AP Physics C - Mechanics - content.njctl.orgcontent.njctl.org/courses/science/ap-physics-c-mechanics/... · AP Physics C - Mechanics Rotational Motion ... What is Rotational Motion?

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AP Physics C - Mechanics

Rotational Motion

2015-12-03

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Table of Contents

· Rotational Kinematics Review

· Rotational Dynamics

· Rotational Kinetic Energy

Click on the topic to go to that section

· Angular Momentum

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Rotational Kinematics Review

Return to Tableof Contents

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What is Rotational Motion?

Linear Kinematics and Dynamics dealt with the motion of "point particles" moving in straight lines. Life is more complex than that.

Objects have an underlying structure, so it cannot be assumed that a force applied anywhere on the object results in the same kind of motion.

In football (American and the rest of the world), if a player is struck from the front, at the ankle, he will fall forward, pivoting about his foot. The same force applied in his midsection will push him backwards.

The player is rotating about his ankle in the first case, and translating (linear) backwards in the second case.

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Axis of Rotation

In the first case, the player was rotated about his axis of rotation - a line parallel to the ground and through his ankle, perpendicular to the force.

Assume that all the objects that we have been dealing with are rigid bodies, that is, they keep their shape and are not deformed in any way by their motion.

"Rotating Sphere". Licensed under Creative Commons Attribution-Share Alike 3.0 via Wikimedia Commons - http://commons.wikimedia.org/wiki/File:Rotating_Sphere.gif#mediaviewer/File:Rotating_Sphere.gif

Here's a sphere rotating aboutits axis of rotation - the verticalred line.

Does the axis of rotation have to be part of the rigid body?

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Axis of Rotation

No - if you were to spin this donut around its center, the axis of rotation would be in the donut hole, pointing out of the page.

"Chocolate dip,2011-11-28" by Pbj2199 - Own work. Licensed under Creative Commons Attribution-Share Alike 3.0 via Wikimedia Commons - http://commons.wikimedia.org/wiki/File:Chocolate_dip,2011-11-28.jpg#mediaviewer/File:Chocolate_dip,2011-11-28.jpg

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Rotational Equivalents to Linear (Translational) Motion

Linear motion is described by displacement, velocity and acceleration.

Rotational motion is described by angular displacement, angular velocity and angular acceleration.

It is also important to define the axis of rotation - objects will have different properties depending on this choice. Once the choice is made, it must remain the same to describe the motion. If the axis is changed, then a different rotational motion occurs.

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Angular Displacement

There are several theories why a circle has 360 degrees. Here's a few - if you're interested, there's a lot of information that can be found on the web.

Some people believe it stems from the ancient Babylonians who had a number system based on 60 instead of our base 10 system.

Others track it back to the Persian or biblical Hebrew calendars of 12 months of 30 days each.

But - it has nothing to do with the actual geometry of the circle. There is a more natural unit - the radian, which is the rotational equivalent of displacement (Δs).

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A

B

Angular Displacement

As a rigid disc (circle) rotates about its center, the angle of rotation, θ, is defined as the ratio of the subtended arc length, s, to the radius of the circle, and is named angular displacement.

Rotational angles are now defined in geometric terms, not based on an old calendar or arbitrary numbering system.

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A

B

RadianAngular displacement is unit less since it is the ratio of two distances. But, we will say that angular displacement is measured in radians (abbreviated "rad"). A point moving around the entire circumference will travel 2πr as an angle of 3600 is swept through.

Using the angular displacement definition:

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A1

B1

A2 A3

B2

B3

Angular DisplacementSomething interesting - look at the three concentric circles drawn on the rigid disc the radii, arc lengths sr and the points Ar and Br.

As the disk rotates, each point A moves to point B, covering the SAME angle θ, but covering a different arc length sr.

Thus, all points on a rotating rigid disc move through the same angle, θ, but different arc lengths s, and different radii, r.

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1 What is the angular displacement for an arc length (s) that is equal to the radius of the circular rigid body?

A 0.5 rad

B 1.0 rad

C 0.5π rad

D 1.0π rad

E 1.5π rad

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1 What is the angular displacement for an arc length (s) that is equal to the radius of the circular rigid body?

A 0.5 rad

B 1.0 rad

C 0.5π rad

D 1.0π rad

E 1.5π rad[This object is a pull tab]

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B

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2 A record spins 4 times around its center. It makes 4 revolutions. How many radians did it pass?

A π rad

B 2π rad

C 4π rad

D 8π rad

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A π rad

B 2π rad

C 4π rad

D 6π rad

E 8π rad



A π rad

B 2π rad

C 4π rad

D 6π rad

E 8π rad [This object is a pull tab]

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E

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4 A circular hoop of radius 0.86 m rotates π/3 rad about its center. A small bug is on the hoop - what distance does it travel (arc length) during this rotation?

A 0.30 m

B 0.90 m

C 1.4 m

D 2.7 m

E 3.2 m


4 A circular hoop of radius 0.86 m rotates π/3 rad about its center. A small bug is on the hoop - what distance does it travel (arc length) during this rotation?

A 0.30 m

B 0.90 m

C 1.4 m

D 2.7 m

E 3.2 m[This object is a pull tab]

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B

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5 There are two people on a merry go round. Person A is 2.3 m from the axis of rotation. Person B is 3.4 m from the axis of rotation. The merry go round moves through an angular displacement of π/4. What linear displacement (arc length) is covered by both people? Compare and contrast these motions.


5 There are two people on a merry go round. Person A is 2.3 m from the axis of rotation. Person B is 3.4 m from the axis of rotation. The merry go round moves through an angular displacement of π/4. What linear displacement (arc length) is covered by both people? Compare and contrast these motions.

[This object is a pull tab]

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The person further from theaxis of rotation moves a greater linear displacement, but the same angular displacement as the personcloser to the axis of rotation.

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Other Angular Quantities

Angular displacement has now been be related to Linear displacement.

