Chapter 7 - Rotational Motion w./ QuickCheck Questions

Chapter 7 - Rotational Motion w./ QuickCheck Questions 2015 Pearson Education, Inc. Anastasia Ierides Department of Physics and Astronomy University of New Mexico October 1, 2015

Review of Last Time Uniform and Nonuniform Circular Motion definitions: period, frequency, speed, acceleration, force Centripetal/Angular/Tangential acceleration Angular velocity, displacement, relations between linear and angular Torque - rotational equivalent to force

Examples of Rotational Motion

QuickCheck Question 7.1 A ball rolls around a circular track with an angular velocity of 4π rad/s. What is the period of the motion? A. 1/2 s B. 1 s C. 2 s D. 1/2π s E. 1/4π s

QuickCheck Question 7.1 A ball rolls around a circular track with an angular velocity of 4π rad/s. What is the period of the motion? A. 1/2 s B. 1 s C. 2 s D. 1/2π s E. 1/4π s Remember: T = (2π rad)/ω = (2π rad)/(4π rad/s) = 1/2 s

QuickCheck Question 7.7 This is the angular velocity graph of a wheel. How many revolutions does the wheel make in the first 4 s? A. 1 rev B. 2 rev C. 4 rev D. 6 rev E. 8 rev

QuickCheck Question 7.7 This is the angular velocity graph of a wheel. How many revolutions does the wheel make in the first 4 s? A. 1 rev B. 2 rev C. 4 rev Remember: θ = ω t D. 6 rev E. 8 rev θ = area under the curve = 1/2 (2rev/s)(2s) + (2 rev/s)(2s) = 2 rev + 4 rev = 6 rev

QuickCheck Question 7.2 Rasheed and Sofia are riding a merry-go-round that is spinning steadily. Sofia is twice as far from the axis as is Rasheed. Sofia s angular velocity is that of Rasheed. A. Half B. The same as C. Twice D. Four times E. We can t say without knowing their radii.

QuickCheck Question 7.3 Rasheed and Sofia are riding a merry-go-round that is spinning steadily. Sofia is twice as far from the axis as is Rasheed. Sofia s speed is that of Rasheed. A. Half B. The same as C. Twice D. Four times E. We can t say without knowing their radii.

QuickCheck Question 7.3 Rasheed and Sofia are riding a merry-go-round that is spinning steadily. Sofia is twice as far from the axis as is Rasheed. Sofia s speed is that of Rasheed. Remember: A. Half v = ωr B. The same as C. Twice D. Four times E. We can t say without knowing their radii. vsofia = ωrsofia = ω(2rrasheed) = 2 (ωrrasheed) = 2 vrasheed

QuickCheck Question 7.4 Two coins rotate on a turntable. Coin B is twice as far from the axis as coin A. A. The angular velocity of A is twice that of B B. The angular velocity of A equals that of B C. The angular velocity of A is half that of B

QuickCheck Question 7.9 Starting from rest, a wheel with constant angular acceleration turns through an angle of 25 rad in a time t. Through what angle will it have turned after time 2t? A. 25 rad B. 50 rad C. 75 rad D. 100 rad E. 200 rad

QuickCheck Question 7.9 Starting from rest, a wheel with constant angular acceleration turns through an angle of 25 rad in a time t. Through what angle will it have turned after time 2t? A. 25 rad Remember: B. 50 rad C. 75 rad D. 100 rad θ = ω t = (α t) t θ t 2 θinitial t 2 E. 200 rad θfinal (2t) 2 4 t 2 4 θinitial

Angular Acceleration Angular acceleration, α, is positive when rotation is speeding up in the counter-clockwise direction or slowing down in the clockwise direction And vice-versa

QuickCheck Question 7.5 The fan blade is slowing down. What are the signs of ω and α? A. ω is positive and α is positive. B. ω is positive and α is negative. C. ω is negative and α is positive. D. ω is negative and α is negative. E. ω is positive and α is zero.

QuickCheck Question 7.5 The fan blade is slowing down. What are the signs of ω and α? Remember: Slowing down and clockwise means that ω and α have opposite signs, and that α is positive. A. ω is positive and α is positive. B. ω is positive and α is negative. C. ω is negative and α is positive. D. ω is negative and α is negative. E. ω is positive and α is zero.

QuickCheck Question 7.5 The fan blade is speeding up. What are the signs of ω and α? A. ω is positive and α is positive. B. ω is positive and α is negative. C. ω is negative and α is positive. D. ω is negative and α is negative.

