Module 07: Rotational Motion Flashcards

Question 1

Q

Lesson 7.1 Angular Quantities, Constant Angular Acceleration, and Rolling

What is translational motion?

Answer

A

Translational Motion: object’s center of mass plus rotational motion about the center

Question 2

Q

Lesson 7.1 Angular Quantities, Constant Angular Acceleration, and Rolling

What is purely rotational motion?

Answer

A

Purely Rotational Motion:

All the points in the object move in a circle

The Center of the circles all lie on one line called the axis of rotation
The Axis of rotation is perpendicular to the center

All the points in an object rotate about a fixed axis moves in a circle

Straight-line drawn from the axis to any point in the object sweeps out the same angle in the same time interval.

Question 3

Q

Lesson 7.1 Angular Quantities, Constant Angular Acceleration, and Rolling

What is the angular position?

Answer

A

Angular Position: How far it rotates.

Specify the Θ of some particular line in the object (with respect to a reference line such as the x-axis)

A point in the object (P) moves through an angle Θ when it travels the distance (l) measured along the circumference of its circular path.

Use radians: Θ = l/r, where r = radius and l = arc length subtended by the angle

if l = r then Θ = 1 radians
Dimensionless, since it is a ratio of two lengths

Question 4

Q

Lesson 7.1 Angular Quantities, Constant Angular Acceleration, and Rolling

How do radians and degrees relate to each other?

Answer

A

1 revolution = 360° = 2π
Radians to Degrees:

x° = rad/π * 180°

Degrees to Radians:

rad = π/180° * x°

Question 5

Q

Lesson 7.1 Angular Quantities, Constant Angular Acceleration, and Rolling

What is angular displacement?

Answer

A

ΔΘ = Θ₂ - Θ₁

Question 6

Q

Lesson 7.1 Angular Quantities, Constant Angular Acceleration, and Rolling

What is the angular velocity (ω)?

Answer

A

Average Angular Velocity:

ω = ΔΘ/Δt

Instantaneous Angular Velocity:

ω = lim_{(Δt - 0)} ΔΘ/Δt

Radians per second (rad/s)
Not all points in a rigid object rotate with the same angular velocity
Angular velocity is always perpendicular to the plane of rotation

Question 7

Q

Lesson 7.1 Angular Quantities, Constant Angular Acceleration, and Rolling

1. (I) Express the following angles in radians: (a) 45.0°, (b) 60.0°, (c) 90.0°, (d) 360.0°, and (e) 445°. Give as numerical values and as fractions of π.

Question 8

Q

Lesson 7.1 Angular Quantities, Constant Angular Acceleration, and Rolling

What is the angular acceleration (α)?

Answer

A

Average Angular Acceleration:

α = Δω/Δt

Instantaneous Angular Acceleration:

α = lim_{(Δt - 0)} Δω/Δt

Acceleration same for all points
Radians per second squared (rad/s²)

Question 9

Q

Lesson 7.1 Angular Quantities, Constant Angular Acceleration, and Rolling

What is linear velocity?

Answer

A

Each point has a linear velocity and acceleration

Can relate the linear quantities to angular quantities for a rigid object rotating about a fixed axis

If an object rotates with angular velocity, any point will have a linear velocity
Direction of linear velocity is tangent to the circular path
Magnitude of linear velocity:

v = Δl/Δt = r ΔΘ/Δt = rω

If the angular velocity of a rotating object changes, the object as a whole has an angular acceleration

Question 10

Q

Lesson 7.1 Angular Quantities, Constant Angular Acceleration, and Rolling

What is linear acceleration?

Answer

A

If the angular velocity of a rotating object changes, the object as a whole has an angular acceleration

Linear acceleration is tangent to the point’s circular path

Question 11

Q

Lesson 7.1 Angular Quantities, Constant Angular Acceleration, and Rolling

What is total linear acceleration?