Next, let's relate angular velocity and angular acceleration to their linear equivalents:

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Define average Angular Velocity (ωavg) as the change in angular displacement over time, and take the limit as Δt approaches zero for the instantaneous angular velocity (ω).

Similarly for average Angular Acceleration (α) and instantaneous angular acceleration (α).

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Since ω and α are both related to θ, and θ is the same for all points on a rotating rigid body, ω and α are also the samefor all points on a rotating rigid body. But the same is not true for linear velocity and acceleration.

Let's relate the linear velocity and acceleration to their angular equivalents.

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Angular and Linear VelocityStart with angular velocity, and substitute in the linear displacement for the angular displacement.

r is constant, so moveit outside the derivative

This confirms what you feel on a merry go round - the further away from the center you move, the faster you feel you're going - you're feeling the linear velocity!

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Angular and Linear AccelerationStart with angular acceleration, and substitute in the linear velocity for the angular velocity.

r is constant, so moveit outside the derivative

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Angular and Linear QuantitiesSummary

Angular Linear Relationship

Displacement

Velocity

Acceleration

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A

B

Angular Velocity sign

We're familiar with how to assign positive and negative values to displacement. Typically if you move to the right or up, we give that a positive displacement (of course, this is arbitrary, but it's a pretty standard convention).

How do we assign a "sign"for angular displacement,velocity or acceleration?

Does "right" or "up" work with rotational motion? Doesit work with a circle?

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Not really. Not at all.

From your math classes, you know that the horizontal axis through a circle is labeled as 00, and angles are measured in a counter clockwise direction.

Once we agree on that, let's look at the definition of angular velocity:

A

B

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As the disc rotates in a counter clockwise fashion, θ increases, so Δθ is positive. Since dt is also positive, dθ/dt must be positive.

Thus, Counter clockwise rotations result in a positive ω.

A

B

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6 Explain why a disc that rotates clockwise leads to a negative value for angular velocity, ω. You can use an example to show your point.

Students type their answers here


6 Explain why a disc that rotates clockwise leads to a negative value for angular velocity, ω. You can use an example to show your point.

Students type their answers here


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When the disc rotates clockwise, the angle its radius makes with the horizontal decreases. Therefore, θf - θ0 = Δθ is negative. Since ω=Δθ/Δt, then ω is negative.

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Angular Acceleration sign

Start with the definition of angular acceleration:

There are four cases here, all similar to the case of linear acceleration. Can you figure them out?

Here's a hint - for linear acceleration, if an object's velocity is increasing in the same direction of its displacement, the acceleration is positive.

If the object's velocity is decreasing in the direction of its displacement, then its acceleration is negative.

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Angular Acceleration sign

Here they are:

ω increases in counter clockwise direction - α is positiveω decreases in the counter clockwise direction - α is negative

ω increases in clockwise direction - α is negativeω decreases in the clockwise direction - α is positive

These conventions are arbitrary, and in this case are based on the axis of rotation, but once chosen, they need to stay consistent for solving problems. There will be problems where the choice is not so obvious, but just make a choice and stay consistent.

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7 What is the angular acceleration of a ball that starts at rest and increases its angular velocity uniformly to 5 rad/s in 10 s?

A 0.5 rad/s2

B 1.0 rad/s2

C 1.5 rad/s2

D 2.0 rad/s2

E 2.5 rad/s2


7 What is the angular acceleration of a ball that starts at rest and increases its angular velocity uniformly to 5 rad/s in 10 s?

A 0.5 rad/s2

B 1.0 rad/s2

C 1.5 rad/s2

D 2.0 rad/s2

E 2.5 rad/s2 [This object is a pull tab]

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A

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Three types of Acceleration

So, we have two types of displacement, velocity and acceleration - linear and rotational.

There is a third type of acceleration - centripetal.

This is the acceleration that a point on a rotating disc experiences when it moves in a circular direction.

The v in the above equation is linear velocity, and for an object moving in a circle, it is the velocity tangent to the circle.

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ac

aT

α

Three types of AccelerationLook at the disc below which shows the linear acceleration that we've defined as well as the centripetal acceleration.

But, centripetal acceleration is also linear! So we will rename the linear acceleration we've been dealing with in this chapter as tangential acceleration.

The angular acceleration, α, is a vector that points out of the page.

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Three types of Acceleration

We now have three accelerations associated with a rigid rotating body: tangential, angular and centripetal, or aT, α, and aC.

They are related to the angular velocity, ω, as follows:

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A

B

DCE

8 At which point is the magnitude of the centripetal acceleration the greatest?

A

B

C

D

E

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ac

aTatotal

Total Linear Acceleration

There are two flavors of linear acceleration - tangential and centripetal. They are perpendicular to each other; using vector addition and the Pythagorean Theorem, the total linear acceleration is found:

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9 A child pushes, with a constant force, a merry-go-round with a radius of 2.5 m, from rest to an angular velocity of 3.0 rad/s in 8.0 s. What is the merry-go-round's tangential acceleration?

A 0.25 m/s2

B 0.35 m/s2

C 0.62 m/s2

D 0.75 m/s2

E 0.94 m/s2


9 A child pushes, with a constant force, a merry-go-round with a radius of 2.5 m, from rest to an angular velocity of 3.0 rad/s in 8.0 s. What is the merry-go-round's tangential acceleration?

