lecture 17

Physics 207: Lecture 17, Pg 1

Lecture 17Goals:Goals:

• Chapter 12Chapter 12 Define center of mass Analyze rolling motion Introduce and analyze torque Understand the equilibrium dynamics of an extended

object in response to forces Employ “conservation of angular momentum” concept

Assignment: HW7 due tomorrow Wednesday, Exam Review


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

A supported object will rotate about its center of mass.

Center of mass: Where the system is balanced ! Building a mobile is an exercise in finding

centers of mass.

m1m2

+m1 m2

+

mobile


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:

M

mN

iii

1CM

rR

(In this case, N = 2)

y

x

r2r1

m1m2

RCM


Sample calculation:

Consider the following mass distribution:

(24,0)(0,0)

(12,12)

m

2m

m

RCM = (12,6)

kji CM CM CM1

CM ZYXM

mN

iii

r

R

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

m at ( 0, 0)

2m at (12,12)

m at (24, 0)


Connection with motion... So a rigid object that has rotation and translation

rotates about its center of mass! And Newton’s Laws apply to the center of mass

For a point p rotating:

nTranslatioRotationTOTALK KK M

mN

iii

1CM

rR

2212

21 )(K ppppR rmvm

p

2CM2

1RotationTOTAL VK MK

VCM

p p p

p p p p


Work & Kinetic Energy:

Recall the Work Kinetic-Energy Theorem: K = WNET

This applies to both rotational as well as linear motion.

What if there is rolling?

NET2

CM2

CM2122

21 )( I WVVmK ifif


Demo Example : A race rolling down an incline

Two cylinders with identical radii and total masses roll down an inclined plane.

The 1st has more of the mass concentrated at the center while the 2nd has more mass concentrated at the rim.

Which gets down first?

Two cylinders with radius R and mass mM

h

M who is 1st ?

M M

M

M

M

A) Mass 1

B) Mass 2

C) They both arrive at same time


Same Example : Rolling, without slipping, Motion

A solid disk is about to roll down an inclined plane.

What is its speed at the bottom of the plane ?

M

h

Mv ?

M M

M

M

M


Rolling without slipping motion

Again consider a cylinder rolling at a constant speed.

VCMCM

2VCM


Motion Again consider a cylinder rolling at a constant speed.

VCM

CM

2VCM

CMVCM

Sliding only

CM

Rotation only

VTang = R

Both with

|VTang| = |VCM |

If acceleration acenter of mass = - R


Example : Rolling Motion A solid cylinder is about to roll down an inclined plane.

What is its speed at the bottom of the plane ? Use Work-Energy theorem

M

h

Mv ?

Disk has radius R

M M

M

M

M

Mgh = ½ Mv2 + ½ ICM 2 and v =R

Mgh = ½ Mv2 + ½ (½ M R2 )(v/R)2 = ¾ Mv2

v = 2(gh/3)½


How do we reconcile force, angular velocity and angular acceleration?


Angular motion can be described by vectors

With rotation the distribution of mass matters. Actual result depends on the distance from the axis of rotation.

Hence, only the axis of rotation remains fixed in reference to rotation.

We find that angular motions may be quantified by defining a vector along the axis of rotation.

We can employ the right hand rule to find the vector direction


The Angular Velocity Vector

• The magnitude of the angular velocity vector is ω.• The angular velocity vector points along the axis

of rotation in the direction given by the right-hand rule as illustrated above.

• As increased the vector lengthens


From force to spin (i.e., ) ?

NET = |r| |FTang| ≡ |r| |F| sin

If a force points at the axis of rotation the wheel won’t turn

Thus, only the tangential component of the force matters With torque the position & angle of the force matters

FTangential

a

r

Fradial

F

A force applied at a distance from the rotation axis gives a torque

Fradial

FTangential

r

=|FTang| sin


Rotational Dynamics: What makes it spin?


Torque is the rotational equivalent of forceTorque is the rotational equivalent of forceTorque has units of kg m2/s2 = (kg m/s2) m = N m

FTangential aa

r

Fradial

FF

A force applied at a distance from the rotation axis

NET = r FTang = r m aTang

= r m r = (m r2)

For every little part of the wheel


For a point mass NET = m r2

This is the rotational version of FNET = ma

Moment of inertia, Moment of inertia, I ≡ imi ri2 , is the rotational is the rotational

equivalent of mass.equivalent of mass. If I is big, more torque is required to achieve a

given angular acceleration.

FTangential aa

r

Frandial

FF

The further a mass is away from this axis the greater the inertia (resistance) to rotation (as wesaw on Wednesday)

NET = I


Rotational Dynamics: What makes it spin?


A constant torque gives constant angular acceleration if and only if the mass distribution and the axis of rotation remain constant.

FTangential a

r

Fradial

F

A force applied at a distance from the rotation axis gives a torque


Torque, like , is a vector quantity

Magnitude is given by (1) |r| |F| sin

(2) |Ftangential | |r|

(3) |F| |rperpendicular to line of action | Direction is parallel to the axis of rotation with respect to the

“right hand rule”

And for a rigid object = I

r

F

F cos(90°) = FTang.

r

FFradial

F

a

r

F

r 90°

r sin line of action


Example : Rolling Motion

Notice rotation CW (i.e. negative) when ax is positive!Combining 3rd and 4th expressions gives f = Max / 2Top expression gives Max + f = 3/2 M ax = Mg sin So ax =2/3 Mg sin

M

Newton’s Laws:

cos0y MgNMay F

fMgMax sinxF

fRMRI CMCM 221

CM

xCM aR Mg

f

N

x dir


Statics

Equilibrium is established when

0 motion nalTranslatio Net F

0 motion RotationalNet

In 3D this implies SIX expressions (x, y & z)


Example

Two children (60 kg and 30 kg) sit on a horizontal teeter-totter. The larger child is 1.0 m from the pivot point while the smaller child is trying to figure out where to sit so that the teeter-totter remains motionless. The teeter-totter is a uniform bar of 30 kg its moment of inertia about the support point is 30 kg m2.

Assuming you can treat both children as point like particles, what is the initial angular acceleration of the teeter-totter when the large child lifts up their legs off the ground (the smaller child can’t reach)?

For the static case:

0 motion RotationalNet


Example: Soln.

Draw a Free Body diagram (assume g = 10 m/s2)

0 = 300 d + 300 x 0.5 + N x 0 – 600 x 1.0

0= 2d + 1 – 4

d = 1.5 m from pivot point

0 UseNet

30 kg

1 m

60 kg30 kg

0.5 m

300 N 300 N 600 N

N


Recap

Assignment: HW7 due tomorrow

Wednesday: review session

lecture 17

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