Answer

A

Total Linear Acceleration of a point is the vector sum of two components:a = a_tan + a_R

a_R is the radial component - the “centripetal” acceleration and direction is towards the center of the circular path

Question 12

Q

Lesson 7.1 Angular Quantities, Constant Angular Acceleration, and Rolling

What is the frequency of rotation?

Answer

A

Number of complete revolutions (rev) per second
One revolution = 2π radians
1 Hz = 1 rev / s

f = ω / 2π

Question 13

Q

Lesson 7.1 Angular Quantities, Constant Angular Acceleration, and Rolling

What is rolling without slipping?

Answer

A

Rolling without slipping depends on static friction between object and ground

Friction is static because the object’s point of constant is at rest at each moment
Depends on both translational and rotational movement

v = rω

(Only is there is no slipping)

Question 14

Q

Lesson 7.1 Angular Quantities, Constant Angular Acceleration, and Rolling

3. (I) A laser beam is directed at the Moon, 380,000 km from Earth. The beam diverges at an angle (Fig. 8–40) of 1.4 × 10^–5 rad. What diameter spot will it make on the Moon?

Question 15

Q

Lesson 7.1 Angular Quantities, Constant Angular Acceleration, and Rolling

5. (II) The platter of the hard drive of a computer rotates at 7200 rpm (rpm = revolutions per minute = rev/min). (a) What is the angular velocity (rad/s) of the platter? (b) If the reading head of the drive is located 3.00 cm from the rotation axis, what is the linear speed of the point on the platter just below it? (c) If a single bit requires 0.50 m of length along the direction of motion, how many bits per second can the writing head write when it is 3.00 cm from the axis?

Question 16

Q

Lesson 7.1 Angular Quantities, Constant Angular Acceleration, and Rolling

7. (II) (a) A grinding wheel 0.35 m in diameter rotates at 2200 rpm. Calculate its angular velocity in rad/s. (b) What are the linear speed and acceleration of a point on the edge of the grinding wheel?

Question 17

Q

Lesson 7.1 Angular Quantities, Constant Angular Acceleration, and Rolling

9. (II) Calculate the angular velocity (a) of a clock’s second hand, (b) its minute hand, and (c) its hour hand. State in rad/s. (d) What is the angular acceleration in each case?

Question 18

Q

Lesson 7.1 Angular Quantities, Constant Angular Acceleration, and Rolling

17. (I) An automobile engine slows down from 3500 rpm to 1200 rpm in 2.5 s. Calculate (a) its angular acceleration, assumed constant, and (b) the total number of revolutions the engine makes in this time.

Question 19

Q

Lesson 7.1 Angular Quantities, Constant Angular Acceleration, and Rolling

19. (I) Pilots can be tested for the stresses of flying high-speed jets in a whirling “human centrifuge,” which takes 1.0 min to turn through 23 complete revolutions before reaching its final speed. (a) What was its angular acceleration (assumed constant), and (b) what was its final angular speed in rpm?

Question 20

Q

Lesson 7.1 Angular Quantities, Constant Angular Acceleration, and Rolling

21. (II) A wheel 31 cm in diameter accelerates uniformly from 240 rpm to 360 rpm in 6.8 s. How far will a point on the edge of the wheel have traveled in this time?

Question 21

Q

Lesson 7.1 Angular Quantities, Constant Angular Acceleration, and Rolling

23. (II) A small rubber wheel is used to drive a large pottery wheel. The two wheels are mounted so that their circular edges touch. The small wheel has a radius of 2.0 cm and accelerates at the rate of 7.2 rad/s², and it is in contact with the pottery wheel (radius 27.0 cm) without slipping. Calculate (a) the angular acceleration of the pottery wheel, and (b) the time it takes the pottery wheel to reach its required speed of 65 rpm.

Question 22

Q

Lesson 7.2 Toque

How does force relate to making an object roll?

Answer

A

Making an object start rolling requires a force
Direction important

F_A: perpendicular to the door

greater the magnitude, the faster it will open

F_B:

not open as quickly
$r_A$ is the lever arm

Question 23

Q

Lesson 7.2 Toque

What is angular acceleration?