A 0.25 m/s2

B 0.35 m/s2

C 0.62 m/s2

D 0.75 m/s2

E 0.94 m/s2 [This object is a pull tab]

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E

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10 A child pushes, with a constant force, a merry-go-round with a radius of 2.5 m, from rest to an angular velocity of 3.0 rad/s in 8.0 s. What is the merry-go-round's centripetal acceleration?A 7.5 m/s2

B 19 m/s2

C 23 m/s2

D 38 m/s2

E 46 m/s2


10 A child pushes, with a constant force, a merry-go-round with a radius of 2.5 m, from rest to an angular velocity of 3.0 rad/s in 8.0 s. What is the merry-go-round's centripetal acceleration?A 7.5 m/s2

B 19 m/s2

C 23 m/s2

D 38 m/s2

E 46 m/s2 [This object is a pull tab]

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C

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11 A child pushes, with a constant force, a merry-go-round with a radius of 2.5 m, from rest to an angular velocity of 3.0 rad/s in 8.0 s. What is the merry-go-round's total linear acceleration?A 5.6 m/s2

B 12 m/s2

C 23 m/s2

D 38 m/s2

E 46 m/s2


11 A child pushes, with a constant force, a merry-go-round with a radius of 2.5 m, from rest to an angular velocity of 3.0 rad/s in 8.0 s. What is the merry-go-round's total linear acceleration?A 5.6 m/s2

B 12 m/s2

C 23 m/s2

D 38 m/s2

E 46 m/s2 [This object is a pull tab]

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C

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Angular Velocity and Frequency

Frequency, f, is defined as the number of revolutions an object makes per second and f = 1/T, where T is the period.

A rotating object moves through an angular displacement, θ = 2πr rad, in one revolution.

The linear velocity of any point on a rotating object is given by the following equation. THESE ARE VERY VALUABLE EQUATIONS for solving rotational motion problems.

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12 A mining truck tire with a radius of 2.1 m rolls with an angular velocity of 8.0 rad/s. What is the frequency of the tire's revolutions? What is the period?

A 1.3 rev/s 0.40 s

B 1.3 rev/s 0.77 s

C 2.6 rev/s 0.39 s

D 2.6 rev/s 0.78 s

E 3.9 rev/s 0.78 s


12 A mining truck tire with a radius of 2.1 m rolls with an angular velocity of 8.0 rad/s. What is the frequency of the tire's revolutions? What is the period?

A 1.3 rev/s 0.40 s

B 1.3 rev/s 0.77 s

C 2.6 rev/s 0.39 s

D 2.6 rev/s 0.78 s

E 3.9 rev/s 0.78 s[This object is a pull tab]

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B

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Translational Kinematics

Now that all the definitions for rotational quantities have been covered, we're ready to discuss the motion of rotating rigid objects - Rotational Kinematics.

First, let's review the equations for translational kinematics - this was one of the first topics that was covered in this course.

*Important* These equations are ONLY valid for cases involving a constant acceleration.

This allows gravitational problems to be solved (a = g, a constant) and makes it possible to solve kinematics equations without advanced calculus. When acceleration changes, the problems are more complex.

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Translational Kinematics

Here are the equations. If you would like, please review their derivations in the Kinematics section of this course.

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Rotational Kinematics

For each of the translational kinematics equations, there is an equivalent rotational equation. We'll derive one of them and just present the others - basically each translational variable is replaced with its rotational analog.

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Rotational Kinematics

Here they are. We've been using Δs for linear displacement, but Δx is just fine. Similar to the translational equations, these ONLY work for constant angular acceleration.

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Rolling Without Slipping

Rolling Motion is a practical application of rotational kinematics. The best example is a tire rotating as the car moves. Another example is a rotating pulley (Atwood's machine).

The key in both examples is there is no slipping - either the tire on the road or the string on the pulley.

In these cases, the equations derived relating angular velocity and acceleration to linear velocity and acceleration hold.

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A

A

A

d


Here's a rotating circle that represents a car tire moving down the road.

The tire makes one complete revolution, so point A on the tire moves a total arc length of s = 2πr. If the tire does not slip on the road, then the linear distance moved on the road during the full revolution, d, is equal to s.

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The tangential velocity of point A is s/t while the linear velocity of the wheel is d/t. Recognize s/t as vT = rω.

That gives us vlinear = vT = rω. By similar reasoning, we havealinear = rα. These equations will come in very handy when solving rolling motion problems.

A

A

A

d

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CAUTION. If the object is rolling with slipping then these two equations cannot be used. If a car tire is spinning on ice, and the car is not moving, then vlinear = 0, and ω can be very large.

Use only for rolling motion without slipping.

A

A

A

d

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The rotational motion equations also work as part of the equations for rolling motion.

Kinematics Equations & Rolling Motion

Rolling motion is simply rotational motion and linear motion combined.

Think of a wheel on a car:

It is rotating, showing rotational motion.

It also moves forward as it rotates, therefore showing linear motion.

Thus, we use the translational and rotational kinematics equations to solve rolling motion problems.

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13 A bicycle wheel with a radius of 0.30 m starts from rest and accelerates at a rate of 4.5 rad/s2 for 11 s. What is its final angular velocity?

A 20 rad/s

B 30 rad/s

C 40 rad/s

D 50 rad/s

E 60 rad/s


13 A bicycle wheel with a radius of 0.30 m starts from rest and accelerates at a rate of 4.5 rad/s2 for 11 s. What is its final angular velocity?

A 20 rad/s

B 30 rad/s

C 40 rad/s

D 50 rad/s

E 60 rad/s[This object is a pull tab]

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D

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14 A bicycle wheel with a radius of 0.30 m starts from rest and accelerates at a rate of 4.5 rad/s2 for 11 s. What is its final linear velocity?

A 25 m/s

B 20 m/s

C 15 m/s

D 10 m/s

E 5 m/s


14 A bicycle wheel with a radius of 0.30 m starts from rest and accelerates at a rate of 4.5 rad/s2 for 11 s. What is its final linear velocity?

A 25 m/s

B 20 m/s

C 15 m/s

D 10 m/s

E 5 m/s[This object is a pull tab]

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C

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15 A bicycle wheel with a radius of 0.300 m starts from rest and accelerates at a rate of 4.50 rad/s2 for 11.0 s. What is its angular displacement?

A 195 rad

B 200 rad

C 230 rad

D 251 rad

E 273 rad


15 A bicycle wheel with a radius of 0.300 m starts from rest and accelerates at a rate of 4.50 rad/s2 for 11.0 s. What is its angular displacement?

A 195 rad

B 200 rad

C 230 rad

D 251 rad

E 273 rad[This object is a pull tab]

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E

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16 A bicycle wheel with a radius of 0.30 m starts from rest and accelerates at a rate of 4.5 rad/s2 for 11 s. How many revolutions did it make?

