In this chapter rotational motion will be discussed. Angular displacement, angular velocity, and angular acceleration will be defined. The first two were discussed in Chapter 5.
Angular Displacement: The symbol generally used for angular displacement is θ pronounced "teta" or "theta." θ is the angle swept by the radius of a circle that points to a rotating mass, M, under study. Below, the symbol ω is pronounced "omega" is used to denote angular velocity.
Example 1: An object travels around a circle10.0 full turns in 2.5 seconds. Calculate (a) the angular displacement, θ in radians, and (b) its average angular speed, ω in (rd/s).
Solution: (a) θ = 10.0 turns ( 6.28 rd / turn ) = 62.8 radians.
(b) ω = Δθ / Δt = 62.8 rd / 2.5s = 25 rd/s.
Angular Velocity (ω): Angular velocity is defined as the change in angular displacement, θ, per unit of time, t.
ω = Δθ/Δt ; [rd/s]
Angular Acceleration (α): Angular acceleration is the change in the angular velocity, ω, per unit of time, t.
α = Δω/Δt ; [rd/s2]
The symbol α is pronounced " Alpha."
Example 2: A car tire is turning at a rate of 5.0 rd / sec as the car travels along a road. The driver increases the car's speed, and as a result, each tire's angular speed increases to 8.0 rd /sec in 6.0 sec. Find the angular acceleration of the tire.
Solution: α = Δω / Δt ; α = (ωf - ωf) /Δt ; α = ( 8.0 rd/s - 5.0 rd/s ) / 6.0s = 0.50 rd/s2.
As can be seen, there is an excellent correspondence between the linear formulas we learned in Chapter 2 and the angular formulas defined here. The following chart, shows the one-to-one correspondence between the linear and angular variables and formulas:
Variables: x, t, v, and a
v = Δx / Δt.
a = Δv / Δt ; a = ( vf - vi ) / Δt.
x = (1/2) a t2 + vi t.
vf2 - vi2 = 2 a x.
θ, t, ω,
ω = Δθ / Δt.
α = Δω / Δt ; α = ( ωf - ωi ) / Δt.
θ = (1/2) α t2 + ωi t.
ωf2 - ωi2 = 2 α θ.
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 at by factor R, the radius of rotation.
x = Rθ ; v = Rω ; at = Rα . (at means tangential acceleration).
This can be easily verified by the following simple mathematics:
Starting with s = Rθ, or x = Rθ and writing 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 ω ; therefore, v = Rω.
If v = Rω is divided through by Δt , yields:
Δv/Δt = R Δω/Δt ; the left side is at and the right side is α ; therefore, at = Rα.
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 push 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.
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 after 3.0s, and (c) the angle every particle of its tires travels during the 3.0-second period.
Solution: (a) Since a linear variable and its corresponding angular variable are given, the radius of rotation can be
calculated. 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.
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 3600.0 rpm in 4.00 seconds. For the pulp, determine (a) the angular acceleration, (b) the angle (radians) it travels 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: (a) First calculate ω in (rd/s) ; ω = 3600 [rev/min] [6.28 rd/rev] [1 min / 60s] = 377 rd /s.
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. (Refer to at in the above figure).
For Part (d), the values for final angular speed, ωf , must be calculated both at t = 2.00s and t = 4.00s.
α = (ωf - ωi) /Δt ; α Δt = ωf - ωi ; ωf = ωi + α Δt ; (ωf)1 = 0 +(94.3rd/s2)(2.0s) = 189 rd/s.
From the problem, at t = 4.00s, ω = 3600 rpm = 377 rd/s or, (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 = 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 (3 sig. figures) ( 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:
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) 18.0 rd/s (b) 18.0 m/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 + (60.0 rd/s)t (b) θ = (60.0rd/s)t (c) θ = ωt.
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 traveled during slowing down, (c) the number of turns made during slowing down, (d) the equation of its rotation, and (e) the angle traveled 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)cos60 =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 = ½ α t2 + ωi t, 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