Answer

A

The angular acceleration is proportional to the (1) magnitude of the force and (2) directly proportional to the perpendicular distance from the axis of rotation to the line along which the force acts

“Distance” is the lever arm or moment arm (r)
Angular Acceleration is proportional to the product of the force times the lever arm
Product is the moment of the force and called the TORQUE (𝜏)

Hence, the angular acceleration is directly proportional to the net applied torque:

α ∝ 𝜏

Analogue with Newton’s Second Law

Question 24

Q

Lesson 7.2 Toque

What is torque in relation to angular acceleration?

Answer

A

Product is the moment of the force and is called the TORQUE (𝜏)

Hence, the angular acceleration is directly proportional to the net applied torque:

α ∝ 𝜏

Question 25

Q

Lesson 7.2 Toque

What is the lever arm and line of action?

Answer

A

Lever Arm: perpendicular distance from the axis of rotation to the line of action of the force (direction of the force)

F_C: line along the direction of Fc (line of action)

Another line perpendicular to line of action
Magnitude of torque: r_CF_C

Line of action of FD passes through the hinge, lever arm = 0

Hence, zero torque & zero angular acceleration

Question 26

Q

Lesson 7.2 Toque

What is the magnitude of the Torque about a Given Axis?

Answer

A

𝜏 = r⊥F

r⊥ is the lever arm (distance from axis is perpendicular to the line of action)

Question 27

Q

Lesson 7.2 Toque

How can torque be resolved through components?

Answer

A

Another way determine to torque is - Resolve the force into components parallel and perpendicular to the line

Question 28

Q

Lesson 7.2 Toque

What happens when there is more than one torque?

Answer

A

α proportional to the net torque

If all the torques are in the same direction: sum of the torques
If the torques are in opposite directions: difference of the torques

Question 29

Q

Lesson 7.2 Toque

What is the direction of torque?

Answer

A

(CW) clockwise is negative
(CCW) counterclockwise is positive

Question 30

Q

Lesson 7.2 Toque

25. (II) Calculate the net torque about the axle of the wheel shown in Fig. 8–42. Assume that a friction torque of 0.60 m • N opposes the motion.

Question 31

Q

Lesson 7.2 Toque

27. (II) Two blocks, each of mass m, are attached to the ends of a massless rod which pivots as shown in Fig. 8–43. Initially the rod is held in the horizontal position and then released. Calculate the magnitude and direction of the net torque on this system when it is first released.

Question 32

Q

Lesson 7.2 Toque

29.* (II) Determine the net torque on the 2.0-m-long uniform beam shown in Fig. 8–45. All forces are shown. Calculate about (a) point C, the CM, and (b) point P at one end.

Question 33

Q

Lesson 7.3 Rotational Dynamics

How do translational motion and acceleration relate?

Answer

A

Angular Acceleration and Net Torque: α ∝ Σ𝜏
Corresponds to Newton’s Second Law: α ∝ ΣF

Translational Motion: Acceleration is proportional to the net force and inversely proportional to the inertia of the object (m): α = ΣF/m

Question 34

Q

Lesson 7.3 Rotational Dynamics

Describe the force of a particle of mass revolved around a circle (r) at the end of a string:

Answer

A

Force tangent to the center
Torque: 𝜏 = Fr
Linear force (Newton’s Second Law): ΣF= ma
Tangential linear acceleration: a_tan =rα

Thus: F = ma = mrα

Torque: 𝜏 = rF = r(mrα) = mr²α

Direct relationship between Angular Acceleration and applied Torque

mr² = rotational inertia of the particle and is the MOMENT of INERTIA

Question 35

Q

Lesson 7.3 Rotational Dynamics

What is the relationship between angular acceleration and torque?

Answer

A

Torque: 𝜏 = rF = r(mrα) = mr²α

Direct relationship between Angular Acceleration and applied Torque
mr² = rotational inertia of the particle and is the MOMENT of INERTIA

Question 36

Q

Lesson 7.3 Rotational Dynamics

What is the sum of torques and moment of inertia of rotating rigid objects?