A 50 rev

B 45 rev

C 43 rev

D 41 rev

E 39 rev


16 A bicycle wheel with a radius of 0.30 m starts from rest and accelerates at a rate of 4.5 rad/s2 for 11 s. How many revolutions did it make?

A 50 rev

B 45 rev

C 43 rev

D 41 rev

E 39 rev[This object is a pull tab]

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C

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17 A bicycle wheel with a radius of 0.30 m starts from rest and accelerates at a rate of 4.5 rad/s2 for 11 s. What is its linear displacement?

A 40 m

B 82 m

C 160 m

D 180 m

E 210 m


17 A bicycle wheel with a radius of 0.30 m starts from rest and accelerates at a rate of 4.5 rad/s2 for 11 s. What is its linear displacement?

A 40 m

B 82 m

C 160 m

D 180 m

E 210 m[This object is a pull tab]

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B

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18 A 50.0 cm diameter wheel accelerates from 5.0 revolutions per second to 7.0 revolutions per second in 8.0 s. What is its angular acceleration?

A π rad/s2

B π/2 rad/s2

C π/3 rad/s2

D π/4 rad/s2

E π/5 rad/s2


18 A 50.0 cm diameter wheel accelerates from 5.0 revolutions per second to 7.0 revolutions per second in 8.0 s. What is its angular acceleration?

A π rad/s2

B π/2 rad/s2

C π/3 rad/s2

D π/4 rad/s2

E π/5 rad/s2 [This object is a pull tab]

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B

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19 A 50.0 cm diameter wheel accelerates from 5.0 revolutions per second to 7.0 revolutions per second in 8.0 s. What is its angular displacement?

A 56π rad

B 82π rad

C 96π rad

D 112π rad

E 164π rad


19 A 50.0 cm diameter wheel accelerates from 5.0 revolutions per second to 7.0 revolutions per second in 8.0 s. What is its angular displacement?

A 56π rad

B 82π rad

C 96π rad

D 112π rad

E 164π rad [This object is a pull tab]

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C

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20 A 50.0 cm diameter wheel accelerates from 5.0 revolutions per second to 7.0 revolutions per second in 8.0 s. What linear displacement, s, will a point on the outside edge of the wheel have traveled during that time?

A 24π rad

B 30π rad

C 36π rad

D 42π rad

E 48π m


20 A 50.0 cm diameter wheel accelerates from 5.0 revolutions per second to 7.0 revolutions per second in 8.0 s. What linear displacement, s, will a point on the outside edge of the wheel have traveled during that time?

A 24π rad

B 30π rad

C 36π rad

D 42π rad

E 48π m [This object is a pull tab]

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E

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Rotational Dynamics


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Rotational Dynamics

Just like there are rotational analogs for Kinematics, there are rotational analogs for Dynamics.

Kinematics allowed us to solve for the motion of objects, without caring why or how they moved.

Dynamics showed how the application of forces causes motion - and this is summed up in Newton's Three Laws.

These laws also apply to rotational motion.

The rotational analogs to Newton's Laws will be presented now.

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TorqueForces underlie translational dynamics. A new term, related to force is the foundation of causing rotational motion. It is called Torque.

http://commons.wikimedia.org/wiki/File%3AAction_meca_equivalence_force_couple_plan.svg

The wrench is about to turna nut by a force applied at point A. The nut is on a bolt that is attached to a wall.Will the nut turn, and in which direction?

Consider the axis of rotation to be in the middle of the nut, and pointed into the page.

What if a force is applied at point B?

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TorqueA force at point A will turn the nut in a clockwise direction - recall that "left" and "right" have no meaning in rotational motion. This is good - a force causes motion from rest which implies an acceleration (but its an angular acceleration, so it's a little new).


A force applied at point Bwill not rotate the nut.

Assuming the forces were the same, why does the force at point B not result in motion?

What is different?

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TorqueThe distance between where the force was applied and the axis of rotation. When the force was applied at the axis of rotation, the nut did not move.

There's just one more variable to consider for torque. Here's a picture of a door (the horizontal line) and its door hinge (the dot).

Let's say you want to open the door by pushing it - rotating it about the hinge. F1 and F2 have the same magnitude.

F1

F2

Which force will cause the door to open quicker (have a greater angular acceleration)?

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Torque

You've probably known this since you first started walking - F1

gives the greater angular acceleration. Pushing perpendicular to a line connecting the force to the axis of rotation results in the greater angular acceleration.

F1

F2

It will also allow a smaller force to have the same impact as a greater force applied at an angle different than 900.

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TorqueThis combination of Force applied at a distance and an angle from the axis of rotation that causes an angular accleration is called Torque, which is represented by the Greek letter tau:

F1

F2

θr

To continue our sign convention, if the object rotates counter clockwise, the torque is positive. Negative values of torque cause clockwise rotations. Torque is a vector, so its direction must be specified along with its magnitude.

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Torque

The sin θ takes into account that the closer the force is applied perpendicular to the line connected to the axis of rotation results in a greater torque - a greater angular acceleration. θ is the angle between the applied force and the line connecting it to the axis of rotation.

F1

F2

θr

When you get to vector calculus, the equation becomes more elegant involving the cross product:

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TorqueThe units of torque are Newton-meters.

Note how this is similar to work and energy - and in those cases, we replace N-m with Joules.

Torque, however is a vector - depending on its direction (positive or negative), it causes a different direction of motion.

Work and energy are scalars. So, to emphasize this difference, torque is never expressed in terms of Joules.


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21 Assume a force is applied perpendicular to the axis of rotation of the wrench-nut system. At which point will the nut experience the greatest torque?

A

B

C

D

E All result in the same torque.http://commons.wikimedia.org/wiki/File%3AAction_meca_equivalence_force_couple_plan.svg

C ADB


21 Assume a force is applied perpendicular to the axis of rotation of the wrench-nut system. At which point will the nut experience the greatest torque?