QuickCheck Question 7.5 The fan blade is speeding up. What are the signs of ω and α? Remember: Speeding up and clockwise means that ω and α have the same signs, not that α is negative. A. ω is positive and α is positive. B. ω is positive and α is negative. C. ω is negative and α is positive. D. ω is negative and α is negative.

QuickCheck Question 7.8 Starting from rest, a wheel with constant angular acceleration spins up to 25 rpm in a time t. What will its angular velocity be after time 2t? A. 25 rpm B. 50 rpm C. 75 rpm D. 100 rpm E. 200 rpm

QuickCheck Question 7.8 Starting from rest, a wheel with constant angular acceleration spins up to 25 rpm in a time t. What will its angular velocity be after time 2t? Remember: A. 25 rpm B. 50 rpm C. 75 rpm D. 100 rpm E. 200 rpm ω = α t t ω t ωinitial t ωfinal 2t 2 ωinitial

QuickCheck Question 7.10 The four forces shown have the same strength. Which force would be most effective in opening the door? A. Force F 1 B. Force F 2 C. Force F 3 D. Force F 4 E. Either F 1 or F 3

QuickCheck Question 7.10 The four forces shown have the same strength. Which force would be most effective in opening the door? A. Force F 1 B. Force F 2 C. Force F 3 D. Force F 4 E. Either F 1 or F 3 Remember: Your intuition likely led you to choose F 1. The reason is that F 1 exerts the largest torque about the hinge. The farther away you apply the force perpendicular to the rigid rod, the less force you hav to apply.

Net Torque Just like the net force, the net torque is the sum of all the torques due to the applied forces, τnet = (r F sin φ)

QuickCheck Question 7.11 Which third force on the wheel, applied at point P, will make the net torque zero?

QuickCheck Question 7.11 Which third force on the wheel, applied at point P, will make the net torque zero? Remember: Since D > d, we know that the force at P has to be larger than the force at D, i.e. FP > FD, and perpendicular to the line segment d for the torques to cancel out d r No torque D φ = 90 A.

Gravitational Torque

Gravitational Torque Each particle in an object experiences torque due to gravity

Gravitational Torque Each particle in an object experiences torque due to gravity This torque is calculated assuming that the net force (weight) acts as a single point called the center of gravity

Example 7.12: The torque on a flagpole A 3.2 kg flagpole extends from a wall at an angle of 25 from the horizontal. Its center of gravity is 1.6 m from the point where the pole is attached to the wall. What is the gravitational torque on the flagpole about the point of attachment?

Example 7.12: The torque on a flagpole A 3.2 kg flagpole extends from a wall at an angle of 25 from the horizontal. Its center of gravity is 1.6 m from the point where the pole is attached to the wall. What is the gravitational torque on the flagpole about the point of attachment? PREPARE The figure shows the situation. For the purpose of calculating torque, we can consider the entire weight of the pole as acting at the center of gravity. Because the moment arm r is simple to visualize here, we ll use Equation 7.11 for the torque.

Example 7.12: The torque on a flagpole SOLVE From the figure, the moment arm is r = (1.6 m) cos 25

Example 7.12: The torque on a flagpole SOLVE From the figure, the moment arm is r = (1.6 m) cos 25 = 1.45 m The gravitational torque on the flagpole, about the point where it attaches to the wall, is thus

Example 7.12: The torque on a flagpole SOLVE From the figure, the moment arm is r = (1.6 m) cos 25 = 1.45 m The gravitational torque on the flagpole, about the point where it attaches to the wall, is thus ASSESS If the pole were attached to the wall by a hinge, the gravitational torque would cause the pole to fall. However, the actual rigid connection provides a counteracting (positive) torque to the pole that prevents this. The net torque is zero.

Center of Gravity An object free to rotate about a pivot will come to rest with the center of gravity below or above that pivot point

Center of Gravity An object free to rotate about a pivot will come to rest with the center of gravity below or above that pivot point There is no torque acting at these positions

QuickCheck Question 7.12 Which point could be the center of gravity of this L-shaped piece?

