" because it is perpendicular to the circular path.


Example 3: 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, and (c) the angle every particle of its tires sweeps during this 3.0-second period.

Solution: (a) Since a linear variable and its corresponding angular variable are given, the radius of rotation can be calculated.    

(a) at = Rα   ;     R = at /α    ;    R = [2.4 m/s2] / [6.86 rd/s2]    ;    R = 0.35m = 14 in.

(b) α =(ωf i)/Δt ;  α Δt = ωf i   ωf = ωi + α Δt;   ωf = 0+(6.86rd/s2)(3.0s) = 21rd/s.

(c) θ = (1/2)α t2 + ωit  ;  θ = (1/2)(6.86 rd/s2)(3.0s)2 + (0) (3.0s) = 31rd.


Example 4: 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,  and (f) the tangential and centripetal force on it at t = 2.00s and t = 4.00s.

Solution: First convert rpm to rd/;  ω = 3600 rev/min = 3600 (6.28rd/60s) = 377 rd/s.

(a) α  = f - ωi)/Δt   ;   α = (377- 0)/4.00s  =  94.3 rd/s2.

(b) θ = (1/2)αt2 + ωi  ;   θ = (1/2)(94.3rd/s2)(4.00s)2 + 0 = 754 rd.

(c) at =  ;   at = (0.0800m)(94.3rd/s2) = 7.54 m/s2.  Refer to at in the above figure.

(d) The values of ωf , must be calculated both at t = 2.00s and t = 4.00s.

α = f i)/Δt ;  αΔt = ωf  ;  ωf = ωi +αΔt ;  f)1 = 0+(94.3rd/s2)(2.0s) = 189 rd/s.   

At  t = 4.00s, ω = 3600 rpm = 377 rd/ ;  f)2  = 0 +(94.3rd/s2)(4.0s) = 377 rd/s.

(d)  v1 = Rf)1    ;    v1 = (0.0800m) (189 rd/s) = 15.1 m/s.

      v2 = Rf)2    ;    v2 = (0.0800m) (377 rd/s) = 30.2 m/s.

(e) (ac)1 = v12 / 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)


Chapter 8 Test Yourself 1:

1) In Fig. 1, the angular displacement of mass M is (a) arc AB = S   (b) ω   (c) angle θ.   click here.

2) In Fig. 1, the linear displacement of mass M is (a) arc AB = S     (b) ω    (c) angle θ.

3) The relation between angle θ and arc S, the arc opposite to it, is (a) S = 2Rθ   (b) S = Rθ   (a) S = πRθ.

4) The symbol for angular speed is (a) θ    (b) ω    (c) S.   click here.

5) The relation between angular speed and linear speed is (a) S = Rθ   (b) V = Rω    (c) S = 2πR.

Problem: The angular speed of a wheel is such that it makes 80.0 turns in 20.0 seconds. The radius of the wheel is 35.0cm.

Answer the following questions:

6) The angular speed of the wheel is (a) 240 rpm   (b) 4.00 turns/s  (c) 25.1 rd/s  (d) a, b, & c.  click here.

7) The linear speed of any point on the outer edge of the wheel is (a) 879 cm/s   (b) 8.79 m/s   (c) a & b.

8) The angular acceleration of the wheel is (a) half of its centripetal acceleration  (b) a nonzero constant   (c) zero, because the angular speed is constant.   click here.

9) The tangential acceleration of the wheel is (a) half of its centripetal acceleration  (b) a nonzero constant  (c) zero, because the linear speed is constant.

10) The centripetal acceleration of any point on the outer edge of the wheel that is at a radius of R = 35.0cm is (a) 25.1 m/s2   (b) 221 m/s2     (c) 221 rd/s2.

11) The variables in uniformly accelerated motion (motion along a straight line at const. velocity) are: x, t, v, and a.  Write down the counterpart variables for rotational motion.  Ans.:.................................... click here to check your answer.

12) Without referring to Table 1 above, try to write down the formulas you know for linear motion from Chapter 2.  Then write down the corresponding angular formulas.  Pay attention to the one-to-one correspondence between the variables and formulas.


 Problem: Make sure you perform all calculations even if they seem obvious to you. 

We all know that our planet completes one revolution about its own axis every 24 hours. 

Answer the following questions by referring to the figure on the right when necessary:





13) The angular speed of the Earth in revolving about its own axis is (a) 1 rev/24h   (b) 2πR/ 86400s    (c) 6.28 rd/86400s   (d) 7.27x10-5 rd/s   (e) a, c, & d.  click here.

14) Any person who lives on the equator is at a radius of rotation of R1 = 3900 miles.  He/she has a linear speed of (a) 0.283mi/s   (b)456m/s   (c) both a & b.

15) Any person who lives exactly at the North Pole or the South Pole where the Earth axis passes through, is at a radius of rotation of  R4 = (a) 3900mi   (b) almost 0   (c) neither a, nor b.

16) Whoever lives exactly at the North Pole or the South Pole, has an angular speed of (a) 7.27x10-5 rd/s   (b) almost 0   (c) neither a, nor b.   click here.

17) A person who lives exactly at the North Pole, has a linear speed of (a) 456 m/s   (b) almost 0   (c) neither a, nor b.

18) The radius of rotation of a person who lives in between the equator and the North Pole is (a) less than 0    (b) more than 3900mi    (c) less than 3900 miles.

19) The radius of rotation for the people who live 45o above the equator is R2 =  (a) [3900/2] miles   (b) [3900/45o] miles    (c) 3900cos45o miles.   click here.