A

B

C

D

E All result in the same torque.http://commons.wikimedia.org/wiki/File%3AAction_meca_equivalence_force_couple_plan.svg

C ADB


Ans

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A

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22 A force of 300 N is applied a crow bar 20 cm from the axis of rotation at an angle of 250 to a line connecting it to the axis of rotation. Calculate the torque.

A 30 N-m

B 25 N-m

C 20 N-m

D 15 N-m

E 10 N-m


22 A force of 300 N is applied a crow bar 20 cm from the axis of rotation at an angle of 250 to a line connecting it to the axis of rotation. Calculate the torque.

A 30 N-m

B 25 N-m

C 20 N-m

D 15 N-m

E 10 N-m [This object is a pull tab]

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B

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Torque ApplicationsWhat if more than one more torque is applied to an object?

Just as in linear dynamics, where multiple forces are added to find the linear acceleration of the object, multiple torques are added (in a vector fashion) to find the angular acceleration of the object as it rotates.

Assign a positive value to the individual torque if it accelerates the object in the counter-clockwise, and a negative value if it accelerates it in the clockwise direction.

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Balance

A B

rA rB

Here is a balance with two masses attached to it. Block A has a mass of 0.67 kg and is located 0.23 m from the pivot point. Block B has a mass of 1.2 kg. Where should it be located so that the balance beam is perfectly horizontal and is not moving?

Is this a translational motion or a rotational motion problem?



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See-Saw (balance)

Winslow Homer, "The See-Saw" 1873

Here's an artistic hint. The two boys on the left represent mass A and the two slightly larger boys on the right represent mass B.

A side note - there are many connections between physics and art - this would be a good project to research.

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Balance

A B

rA rB

This is a rotational motion problem, so in order to have the beam not rotate, the sum of the torques applied to it must equal zero (Newton's Second Law).

Note the signs - mass A is trying to rotate the beam counterclockwise - so its torque is positive. Mass B is trying to rotate the beam clockwise - negative torque.

Slide 73 / 130

Balance and See-Saw

A B

rA rBAgain, many elementary school students already know this! If you're on a see-saw and you have to balance several students on the other side, you position yourself further from the fulcrum than they are.

Winslow Homer, "The See-Saw" 1873

Slide 74 / 130

23 John and Sally are sitting the on the opposite sides of a See-Saw, both of them 5.0 m distant from the fulcrum. Sally has a mass of 45 kg. What is the gravitational force on John if the See-Saw is not moving? Use g = 10 m/s2.

A 450 N

B 400 N

C 350 N

D 300 N

E 250 N


23 John and Sally are sitting the on the opposite sides of a See-Saw, both of them 5.0 m distant from the fulcrum. Sally has a mass of 45 kg. What is the gravitational force on John if the See-Saw is not moving? Use g = 10 m/s2.

A 450 N

B 400 N

C 350 N

D 300 N

E 250 N [This object is a pull tab]

Ans

wer


A

Slide 75 / 130

24 A student wants to balance two masses of 2.00 kg and 0.500 kg on a meter stick. The 0.500 kg mass is 0.500 m from the fulcrum in the middle of the meter stick. How far away should she put the 2.00 kg mass from the fulcrum?

A 0.500 m

B 0.333 m

C 0.250 m

D 0.125 m

E 0.075 m


24 A student wants to balance two masses of 2.00 kg and 0.500 kg on a meter stick. The 0.500 kg mass is 0.500 m from the fulcrum in the middle of the meter stick. How far away should she put the 2.00 kg mass from the fulcrum?

A 0.500 m

B 0.333 m

C 0.250 m

D 0.125 m

E 0.075 m [This object is a pull tab]

Ans

wer


D

Slide 76 / 130

25 Sally wants to lift a rock of 105 kg off of the ground. She has a 3.00 m long metal bar. She positions the bar on a fulcrum (actually, a thick tree log 1.00 m away from the rock) and wedges it under the rock. She can exert a force of 612 N with her arms. How far away from the fulcrum should she exert this force to move the rock? Use g = 10 m/s2.A 0.17 m

B 0.85 m

C 1.7 m

D 2.0 m

E 2.7 m


25 Sally wants to lift a rock of 105 kg off of the ground. She has a 3.00 m long metal bar. She positions the bar on a fulcrum (actually, a thick tree log 1.00 m away from the rock) and wedges it under the rock. She can exert a force of 612 N with her arms. How far away from the fulcrum should she exert this force to move the rock? Use g = 10 m/s2.A 0.17 m

B 0.85 m

C 1.7 m

D 2.0 m

E 2.7 m[This object is a pull tab]

Ans

wer


C

Slide 77 / 130

The ladder against the wall

For an object to be in equilibrium (not moving), there are two requirements that need to be satisfied. The sum of the torques on the object needs to be zero and the sum of the forces needs to be zero.

This is a branch of engineering called statics - as nothing is moving, which is a nice thing to happen if you're planning a bridge or building.

Let's apply this to a classic physics problem - a ladder leaning against a wall.

Slide 78 / 130


A ladder of length L and mass m, is leant against a wall. The coefficient of static friction between the ladder and the ground is 0.6 (μsg). The coefficient of static friction between the ladder and the wall is 0.3 (μsw) . At what minimum angle, θ, from the ground should the ladder be placed so that it doesn't slip?.

θ

First step - draw a coordinatesystem at the ladder/groundinterface and sketch the forcesacting on the ladder. L

Slide 79 / 130


θx

y FNw

fwl

mg

fgl

FNg

fwl ≡ static friction force of wall on ladderFNw ≡ normal force of wall on ladder

fgl ≡ static friction force of ground on ladderFNg ≡ normal force of ground on ladder

mg ≡ gravitational force on ladder

Slide 80 / 130


θx

y FNw

fwl

mg

fgl

FNg

Choose a point about which the torques will be calculated. Normally, a point is chosen to minimize the amount of torques created by the forces. But since this problem is asking for θ, we'll choose the center of mass of the ladder. This will remove the mass from the torque equation.