Answer

A

Rotating rigid object (Wheel as consisting of many particles located at various distances from the axis of rotation):

Sum of various torques: Σ𝜏 = (Σmr²)α

α - the same for all the particles
Σmr²- the sum of the masses of each particle in the object * square of the distance
Σmr²= m₁r₁² + m₂r₂² + m₃r₃² + …

Thus, Moment of Inertia (rotational inertia) I of the object:

I = Σmr²= m₁r₁² + m₂r₂² + m₃r₃² + …

Combining the two equations:

Σ𝜏 =Iα

The rotational equivalent of Newton’s Second Law
Valid of a fixed axis

Question 37

Q

Lesson 7.3 Rotational Dynamics

31. (I) Estimate the moment of inertia of a bicycle wheel 67 cm in diameter. The rim and tire have a combined mass of 1.1 kg. The mass of the hub (at the center) can be ignored (why?).

Question 38

Q

Lesson 7.3 Rotational Dynamics

33. (II) An oxygen molecule consists of two oxygen atoms whose total mass is 5.3 × 10^–26 kg and whose moment of inertia about an axis perpendicular to the line joining the two atoms, midway between them, is 1.9 × 10^–46 kg • m². From these data, estimate the effective distance between the atoms.

Question 39

Q

Lesson 7.3 Rotational Dynamics

35. (II) The forearm in Fig. 8–46 accelerates a 3.6-kg ball at 7.0 m/s2 by means of the triceps muscle, as shown. Calculate (a) the torque needed, and (b) the force that must be exerted by the triceps muscle. Ignore the mass of the arm.

Question 40

Q

Lesson 7.3 Rotational Dynamics

37. (II) A softball player swings a bat, accelerating it from rest to 2.6 rev/s in a time of 0.20 s. Approximate the bat as a 0.90-kg uniform rod of length 0.95 m, and compute the torque the player applies to one end of it.

Question 41

Q

Lesson 7.3 Rotational Dynamics

39. (II) Calculate the moment of inertia of the array of point objects shown in Fig. 8–47 about (a) the y axis, and (b) the x axis. Assume m = 2.2 kg, M = 3.4 kg, and the objects are wired together by very light, rigid pieces of wire. The array is rectangular and is split through the middle by the x axis. (c) About which axis would it be harder to accelerate this array?

Question 42

Q

Lesson 7.3 Rotational Dynamics

41. (II) A dad pushes tangentially on a small hand-driven merry-go-round and is able to accelerate it from rest to a frequency of 15 rpm in 10.0 s. Assume the merry-go-round is a uniform disk of radius 2.5 m and has a mass of 560 kg, and two children (each with a mass of 25 kg) sit opposite each other on the edge. (a) Calculate the torque required to produce the acceleration, neglecting frictional torque. (b) What force is required at the edge?

Question 43

Q

Lesson 7.3 Rotational Dynamics

43. (II) Let us treat a helicopter rotor blade as a long thin rod, as shown in Fig. 8–49. (a) If each of the three rotor helicopter blades is 3.75 m long and has a mass of 135 kg, calculate the moment of inertia of the three rotor blades about the axis of rotation. (b) How much torque must the motor apply to bring the blades from rest up to a speed of 6.0 rev/s in 8.0 s?

Question 44

Q

Lesson 7.4 Rotational Kinetic Energy

What is rotational kinetic energy?

Answer

A

Rotational Objects (rigid rotating objects)
Joules
Total kinetic energy is the sum of the energies:

KE = Σ(1/2mv²) = Σ(1/2mr²ω²) = 1/2(Σmv²)ω²

Since I = Σmr²

KE_r = 1/2Iω²

Question 45

Q

Lesson 7.4 Rotational Kinetic Energy

What type of kinetic energy does an object while its CM undergoes translational motion?

Answer

A

Object while its CM undergoes translational motion will have both translational and rotational kinetic energy:

KE = 1/2Mv²_CM + 1/2I_CMω²

M = total mass
v²_CM = linear velocity of the center of the mass
I_CM= Moment of Inertia
ω = Angular velocity

Question 46

Q

Lesson 7.4 Rotational Kinetic Energy

What is to work done by torque?

