Chapter 8- Rotational Motion

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Chapter 8

• Angular Quantities • Constant Angular Acceleration • Rolling Motion (Without Slipping) • Centripetal Forces • Torque • Rotational Dynamics; Torque and Rotational Inertia • Rotational Kinetic Energy • Angular Momentum and Its Conservation

Recalling Last Lecture

Momentum and Its Relation to Force

Momentum is a vector symbolized by the symbol p, and is defined as (7-20) The momentum of an object tells how hard (or easy) is to change its state of motion.

Momentum and Its Relation to Force

(6-22)

Eq. 6-22 is another way of expressing Newton’s second law. However, it is a more general definition because it introduces the situation where the mass may change.

Collision and Impulse We can use eq. 6-22 and define the impulse on an object as:

(6-23)

Conservation of Momentum

(6-25)

Equation 6-25 tells that the total momentum of the system (the sum of the momentum of the two balls) before the collision is equal to the total momentum of the system after the collision IF the net force acting on the system is zero
isolated system. This is known as Conservation of Total Momentum. The above equation can be extended to include any number of objects such that the only forces are the interaction between the objects in the system.

Conservation of Energy and Momentum in Collisions In general, we can identify two different types of collisions: 1. Elastic collision 2. Inelastic collision In elastic collisions the total kinetic energy of a system is conserved.

An example is the collision between the two billiard balls discussed in the previous slides: (6-26)

In inelastic collision, there is NO conservation of kinetic energy.

Conservation of Energy and Momentum in Collisions Note: Also…

The total energy (the sum of all energies) in a closed (isolated system) is ALWAYS conserved.

Conservation of Energy and Momentum in Collisions Collision in two or more dimensions We then have: (i) (ii)

It follows then, using the above expressions into (i) and (ii), that: ⇒ This gives a system of two equations and three variables. If you can measure any of these variables, the other two can be calculated from system of equations.

Today

Linear Momentum Problem 7-34 (textbook) An internal explosion breaks an object, initially at rest, into two pieces, one of which has 1.5 times the mass of the other. If 7500 J were released in the explosion, how much kinetic energy did each piece acquire?

Linear Momentum Problem 7-34 Use conservation of momentum in one dimension, since the particles will separate and travel in opposite directions. Call the direction of the heavier particle’s motion the positive direction. Let A represent the heavier particle, and B represent the lighter particle. We have

m A = 1.5 m B

vA = vB = 0

p initial = p final

→

0 = m A v A′ + m B v B′

→ v A′ = −

m B v B′ mA

= − 23 v B′

The negative sign indicates direction. Since there was no mechanical energy before the explosion, the kinetic energy of the particles after the explosion must equal the energy added.

E added = KE A′ + KE B′ = 12 m A v ′A2 + 12 mB v B′ 2 = KE B′ = 53 E added = Thus:

3 5

( 7500 J ) = 4500 J

1 2

(1.5mB ) ( 23 vB′ )

2

+ 12 m B v B′ 2 =

5 3

(

1 2

)

m B v B′ 2 = 53 KE B′

KE A′ = Eadded − KE B′ = 7500 J − 4500 J = 3000 J

K E A′ = 3.0 × 10 3 J K E B′ = 4.5 × 10 3 J

Angular Quantities We have extensively discussed translational (linear) motion of an object in terms of its kinematics (displacement), dynamics (forces), momentum and energy.

But we also know that objects can also move following some circular path.

This is called rotational motion.

The discussions on rotational motion can be based on what have been discussed so far concerning translational motion.

So, I will use the definitions introduced in the previous chapters and apply then to introduce you to rotational motion. We will consider only rigid objects
in other words, objects that do not change shape (or the distances between points that compounds an object do not change)

Angular Quantities
In purely rotational motion, all points on the object move in circles around the axis of rotation (“O”) which is perpendicular to this slide.
The radius of the circle is r.
All points on a straight line drawn through the axis move through the same angle in the same interval of time.
The angle θ in radians is defined: (8-1)

Where

r = radius of the circle l = arc length covered by the angle θ

The angular displacement is what characterizes the rotational motion.

Angular Quantities For mathematical reasons, it is more convenient to define angle not in degrees but in radians. One radian is defined such that it corresponds to an arc of circle equal to the radius of the circle. Or, if we use eq. 8.1:

Note that radians are dimensionless. Radians can be related to degrees by observing that the full length of a circle corresponds to the maximum arc length, or 2πr. It comprises an angle of 3600. Using eq. 8.1, we have:

(8-2)

Angular Quantities We can also define an object revolution as the length in radians that the object has travelled.

A complete revolution will correspond to the total length of the circle, or:

(8-3)

Angular Quantities

As mentioned before, the object rotational displacement is defined in terms of the angle θ.