And, as it will turn out, we won't need the sum of the forces in the y direction to solve the problem.

reference point

Slide 81 / 130


θx

y FNw

fwl

mg

fgl

FNg

θ

The sum of the torques about the center of mass equal zero. mg does not contribute a torque as it acts at the centerof mass.

The sum of the forces in the x and y direction equal zero.

Slide 82 / 130


θx

y FNw

fwl

mg

fgl

FNg

To find the minimum angle at which the ladderwill not slip, the static friction forces will be attheir maximum value:

Use the forces in the x direction:

Slide 83 / 130

Slide 84 / 130


θx

y FNw

fwl

mg

fgl

FNg

The angle is independent ofthe mass of the ladder - muchlike a mass sliding down an incline.

Slide 85 / 130

Torque

Torque is the rotational analog to Force.

Newton's Second Law states that F = ma. Time to derive the angular version of this law. Newton's Second Law has been applied assuming it was dealing with point particles that don't rotate. We now have an extended rigid body, but let's start with the external force on one small piece of the rigid body as it's rotating:

since ai = riα, where ri is the distance between the small piece of the object and the axis of rotation

Slide 86 / 130

TorqueAssume that the external force is applied perpendicular to the line connected to the axis of rotation (sinθ = 1):

Now sum up the torques due to the external force acting on all the little pieces of the rigid body:

α is factored out of the summation because it is constant for all points on a rigid object.

from the previous page

Why are we not including the torques due to the internal forces?

Slide 87 / 130

Moment of Inertia

From linear dynamics, we know that internal forces form action - reaction pairs and the net force is equal to zero.

Similar logic applies to the sum of the internal torques - their forces are equal and opposite and act along the same displacement r, so they all cancel.

Define the net external torque as:

Where I is the moment of inertia of the rigid body:

Slide 88 / 130

Moment of Inertia

This looks very similar to Newton's Second Law in translational motion. ΣF has been replaced by Στ, a has been replaced by α, and m has been replaced by I.

The mass of a given rigid object is always constant.

Can the same be said for the moment of inertia?

Slide 89 / 130

Slide 90 / 130

26 A massless rod is connected to two spheres, each of mass m. Assume the spheres act as point particles. What is the moment of inertia of this configuration if the axis of rotation is located at one of the masses? What is the moment of inertia if the axis of rotation is located at the midpoint of the rod? Which is easier to rotate? Why?

L


26 A massless rod is connected to two spheres, each of mass m. Assume the spheres act as point particles. What is the moment of inertia of this configuration if the axis of rotation is located at one of the masses? What is the moment of inertia if the axis of rotation is located at the midpoint of the rod? Which is easier to rotate? Why?

L


Ans

wer


The rotation about the midpoint is easier, since the moment of inertia is smaller - less resistance to rotation.

Slide 91 / 130

Moment of InertiaThe moment of inertia was easily calculated when there were a limited number of objects that could be represented as point particles.

For rigid objects of given shapes, we will need to add an infinite amount of small masses times their distance from the axis of rotation. For objects with symmetries, we can use integral calculus to calculate their moments of inertia.

Each small mass will rotate about the axis of rotation due to an external tangential force providing a tangential acceleration.

Slide 92 / 130

Moment of InertiaLabel each small mass, dm, that it is acted on by an external tangential force, dFT, producing a tangential acceleration aT. Working with the magnitudes and Newton's Second Law:

The tangential force provides a torque about the axis of rotation:

The moment of inertia can be calculated now:

since

α is constant at all pointson the rigid body

Slide 93 / 130

r

dr

LR

z

Moment of Inertia CalculationFind the moment of inertia of a solid cylinder of radius R, length L, and mass M rotating about the z axis through its center of mass. Express dm as ρdV, where ρ is the volume mass density and dV is a small piece of volume, as it is easier to work with volumes than masses.

We'll need to use the cylindrical symmetry to substitute for dm.

Slide 94 / 130

Moment of Inertia CalculationDraw a thin shell of thickness dr that is centered on, and a distance r from the axis of rotation. It extends from the top of the cylinder to the bottom, so its length is L. The cross sectional area of this shell is 2πrdr. The volume, dV, is 2πrLdr.

r

dr

LR

z

Slide 95 / 130 Slide 96 / 130

Moment of Inertia

Here are various momentsof inertia for other symmetrical objects.

All of the objects have thesame mass, M, but differentshapes and different axes ofrotation.

Take some time to see howand why the moments of inertia change.

Slide 97 / 130

Moment of InertiaJust as mass is defined as the resistance of an object to accelerate due to an applied force, the moment of inertia is a measure of an object's resistance to angular acceleration due to an applied torque.

The greater the moment of inertia of an object, the less it will accelerate due to an applied torque.

The same torque is applied to a solid cylinder and a rectangular plate.

Which object will have a greater angular acceleration? You can go back a slide to find their moments of inertia.

Slide 98 / 130

Moment of InertiaSince the rectangular plate has a smaller moment of inertia, it will have a greater angular acceleration.

The angular acceleration can be different for the same object, if a different axis of rotation is chosen.

A long thin rod is easier to rotate if it is held in the middle. By rotating it about that axis, as compared to rotating it at the end, it has a smaller moment of inertia, so for a given torque, it will have a greater angular acceleration - it will be easier to rotate.

This was solved earlier in the chapter with a rod with a mass at either end. It's nice that physics is consistent.

Slide 99 / 130

Parallel Axis TheoremIf the moment of inertia about the center of mass of a rigid body is known, the moment of inertia about any other axis of rotation of the body can be calculated by use of the Parallel Axis Theorem. Often, this simplifies the mathematics.

This theorem will be presented without proof:

where D is the distance between the new axis of rotation from the axis of rotation through the center of mass (Icm), and M is the total mass of the object.

Slide 100 / 130

27 A child rolls a 0.02 kg hoop with an angular acceleration of 10 rad/s2. If the hoop has a radius of 1 m, what is the net torque on the hoop?

A 0.6 N-m

B 0.5 N-m

C 0.4 N-m

D 0.3 N-m

E 0.2 N-m


27 A child rolls a 0.02 kg hoop with an angular acceleration of 10 rad/s2. If the hoop has a radius of 1 m, what is the net torque on the hoop?

