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Lecture 16. Chapter 12 Extend the particle model to rigid-bodies U nderstand the equilibrium of an extended object. Analyze rolling motion Understand rotation about a fixed axis. Employ “conservation of angular momentum” concept. Goals:. Assignment: HW7 due March 25 th

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Lecture 16

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Lecture 16 l.jpg

Lecture 16

  • Chapter 12

    • Extend the particle model to rigid-bodies

    • Understand the equilibrium of an extended object.

    • Analyze rolling motion

    • Understand rotation about a fixed axis.

    • Employ “conservation of angular momentum” concept

Goals:

Assignment:

  • HW7 due March 25th

  • After Spring Break Tuesday:

    Catch up


Rotational dynamics a child s toy a physics playground or a student s nightmare l.jpg

Rotational Dynamics: A child’s toy, a physics playground or a student’s nightmare

  • A merry-go-round is spinning and we run and jump on it. What does it do?

    What principles would apply?

  • We are standing on the rim and our “friends” spin it faster. What happens to us?

  • We are standing on the rim a walk towards the center. Does anything change?


Rotational variables l.jpg

Rotational Variables

  • Rotation about a fixed axis:

    • Consider a disk rotating aboutan axis through its center:

  • How do we describe the motion:

    (Analogous to the linear case )


Rotational variables4 l.jpg

v = w R

x

R

Rotational Variables...

  • Recall: At a point a distance R away from the axis of rotation, the tangential motion:

    • x =  R

    • v = R

    • a =  R


Comparison to 1 d kinematics l.jpg

Comparison to 1-D kinematics

AngularLinear

And for a point at a distance R from the rotation axis:

x = R v =  R aT =  R

Here aT refers to tangential acceleration


System of particles distributed mass l.jpg

System of Particles (Distributed Mass):

  • Until now, we have considered the behavior of very simple systems (one or two masses).

  • But real objects have distributed mass !

  • For example, consider a simple rotating disk and 2 equal mass m plugs at distances r and 2r.

  • Compare the velocities and kinetic energies at these two points.

w

1

2


System of particles distributed mass7 l.jpg

1 K= ½ m v2 = ½ m (w r)2

w

2 K= ½ m (2v)2 = ½ m (w 2r)2

System of Particles (Distributed Mass):

  • Twice the radius, four times the kinetic energy

  • The rotation axis matters too!


A special point for rotation system of particles center of mass cm l.jpg

+

+

m2

m1

m1

m2

A special point for rotationSystem of Particles: Center of Mass (CM)

  • If an object is not held then it will rotate about the center of mass.

  • Center of mass: Where the system is balanced !

    • Building a mobile is an exercise in finding

      centers of mass.

mobile


System of particles center of mass l.jpg

System of Particles: Center of Mass

  • How do we describe the “position” of a system made up of many parts ?

  • Define the Center of Mass (average position):

    • For a collection of N individual point like particles whose masses and positions we know:

RCM

m2

m1

r2

r1

y

x

(In this case, N = 2)


Sample calculation l.jpg

RCM = (12,6)

(12,12)

2m

m

m

(0,0)

(24,0)

Sample calculation:

  • Consider the following mass distribution:

XCM = (m x 0 + 2m x 12 + m x 24 )/4m meters

YCM = (m x 0 + 2m x 12 + m x 0 )/4m meters

XCM= 12 meters

YCM= 6 meters


System of particles center of mass11 l.jpg

System of Particles: Center of Mass

  • For a continuous solid, convert sums to an integral.

dm

r

y

where dmis an infinitesimal

mass element.

x


Connection with motion l.jpg

VCM

Connection with motion...

  • So for a rigid object which rotates about its center of mass and whose CM is moving:

    For a point p rotating:


Rotation kinetic energy l.jpg

Rotation & Kinetic Energy

  • Consider the simple rotating system shown below. (Assume the masses are attached to the rotation axis by massless rigid rods).

  • The kinetic energy of this system will be the sum of the kinetic energy of each piece:

  • K = ½m1v12 + ½m2v22 + ½m3v32 + ½m4v42

m4

m1

r1

r4

r2

m3

r3

m2


Rotation kinetic energy14 l.jpg

m4

m1

r1

r4

m3

r2

r3

m2

Rotation & Kinetic Energy

  • Notice that v1 = w r1 , v2 = w r2 , v3 = w r3 , v4 = w r4

  • So we can rewrite the summation:

  • We recognize the quantity, moment of inertia orI, and write:


Calculating moment of inertia l.jpg

Calculating Moment of Inertia

where r is the distance from the mass to the axis of rotation.

Example:Calculate the moment of inertia of four point masses

(m) on the corners of a square whose sides have length L,

about a perpendicular axis through the center of the square:

m

m

L

m

m


Calculating moment of inertia16 l.jpg

Calculating Moment of Inertia...

  • For a single object, Idepends on the rotation axis!

  • Example:I1 = 4 m R2 = 4 m (21/2 L / 2)2

I1= 2mL2

I2= mL2

I= 2mL2

m

m

L

m

m


Moments of inertia l.jpg

dr

r

R

L

dm

r

Moments of Inertia

  • For a continuous solid object we have to add up the mr2contribution for every infinitesimal mass element dm.

    • An integral is required to find I:

Solid disk or cylinder of mass M and radius R, about perpendicular axis through its center.

I = ½ M R2

  • Some examples of I for solid objects:

Use the table…


Exercise rotational kinetic energy l.jpg

Ball 1

Ball 2

Exercise Rotational Kinetic Energy

  • ¼

  • ½

  • 1

  • 2

  • 4

  • We have two balls of the same mass. Ball 1 is attached to a 0.1 m long rope. It spins around at 2 revolutions per second. Ball 2 is on a 0.2 m long rope. It spins around at 2 revolutions per second.

  • What is the ratio of the kinetic energy

    of Ball 2 to that of Ball 1 ?


Exercise rotational kinetic energy19 l.jpg

Ball 1

Ball 2

Exercise Rotational Kinetic Energy

  • K2/K1 = ½ m wr22 / ½ m wr12 = 0.22 / 0.12 = 4

  • What is the ratio of the kinetic energy of Ball 2 to that of Ball 1 ?

    (A) 1/4 (B) 1/2 (C) 1 (D) 2 (E) 4


Exercise work energy l.jpg

Exercise Work & Energy

  • Strings are wrapped around the circumference of two solid disks and pulled with identical forces, F, for the same linear distance, d. Disk 1 has a bigger radius, but both are identical material (i.e. their density r = M / V is the same). Both disks rotate freely around axes though their centers, and start at rest.

    • Which disk has the biggest angular velocity after the drop?

W =F d = ½ I w2

(A)Disk 1

(B)Disk 2

(C)Same

w2

w1

F

F

start

d

finish


Exercise work energy21 l.jpg

Exercise Work & Energy

  • Strings are wrapped around the circumference of two solid disks and pulled with identical forces for the same linear distance. Disk 1 has a bigger radius, but both are identical material (i.e. their density r = M/V is the same). Both disks rotate freely around axes though their centers, and start at rest.

    • Which disk has the biggest angular velocity after the drop?

W =F d = ½ I1w12 = ½ I2w22

w1 = (I2 / I1)½w2 and I2 < I1

(A)Disk 1

(B) Disk 2

(C)Same

w2

w1

F

F

start

d

finish


Lecture 1622 l.jpg

Lecture 16

Assignment:

  • HW7 due March 25th

  • For the next Tuesday:

    Catch up


Lecture 1623 l.jpg

Lecture 16

Assignment:

  • HW7 due March 25th

  • After Spring Break Tuesday: Catch up


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