Answer

A

Work done on an object rotating about a fixed axis
Work: W = FΔl
Small Angle: ΔΘ = Δl/r

Hence, W = FΔl = W = FrΔΘ

Torque: 𝜏 = rF

Then, W = 𝜏ΔΘ

Power(Rate of Work): P = W/Δt = 𝜏ΔΘ/Δt = 𝜏ω

Similar to the translational version (P=Fv)

Question 47

Q

Lesson 7.4 Rotational Kinetic Energy

49. (I) An automobile engine develops a torque of 265 m • N at 3350 rpm. What is the horsepower of the engine?

Question 48

Q

Lesson 7.4 Rotational Kinetic Energy

51. (I) Calculate the translational speed of a cylinder when it reaches the foot of an incline 7.20 m high. Assume it starts from rest and rolls without slipping.

Question 49

Q

Lesson 7.4 Rotational Kinetic Energy

53. (II) Estimate the kinetic energy of the Earth with respect to the Sun as the sum of two terms, (a) that due to its daily rotation about its axis, and (b) that due to its yearly revolution about the Sun. [Assume the Earth is a uniform sphere with mass = 6.0 × 10²⁴ kg, radius = 6.4 × 10⁶ m, and is 1.5 × 10⁸ km from the Sun.]

Question 50

Q

Lesson 7.4 Rotational Kinetic Energy

55. (II) A merry-go-round has a mass of 1440 kg and a radius of 7.50 m. How much net work is required to accelerate it from rest to a rotation rate of 1.00 revolution per 7.00 s? Assume it is a solid cylinder.

Question 51

Q

Lesson 7.4 Rotational Kinetic Energy

57. (II) A ball of radius r rolls on the inside of a track of radius R (see Fig. 8–53). If the ball starts from rest at the vertical edge of the track, what will be its speed when it reaches the lowest point of the track, rolling without slipping?

Question 52

Q

Lesson 7.5 Angular Momentum

What is Angular Momentum (L)?

Answer

A

Linear momentum: p=mv

Symmetrical object rotating about a fixed axis through the CM: L=Iω

SI units: kg * m²/s

Question 53

Q

Lesson 7.5 Angular Momentum

What is Newton’s Second Law for Rotation?

Answer

A

Σ𝜏 = ΔL / Δt

Σ𝜏 = net torque
ΔL = Change in Angular Momentum
Σ𝜏 = Iα is a special case where the moment of inertia is constant

Question 54

Q

Lesson 7.5 Angular Momentum

What is the conservation of angular momentum?

Answer

A

The total angular momentum of a rotating object remains constant if the net torque acting on it is zero

If there is zero torque, the object is rotating about a fixed axis: Iω = I₀ω₀ = constant

Question 55

Q

Lesson 7.5 Angular Momentum

61. (I) (a) What is the angular momentum of a 2.8-kg uniform cylindrical grinding wheel of radius 28 cm when rotating at 1300 rpm? (b) How much torque is required to stop it in 6.0 s?

Question 56

Q

Lesson 7.5 Angular Momentum

63. (II) A nonrotating cylindrical disk of moment of inertia I is dropped onto an identical disk rotating at angular speed w. Assuming no external torques, what is the final common angular speed of the two disks?

Question 57

Q

Lesson 7.5 Angular Momentum

65. (II) A figure skater can increase her spin rotation rate from an initial rate of 1.0 rev every 1.5 s to a final rate of 2.5 rev/s. If her initial moment of inertia was 4.6 kg • m², what is her final moment of inertia? How does she physically accomplish this change?

Question 58

Q

Lesson 7.5 Angular Momentum

67. (II) A person of mass 75 kg stands at the center of a rotating merry-go-round platform of radius 3.0 m and moment of inertia 820 kg • m². The platform rotates without friction with angular velocity 0.95 rad/s. The person walks radially to the edge of the platform. (a) Calculate the angular velocity when the person reaches the edge. (b) Calculate the rotational kinetic energy of the system of platform plus person before and after the person’s walk.