A 0.6 N-m

B 0.5 N-m

C 0.4 N-m

D 0.3 N-m

E 0.2 N-m [This object is a pull tab]

Ans

wer


E

Slide 101 / 130

28 A construction worker spins a square sheet of metal of mass 0.040 kg with an angular acceleration of 10.0 rad/s2 on a vertical spindle (pin). What are the dimensions of the sheet if the net torque on the sheet is 100.0 N-m?

A 42 m x 42 m

B 42 m x 39 m

C 39 m x 39 m

D 30 m x 39 m

E 30 m x 42 m


28 A construction worker spins a square sheet of metal of mass 0.040 kg with an angular acceleration of 10.0 rad/s2 on a vertical spindle (pin). What are the dimensions of the sheet if the net torque on the sheet is 100.0 N-m?

A 42 m x 42 m

B 42 m x 39 m

C 39 m x 39 m

D 30 m x 39 m

E 30 m x 42 m [This object is a pull tab]

Ans

wer


C

Slide 102 / 130

29 A massless rod is connected to two spheres, each of mass m. Assume the spheres act as point particles. The moment of inertia about the center of mass of the system is 1/2 mL2. Use the parallel axis theorem to find the moment of inertia of the system if it is rotated about one of the masses.

L


29 A massless rod is connected to two spheres, each of mass m. Assume the spheres act as point particles. The moment of inertia about the center of mass of the system is 1/2 mL2. Use the parallel axis theorem to find the moment of inertia of the system if it is rotated about one of the masses.

L


Ans

wer


Slide 103 / 130

Rotational Kinetic Energy


Slide 104 / 130

Rotational Kinetic EnergyAfter Kinematics and Dynamics were covered, your physics education continued with the concept of Energy. Energy is hard to define, but since it is a scalar quantity, it is great for solving problems.

By now, you should be ready for the concept of a rotational version of the linear (also called translational) kinetic energy.

Given that KE = 1/2 mv2, what substitution should be made?

Slide 105 / 130

Rotational Kinetic EnergyUse the rotational equivalent of velocity and substitute it into the Kinetic Energy equation. The second line shows where one piece of a rotating object is being considered, and it is now a rotational energy since v is replaced by ω.

Add up all the KER's for each small mass

page40svg

Slide 106 / 130

Work done by TorqueThe concept of Work was introduced in parallel with Energy, where .

A shortcut will now be taken to determine how much work is done by an applied torque - the rotational equivalents of force and displacement will just be substituted into the above equation:

Since torque is always perpendicular to the angular displacement, the vector dot product reduces to multiplying the values for torque and angular displacement (in radians).

Slide 107 / 130

Slide 108 / 130

Total Kinetic EnergyA common rotational motion problem is having two different objects, each with the same mass and determining which object will get to the bottom of an incline first.

If they were just blocks, or even spheres, that slid down without rotating, they would both get to the bottom at the same time, since objects fall at the same rate.

By using an extended mass (not a point particle), and rotation, it's quite different.

Why? Think about Energy. At the top of the ramp, both objects have the same energy - it's all potential. How about at the bottom?

Slide 109 / 130

Total Kinetic EnergyFor a non rotating object, all of the gravitational potential energy is transferred into translational kinetic energy. If the object rotates, then part of that energy goes into rotational kinetic energy.

Rotational kinetic energy depends on the shape of the object - which affects the moment of inertia.

Here's an example. A sphere and a hoop of the same mass and radius both start at rest, at the top of a ramp. Assume that both objects roll without slipping, which one gets to the bottom first?

Slide 110 / 130

Total Kinetic Energy

Hoop Sphere

h

Here's the sketch.

What is the equation, using the Conservation of Energy, that will give the velocity of the objects at the bottom of the ramp? Solve the equation. Qualitatively, how can that be used to determine which object reaches the bottom first?

Slide 111 / 130


Hoop Sphere

h

This equation works for both the hoop and the sphere.

Slide 112 / 130


Hoop Sphere

h

Did you question why it was stated that the sphere and the hoop roll without slipping? When that occurs, then vlinear = v = rω. We will use this fact to replace ω in the above equation with v/r.

Slide 113 / 130

Total Kinetic EnergyHoop Sphere

h

Hoop Sphere

Slide 114 / 130


h

The velocity of the sphere is 1.2 times as much as the velocity of the hoop. The sphere gets to the bottom of the ramp first - because at any point on the incline, the sphere has a greater linear velocity (h is just the height above ground). Solving the kinematics equations would also give the same answer.

One more question - why did this happen?

Slide 115 / 130


h

Both the sphere and the hoop started with the same GPE. And, at the bottom of the ramp, both had zero GPE and a maximum KE - again, both the same.

But - the hoop, with its greater moment of inertia, took a greater amount of the total kinetic energy to make it rotate - so there was less kinetic energy available for linear motion.

Hence, the sphere has a greater linear velocity than the hoop and reaches the bottom first.

Slide 116 / 130

30 A spherical ball with a radius of 0.50 m rolls down a 10.0 m high hill. What is the velocity of the ball at the bottom of the hill? Use g = 10 m/s2.

A 10 m/s

B 11 m/s

C 12 m/s

D 13 m/s

E 14 m/s


30 A spherical ball with a radius of 0.50 m rolls down a 10.0 m high hill. What is the velocity of the ball at the bottom of the hill? Use g = 10 m/s2.

A 10 m/s

B 11 m/s

C 12 m/s

D 13 m/s

E 14 m/s [This object is a pull tab]

Ans

wer


C

Slide 117 / 130

31 A solid cylinder with a radius of 0.50 m rolls down a hill. What is the height of the hill if the final velocity of the block is 10.0 m/s? Use g = 10 m/s2.

A 7.0 m

B 7.5 m

C 8.0 m

D 8.5 m

E 9.0 m


31 A solid cylinder with a radius of 0.50 m rolls down a hill. What is the height of the hill if the final velocity of the block is 10.0 m/s? Use g = 10 m/s2.