QuickCheck Question 7.12 Which point could be the center of gravity of this L-shaped piece? (a)

Finding the Center of Gravity The torque due to gravity when the pivot is at the center of gravity is 0

Finding the Center of Gravity The torque due to gravity when the pivot is at the center of gravity is 0 To calculate the position of the center of gravity we find the torque on each side of the pivot

Finding the Center of Gravity The torque due to gravity when the pivot is at the center of gravity is 0 To calculate the position of the center of gravity we find the torque on each side of the pivot The position of the center of gravity is denoted by xcg

Finding the Center of Gravity The torque on the left side of the pivot due to m1 is

Finding the Center of Gravity The torque on the left side of the pivot due to m1 is The torque on the right side of the pivot due to m2 is

Finding the Center of Gravity The total torque then is due to both m1 and m2 is

Finding the Center of Gravity The total torque then is due to both m1 and m2 is Solving for the position of the center of gravity, xcg, we have

Finding the Center of Gravity We see from the relation of the center of gravity to the masses

Finding the Center of Gravity We see from the relation of the center of gravity to the masses that it should lie closer to the heavier mass that makes up the object

Finding the Center of Gravity We see from the relation of the center of gravity to the masses that it should lie closer to the heavier mass that makes up the The center of mass or pivot point is closer to the heavier mass object

Finding the Center of Gravity

Rotation Dynamics Torque causes angular acceleration

Rotation Dynamics Torque causes angular acceleration Angular and tangential accelerations are defined as

Rotation Dynamics Comparing the torque with the acceleration, a relationship is found

Rotation Dynamics Comparing the torque with the acceleration, a relationship is found Yielding a torque of

Newton s Second Law for Rotational Motion All objects are composed of many particles

Newton s Second Law for Rotational Motion All objects are composed of many particles All particles on an object rotating about a fixed axis exhibit the same angular acceleration

Newton s Second Law for Rotational Motion Each particle s torque contributes to the net torque on the object

Newton s Second Law for Rotational Motion The quantity of Σmr 2 is called the moment of inertia, denoted by I

Newton s Second Law for Rotational Motion The quantity of Σmr 2 is called the moment of inertia, denoted by I It has units of kg m 2 and depends on the axis of rotation

Newton s Second Law for Rotational Motion

Newton s Second Law for Rotational Motion Net torque is the cause of angular acceleration!

Interpreting the Moment of Inertia It is the rotational equivalent to mass

Interpreting the Moment of Inertia It is the rotational equivalent to mass It depends on the object s mass and on how the mass is distributed around the rotation axis

Interpreting the Moment of Inertia

Example 7.15: Calculating the Moment of Inertia An abstract sculpture that consists of three small, heavy spheres attached by very lightweight 10-cm-long rods is shown. The spheres have masses m 1 = 1.0 kg, m 2 = 1.5 kg, and m 3 = 1.0 kg. What is the object s moment of inertia if it is rotated about axis A? About axis B?

Example 7.15: Calculating the Moment of Inertia An abstract sculpture that consists of three small, heavy spheres attached by very lightweight 10-cm-long rods is shown. The spheres have masses m 1 = 1.0 kg, m 2 = 1.5 kg, and m 3 = 1.0 kg. What is the object s moment of inertia if it is rotated about axis A? About axis B? PREPARE We ll use for the moment of inertia: I = m 1 r 1 2 + m 2 r 2 2 + m 3 r 3 2 In this expression, r 1, r 2, and r 3 are the distances of each particle from the axis of rotation, so they depend on the axis chosen.

Example 7.15: Calculating the Moment of Inertia Particle 1 lies on both axes, so r 1 = 0 cm in both cases. Particle 2 lies 10 cm (0.10 m) from both axes. Particle 3 is 10 cm from axis A but farther from axis B. We can find r 3 for axis B by using the Pythagorean theorem, which gives r 3 = 14.1 cm. These distances are indicated in the figure.

Example 7.15: Calculating the Moment of Inertia SOLVE For each axis, we can prepare a table of the values of r, m, and mr 2 for each particle, then add the values of mr 2. For axis A we have

Example 7.15: Calculating the Moment of Inertia SOLVE For axis B we have ASSESS We ve already noted that the moment of inertia of an object is higher when its mass is distributed farther from the axis of rotation. Here, m 3 is farther from axis B than from axis A, leading to a higher moment of inertia about that axis.

Moment of Inertia of Common Shapes

Using Newton s 2 nd Law of Rotation

Example 7.18: Starting an airplane engine The engine in a small air-plane has a torque of 500 N m. This engine drives a 2.0-m-long, 40 kg singleblade propeller. On start-up, how long does it take the propeller to reach 2000 rpm?