Using the coordinate system depicted in the figure, The displacement of a certain point P on the object can be given by:

(8-4)

Angular Quantities Similarly as for translational motion, we can define the average angular velocity and instantaneous angular velocity of this point as: (8-5)

(8-6)

Also, the average angular acceleration and instantaneous angular acceleration can be defined as: (8-7)

(8-8)

Note that given the fact that each point on the object will be displaced by the same angle in the same interval of time,
then, both the velocity and acceleration are the same for any point on the object.

Angular Quantities The rotational motion of an object or a point on the object can be related to its translational motion. For realizing that, you should observe that a point rotating around a circle will also be subjected to a translational motion as depicted in the figure.

∆ ∆

At each angular position, this point will have a linear velocity whose directions are tangent to its circular path.

Important note: the direction of the linear velocity changes as the point undergoes a rotational motion.
This is due to the so called centripetal acceleration. We will come back to this acceleration in two slides from now.

Angular Quantities Back to the linear velocity, the figure shows that a change in the rotation angle ∆θ corresponds to a linear distance traveled ∆l.

∆

With the help of eq. 8-1: ∆

(8-9)

Note: Here we assume a very small Angular displacement such the arc of length corresponding to the angular displacement can approximately be assumed is a linear displacement.

Angular Quantities (8-9) Eq. 8-9 says that although ω is the same for every point in the rotating object, the linear velocity changes with the distance r of the point to the axis of rotation.
Therefore, objects farther from the axis of rotation will move faster. If there is an angular acceleration α, the angular velocity ω changes.
Therefore, there is also a change in the linear velocity and thus a linear acceleration involved in the process. This acceleration is in the direction of the velocity and is called tangential linear acceleration:

(8-10)

Angular Quantities As already mentioned a couple of slides ago, the vector velocity changes direction. But the tangential acceleration is parallel to the velocity, so it is not responsible for the change in the velocity’s direction. It turns out that the responsible is the so called radial acceleration, or centripetal acceleration. It is always perpendicular to the direction of the velocity. The magnitude of the centripetal acceleration is given by (see textbook, page 107 for details): (8-11)

(8-12) Therefore, objects farther from the axis of rotation will have greater centripetal acceleration.

Angular Quantities The total acceleration of the point at a distance r from the axis of rotation of the object is the vector sum of the radial (centripetal) and linear (tangential) accelerations:

(8-13)

This discussion can so far be summarized using the following table:

Angular Quantities Another important quality in rotational motion is the frequency of rotation of an object. The frequency is the number of complete revolutions per second: (8-14)

Frequencies are measured in Hertz.

The time required for a complete revolution is called period, or in other words: period is the time one revolution takes: (8-15)

Constant Angular Acceleration The equations of motion for constant angular acceleration are the same as those for linear motion, with the substitution of the linear quantities by the corresponding angular ones.

Linear Momentum Problem 8-8 (textbook) A rotating merry-go-round makes one complete revolution in 4.0 s (Fig. 8–38). (a) What is the linear speed of a child seated 1.2 m from the center? (b) What is her acceleration (give components)?

Linear Momentum Problem 8-8 The angular speed of the merry-go-round is

2π rad 4.0 s = 1.57 rad s (a)

v = ω r = (1 .5 7 r a d s e c ) (1 .2 m

)=

1 .9 m s

(b) Ignoring air our other resistance, there is no tangential acceleration (no tangential forces are applied).Therefore, the acceleration is purely radial.

a R = ω 2 r = (1.57 rad sec ) (1.2 m ) = 3.0 m s 2 towards the center 2

and atan = 0

Linear Momentum Problem 8-13 (textbook) A turntable of radius R1 is turned by a circular rubber roller of radius R2 in contact with it at their outer edges. What is the ratio of their angular velocities, ω1/ ω2.

Linear Momentum Problem 8-13

The tangential speed of the turntable must be equal to the tangential speed of the roller, if there is no slippage.

v1 = v 2

→

ω 1 R1 = ω 2 R 2

→

ω 1 ω 2 = R 2 R1

Linear Momentum Problem 8-19 (textbook) A cooling fan is turned off when it is running at 850 rev/min. It turns 1500 revolutions before it comes to a stop. (a) What was the fan’s angular acceleration, assumed constant?

(b) How long did it take the fan to come to a complete stop?

Linear Momentum Problem 8-19 (a) The angular acceleration can be found from

ω

2

= ω o2 + 2 α θ

ω 2 − ω o2 0 − ( 850 rev min ) rev   2π rad   1 min  rad  α = = =  −241 = − 0.42    2θ 2 (1500 rev ) min 2   1 rev   60 s  s2  2

(b) The time to come to a stop can be found from

θ =

1 2

(ω o

+ω

)t

2 (1 5 0 0 re v )  6 0 s  t = =   = 210 s ωo + ω 8 5 0 re v m in  1 m in  2θ

2