Question 59

Q

Lesson 7.5 Angular Momentum

69. (II) A 4.2-m-diameter merry-go-round is rotating freely with an angular velocity of 0.80 rad/s. Its total moment of inertia is 1360 kg • m². Four people standing on the ground, each of mass 65 kg, suddenly step onto the edge of the merry-go-round. (a) What is the angular velocity of the merry-go-round now? (b) What if the people were on it initially and then jumped off in a radial direction (relative to the merry-go-round)?

Answer

A

b. The angular velocity of the merry-go-round will stay the same because the people are jumping in the radial direction, so along or “with” the curved path of the merry-go-round, so they will not provide any resistant force against the movement of the merry-go-round, so its angular velocity will remain the same as before the people stepped onto the merry-go-round.

Question 60

Q

Quiz 7.2

A car is traveling along a freeway at 65 mph. What is the linear speed, relative to the highway, of each of the following points on one of its tires?

*(a)** the highest point on the tire
*(b)** the lowest point on a tire
*(c)** the center of the tire

Answer

A

(a) v = 2rω = 130 mph (r = 2R)

(b) v = 0 = 65 mph (r = R)

(c) v = rω = (r = 0)

Question 61

Q

Quiz 7.2

A chicken is running in a circular path with an angular speed of 1.52 rad/s. How long does it take the chicken to complete one revolution?

4.13 s
118 s
4.77 s
8.26 s
2.07 s

Question 62

Q

Quiz 7.2

Suppose a uniform solid sphere of mass M and radius R rolls without slipping down an inclined plane starting from rest. The linear velocity of the sphere at the bottom of the incline depends on

the radius of the sphere.
both the mass and the radius of the sphere.
the mass of the sphere.
neither the mass nor the radius of the sphere.

Answer

A

neither the mass nor the radius of the sphere

Question 63

Q

Quiz 7.2

At a certain instant, a compact disc is rotating at 210 rpm. What is its angular speed in rad/s?

660 rad/s
69 rad/s
22 rad/s
45 rad/s
11 rad/s

Question 64

Q

Quiz 7.2

An electrical motor spins at a constant 2695.0. If the rotor radius is 7.165 cm, what is the linear acceleration of the edge of the rotor?

281.6 m/s²
5707 m/s²
28.20 m/s²
572,400 m/s²

Answer

A

5707 m/s²

Answer 35

A

85.14 rad/s

Answer 36

A

The distance will be 40 cm.

Answer 37

A

the first force (applied perpendicular to the door)

Answer 38

A

The angular speed is 1.1×10^-3 rad/s

Answer 39

A

Translational kinetic energy is larger

Answer 40

A

Part B:

Linear acceleration = a_tan+ a_radial

a_tan = Since the tangential speed is constant, the tangential acceleration is zero

a_radial = w²r = 1.18 rad/s²

Therefore, Linear acceleration = 0 + 1.18 rad/s² = 1.18 m/s²

Answer 41

A

to your left

Answer 42

A

=(9.28×10^-3rad)(1.5×10¹¹m)

=1.4×10⁹m

Answer 43

A

3.5 ^rad/_sec

Answer 44

A

(a) The angular acceleration is approximately 230rad/s²

(b) The centrifuge takes approximately 1.7 seconds to come to rest

Answer 45

A

0.51 J

Answer 46

A

1.97 m/s

Answer 47

A

2.36 m/s

W_total = KE

Fx = 1/2 Iw²*
v = wr*

Answer 48

A

2.3 m/s

Answer 49

A

(c) In part A, the object is not slipping meaning there is static friction and the motion depends on both translational and rotational movement. Therefore, the kinetic energy in part A is both translational and rotational. Comparatively, when there is slipping and the force of friction is zero, there is only translational motion, so the speeds are different in each case because the kinetic energy is made up of different components depending on how friction influences the movement of the ball.