20) The linear speed of the people who live 45o above the equator is (a) 323 m/s    (b) 0.200 mi/s    (c) both a & b.

21) The tangential acceleration of any person on this planet because of Earth motion about its own axis is (a) 9.8 m/s2   (b) 0    (c) 456 m/s2. 

22) The centripetal acceleration of a person living on the equator because of the Earth's rotation about its own axis is (a) 9.8 m/s2   (b) 0   (c) 0.0331 m/s2.   click here.

23) Because of the Earth motion about its own axis, the direction of the centripetal acceleration vector for those who live on the equator is (a)  parallel to the direction of g   (b) perpendicular to the direction of g   (c) neither a, nor b.

24) The direction of the centripetal acceleration vector for those who live 45o above the equator is (a)  parallel to the direction of g  (b) perpendicular to the direction of g   (c) neither a, nor b.

25) The direction of the centripetal acceleration vector for those who live close to the North Pole (a)  parallel to the direction of g  (b) almost perpendicular to the direction of g   (c) neither a, nor b.   click here.

Problem: Again, make sure you perform all calculations even if they look obvious to you.  We all know that our planet completes one revolution about the Sun every year.  Assume circular orbit for simplicity.  Answer the following questions:

26) The angular speed of the Earth in its rotation around the Sun is (a) 2πR/yr   (b) 6.28 rd/yr   (c)1.99x10-7 rd/s  (d) b&c.     

27) Knowing that the average Earth-Sun distance is 150,000,000km,  the linear speed of the Earth in its motion around the Sun is (a) 30mi/h   (b) 30 km/s   (c) 19 mi/s  (d) b & c.  click here.

Problem:  In the TV game "Price is right", suppose a person gives an initial angular speed of 2.0 rd/s to the wheel and the wheel comes to stop after 1.5 turns.  Answer the following questions:

28) The final angular speed is (a) 2.0 rd/s   (b) 0   (c) 1.5 rd/s.  click here.

29) The angular displacement, θ, before the wheel comes to stop is (a) 9.42 rd    (b) 0.80rd  (c) 6.28 rd.

30) Since time is not given, one good way to solve for the angular acceleration, α , is to use the equation.............................. .  Write the equation first, and then check your answer.  For answer, click here.

31) The value of α  is (a) 0.21 rd/s2    (b) -0.21 rd/s2    (c) -0.21 rd/s.

32) The elapsed time is (a) 2.8s    (b) 1.4s    (c) 9.5s.

Problem: A car is traveling at a constant speed of 18.0m/s.  The radius of its tires is 30.0cm.  Answer the following:

33) The linear speed of every point on the outer edge of its tires that perform circular motion is (a) 18.0rd/s   (b) 18.0m/s  (c) neither a, nor b.

34) The angular speed ω of each tire is (a) 60 rd/s    (b) 30 rd/s   (c) 30 cm/s.

35) The angular acceleration of each tire is (a) 540 rd/s2   (b) 18 rd/s2   (c) 0.

36) The equation of its angular motion is (a) θ = αt2 + (60 rd/s)t   (b) θ =(60 rd/s)t       (c) θ = αt.

37) The angle each tire rotates in 45 seconds is (a) 2700 turns   (b) 2700o   (c) 2700 rd.


1) Calculate (a) the average angular speed of the Moon about the Earth that completes each turn in about 29 days, and (b) its average linear speed in its motion about the Earth.  The average distance from here to the Moon is 384,000km.

2) A juicer reaches its maximum angular speed of 3600rpm, 2.00s after start.  Find (a) its angular acceleration, (b) maximum linear speed of its porous cylinder wall if its radius is 12.0cm, (c) the centripetal acceleration of points on its inner cylinder wall when at maximum speed, and (d) the angular displacement of any point on its cylinder during the acceleration period.

3) A car tire is spinning at 377 rd/s in a tire balancing equipment.  If it is slowed down to 251 rd/s in 3.00s, calculate (a) its angular acceleration, (b) the angle swept during slowing down, (c) the number of turns made during slowing down, (d) the equation of its rotation, and (e) the angle swept by any radius during the 3rd second.

4) Starting from rest, a mother pushes her daughter in a merry-go-round uniformly for 1/4 of a turn where she reaches a running speed of 6.0m/s.  If the daughter's seat is at an average radius of 5.0m, calculate (a) her initial and final angular speeds, (b) her angular acceleration within the 1/4 turn, (c) the elapsed time, and (d) her tangential and centripetal acceleration when at 1/4 turn position.

5) For rotation about the axis of the Earth, find (a) the angular speed, (b) the linear speed, (c) the angular acceleration, (d) the tangential acceleration, and (e) the centripetal acceleration of the people who live at the 60.0o latitude above the Equator.  Draw a sphere, select a point at the 60o latitude,  and show both of centripetal acceleration and the gravity acceleration vectors at that point.  Note that people at the Equator are at 0o latitude and the people at the North pole are at +90o latitude. The radius of rotation at the Equator is 6280km (the radius of the Earth), and the radius of rotation at each of the poles is zero.  The radius of rotation about the Earth's axis at 60o latitude is (6280km)cos60o =3140km.

Answers:  1) 2.5x10-6 rd/s, 960m/s   2) 188 rd/s2, 45.2m/s, 1.70x104m/s2, 377 rd   

3) -42.0 rd/s2,    942 rd,    150 turns,    θ = -21.0t2 + 377t,    272 rd  

4) 0 and 1.2rd/s,    0.46 rd/s2,    2.6s,     2.3m/s2  and  7.2m/s2     

5) 7.27x10-5 rd/s,    228m/s,    0,     0,     0.0166m/s2