A 7.0 m

B 7.5 m

C 8.0 m

D 8.5 m

E 9.0 m [This object is a pull tab]

Ans

wer


B

Slide 118 / 130

32 Consult a table that lists the moments of inertia of various symmetric objects. If each of the following have the same mass and radius, rotate as described, and are let go from the top of an incline, which object reaches the bottom first?A Thin walled cylinder about its central axis

B Thick walled cylinder about its central axis

C Solid sphere about its center

D Hollow sphere about its center

E Solid cylinder about its central axis


32 Consult a table that lists the moments of inertia of various symmetric objects. If each of the following have the same mass and radius, rotate as described, and are let go from the top of an incline, which object reaches the bottom first?A Thin walled cylinder about its central axis

B Thick walled cylinder about its central axis

C Solid sphere about its center

D Hollow sphere about its center

E Solid cylinder about its central axis[This object is a pull tab]

Ans

wer


C

Slide 119 / 130

Angular Momentum


Slide 120 / 130

Angular Momentum

As was done in the last chapter, we won't derive angular momentum from linear momentum - we will substitute in the angular values for the linear (translational) ones.

Linear momentum (p):

Angular momentum (L):

Angular momentum is also described in terms of linear momentum as follows:

page40svg

Slide 121 / 130

Conservation of Angular MomentumThe conservation of (linear) momentum states that in the absence of external forces, momentum is conserved. This came from the original statement of Newton's Second Law:

If there are no external forces, then:

Thus there is no change in momentum; it is conserved.

Slide 122 / 130

Conservation of Angular MomentumStarting with Newton's Second Law for rotational motion, the conservation of angular momentum will be derived.

If there are no external torques, then:

Thus there is no change in angular momentum; it is conserved.

Slide 123 / 130

Conservation of Angular MomentumThe standard example used to illustrate the conservation of angular momentum is the ice skater. Start by assuming a frictionless surface - ice is pretty close to that - although if it were totally frictionless, the skater would not be able to stand up and skate!

The skater starts with his arms stretched out and spins around in place. He then pulls in his arms.

What happens to his rotational velocity after he pulls in his arms?

Click on the below site and observe what happens.

https://www.youtube.com/watch?v=MjYk5TRpOlE

Slide 124 / 130

Conservation of Angular MomentumAssuming a nearly frictionless surface implies there is no net external torque on the skater, so the conservation of angular momentum can be used.

The human body is not a rigid sphere or a hoop or anything simple. But, knowing what the definition of moment of inertia is, at what point in the spinning does the skater have a greater moment of inertia?

Slide 125 / 130

Conservation of Angular MomentumWhen the skater's arms are stretched out, more of his mass is located further from his axis of rotation (the skates on the ice), so he has a greater moment of inertia then when he tucks his arms in. In the video, is greater than . Thus, ωf is greater than ω0 and the skater rotates faster with his arms tucked in.

Slide 126 / 130

33 A 2.0 kg solid spherical ball with a radius of 0.20 m rolls with an angular velocity of 5.0 rad/s down a street. Magneto shrinks the radius of the ball down to 0.10 m, but does not change the ball's mass and keeps it rolling. What is the new angular velocity of the ball?A 10 rad/s

B 15 rad/s

C 20 rad/s

D 25 rad/s

E 30 rad/s

http://njc.tl/12k


33 A 2.0 kg solid spherical ball with a radius of 0.20 m rolls with an angular velocity of 5.0 rad/s down a street. Magneto shrinks the radius of the ball down to 0.10 m, but does not change the ball's mass and keeps it rolling. What is the new angular velocity of the ball?A 10 rad/s

B 15 rad/s

C 20 rad/s

D 25 rad/s

E 30 rad/s[This object is a pull tab]

Ans

wer


C

Slide 127 / 130

34 An ice skater extends her 0.5 m arms out and begins to spin. Then she brings her arms back to her chest. Which statement is true in this scenario?

A Her angular velocity increases.

B Her angular velocity decreases.

C Her moment of inertia increases.

D Her angular acceleration decreases.

E Her angular acceleration is zero.


34 An ice skater extends her 0.5 m arms out and begins to spin. Then she brings her arms back to her chest. Which statement is true in this scenario?

A Her angular velocity increases.

B Her angular velocity decreases.

C Her moment of inertia increases.

D Her angular acceleration decreases.

E Her angular acceleration is zero.[This object is a pull tab]

Ans

wer


A

Slide 128 / 130

Kepler's Second LawKepler's Second Law states that a radius vector from the sun to one of its planets (like earth) sweeps out an equal area in an equal time during the planet's orbit. Since the orbits are ellipses with the sun at one of the foci (Kepler's First Law), the planets move faster when they are further away from the sun.

In the Energy unit of this course,the Work Energy Equation was usedto explain the speed change.Conservation of Angular Momentum can also be used.

By Kepler2.gif: Illustration by RJHall using Paint Shop Pro derivative work: Talifero (Kepler2.gif) [CC BY-SA 2.0 at (http://creativecommons.org/licenses/by-sa/2.0/at/deed.en)], via Wikimedia Commonshttps://commons.wikimedia.org/wiki/File%3AKepler's_law_2_en.svg

Slide 129 / 130

Kepler's Second LawIn orbit, there are no external non conservative forces acting on the earth. The gravitational force is an internal force to the system and is accounted for with its potential energy. Thus, the Conservation of Angular Momentum applies.

Select points A and B, recognizing that rA > rB means that IA > IB

and using the moment of Inertia formulation:

The earth has a smaller angular velocity at point A.


A B

Slide 130 / 130

Kepler's Second LawUsing the formula that relates angular momentum to linear momentum and recognizing that rA > rB:

The earth has a smaller linear(tangential) momentum, andhence, linear velocity at point A.Since v = rw, it is a good thingthat we get the same resultsusing both equations.


A B

https://commons.wikimedia.org/wiki/File%3AKepler's_law_2_en.svg



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