Example 7.18: Starting an airplane engine The engine in a small air-plane has a torque of 500 N m. This engine drives a 2.0-m-long, 40 kg singleblade propeller. On start-up, how long does it take the propeller to reach 2000 rpm? PREPARE The propeller can be modeled as a rod that rotates about its center. The engine exerts a torque on the propeller. The figure shows the propeller and the rotation axis.

Example 7.18: Starting an airplane engine SOLVE The moment of inertia of a rod rotating about its center is found in Table 7.1:

Example 7.18: Starting an airplane engine SOLVE The moment of inertia of a rod rotating about its center is found in Table 7.1: The 500 N m torque of the engine causes an angular acceleration of

Example 7.18: Starting an airplane engine The time needed to reach is ω f = 2000 rpm = 33.3 rev/s = 209 rad/s

Example 7.18: Starting an airplane engine ASSESS We ve assumed a constant angular acceleration, which is reasonable for the first few seconds while the propeller is still turning slowly. Eventually, air resistance and friction will cause opposing torques and the angular acceleration will decrease. At full speed, the negative torque due to air resistance and friction cancels the torque of the engine. Then and the propeller turns at constant angular velocity with no angular acceleration.

Constraints Due to Ropes and Pulleys Assuming a nonslipping rope, the pulley rim speed and acceleration is equal to the speed and acceleration of the rope and object attached to the rope

Rolling Motion A combination or rotational and translational motion

Rolling Motion A combination or rotational and translational motion Straight line trajectory

Rolling Motion In one revolution, the center moves forward by exactly one circumference (Δx = 2πR), with velocity

Rolling Motion Since the angular velocity, ω = 2π/T, we see that the velocity is equivalent to

Rolling Motion Since the angular velocity, ω = 2π/T, we see that the velocity is equivalent to This is the rolling constraint

Rolling Motion The bottom has translational and rotational velocities in the opposing directions, canceling each other out

Rolling Motion The bottom has translational and rotational velocities in the opposing directions, canceling each other out The point at the point is instantaneously at rest

Example 7.20: Rotating your Tires The diameter of your tires is 0.60 m. You take a 60 mile trip at a speed of 45 mph. a. During this trip, what was your tires angular speed? b. How many times did they revolve?

Example 7.20: Rotating your Tires The diameter of your tires is 0.60 m. You take a 60 mile trip at a speed of 45 mph. a. During this trip, what was your tires angular speed? b. How many times did they revolve? PREPARE The angular speed is related to the speed of a wheel s center by ν = ωr. Because the center of the wheel turns on an axle fixed to the car, the speed v of the wheel s center is the same as that of the car. We prepare by converting the car s speed to SI units:

Example 7.20: Rotating your Tires PREPARE The angular speed is related to the speed of a wheel s center by ν = ωr. Because the center of the wheel turns on an axle fixed to the car, the speed v of the wheel s center is the same as that of the car. We prepare by converting the car s speed to SI units: Once we know the angular speed, we can find the number of times the tires turned from the rotational-kinematic equation Δθ = ω Δt. We ll need to find the time traveled Δt from ν = Δx/Δt.

Example 7.20: Rotating your Tires SOLVE a. From the equation, ν = ωr, we have

Example 7.20: Rotating your Tires SOLVE a. From the equation, ν = ωr, we have b. The time of the trip is

Example 7.20: Rotating your Tires Thus the total angle through which the tires turn is

Example 7.20: Rotating your Tires Thus the total angle through which the tires turn is Because each turn of the wheel is 2π rad, the number of turns is

Example 7.20: Rotating your Tires ASSESS You probably know from seeing tires on passing cars that a tire rotates several times a second at 45 mph. Because there are 3600 s in an hour, and your 60 mile trip at 45 mph is going to take over an hour say, 5000 s you would expect the tire to make many thousands of revolutions. So 51,000 turns seems to be a reasonable answer. You can see that your tires rotate roughly a thousand times per mile. During the lifetime of a tire, about 50,000 miles, it will rotate about 50 million times!

Summary

Things that are due Homework #6 Due October 8, 2015 by 11:59 pm Exam #2 October 6, 2015 at 5:00 pm

EXAM #2 Covers Chapters 4-7 & Lectures 7-13 Tuesday, October 6, 2015 Bring a calculator and cheat sheet (turn in with exam, one side of 8.5 x11 piece of paper) Practice exam will be available on website along with solutions

QUESTIONS?