Rolling, Torque and Angular

Momentum Mathy AP Physics C

How did they move 30 ton statues 7 km on Easter Island 400 years ago?

The physics and engineering of rolling has been around a long time, but as self correcting Segways have shown, there will always be something new !

Rolling as Translation & Rotation The center of mass (O) of a rolling wheel moves a distance s at while the wheel rotates through angle θ. The point P at which the wheel makes contact with the surface over which the wheel rolls also moves a distance s to the next point P! n  During any time interval s= Rθ n  Differentiating this equation with respect to time

€

v v com

€

vcom =ωR€

st

=Rθt

↓

Equation for smooth rolling motion

Fig 11-3

Rotation vs Translation vs Rolling !

n  All points move with ω n  Points on edge move with v =

€

v v com

n  All points move to the right with the same a linear velocity of v =

€

v v com

n  The rolling motion of a wheel is a combination of (a) & (b) §  The portion of the wheel at the bottom (P) is stationary §  The top is moving at 2 !

€

v v com

Consider (c ) to be an overlay of (a) on top of (b) ! �

Fig 11.4

The Kinetic Energy of Rolling Object

n  KE consists of rotational about the axis and translation of a moving COM !

€

K =1/2Iω 2 +1/2Mv 2com

Sample Problem 1.0

n  The land speed record of 1233 km /hr was set by jet car that had disc wheels (uniform) that had a mass of 170 kg! Calculate the K of one wheel .

€

K =1/2Iω 2 +1/2Mv 2com

€

vcom =ωR

K of 1 stick of dynamite 2.1x106 J

€

K =1/2(1/2Mr2) (vcom )(r)2

2

+1/2Mv 2com

K = 3/4mv 2

K

Example 2.0

Consider a ball rolling down a ramp. Calculate the translational acceleration of the ball's center of mass as the ball rolls down. Find the angular acceleration as well. Assume the ball is a solid sphere.

Let’s first look at the ball’s F.B.D

θ

mg

Fn

Ff The key word here is “rolling”. Up to this point we have always dealt with objects sliding down inclined planes. The term “rolling” tells us that FRICTION is causing the object to rotate (by applying a torque to the ball).

Example cont’

mamgFmaFmg

maF

f

f

net

−=

=−

=

θ

θ

sin

sin

maFRamRRF

mRIraIRF

IFr

f

f

CMspheref

32

)(32

32,

sin

2

2@

=

=

===

==

αα

ατθ

θ

mg

Fn

Ff

θ mgcosθ

mgsinθ

Rg

Raga

mamg

mamamg

maFmamg f

5sin3

,5sin3

35sin32sin

32sin

θα

αθ

θ

θ

θ

=

==

=

+=

==−

Angular Momentum Translational momentum is defined as inertia in motion. It too has an angular

counterpart. Consider the following disc that is spinning, and a small mass point at the outside of the rim.

r m

F

v

€

Consider Newton's Second Law

F =ΔρΔt

=ΔmvΔt

Multiply both sides by r, and we introduce torque

Fr = ΔrmvΔt

Force = rate of change of linear momentum Torque = rate of change of angular momentum

€

L€

rmv is a special variable in physics

€

L = r(mv)

€

L = Iω

2 ways to find the angular momentum Rotational relationship

ω

ω2mRL

IL=

=

R ω v

ω

ω

θ

2

90,

mRLRvmvRL

rpL

=

==

=⊗=

Translational relationship

mass

In the case for a mass moving in a circle.

In both cases the angular momentum is the same equation.

Don’t forget

Just like TORQUE, angular momentum is a cross product. That means the direction is always on a separate axis from the 2 variables you are crossing. In other words, if you cross 2 variables in the X/Y plane the cross product’s direction will be on the “Z” axis

The Torque as a Cross Product

nFr ̂sinφτ rF=×=The torque is defined relative to some point of origin.

Angular Momentum

prL×=nL ̂sin)( φmvr=

Angular Momentum for Circular Motion

n  For a particle moving in a circle, the angular momentum about the center of the circle is

kvrprL ̂90sinrmvm=×=×=

ωωkL Imrrmr=== 2 ̂)(ω

Angular Momentum for a System of Particles

n  For a single particle, the formula L=Iω only holds for circular motion where the origin is the center of the circle.

n  The formula also holds for a system of particles rotating about an axis of symmetry.

Torque and Angular Momentum

n  Torque has the same relationship to angular momentum as force has to linear momentum.

dtd

EXTPF =∑ ∑ =

dtdL

EXTτ

Conservation of Angular Momentum

n  If the net external torque acting on a system is zero, the total angular momentum of the system is constant.

Angular Momentum is also conserved Here is what this says: IF THE NET TORQUE is equal to ZERO the CHANGE ANGULAR MOMENTUM is equal to ZERO and thus the ANGULAR MOMENTUM is CONSERVED. Here is a common example. An ice skater begins a spin with his arms out. His angular velocity at the beginning of the spin is 2.0 rad/s and his moment of inertia is 6 kgm2. As the spin proceeds he pulls in his arms decreasing his moment of inertia to 4.5 kgm2. What is the angular velocity after pulling in his arms?

=

=

=

=

ω

ω

ωω

)5.4()2)(6(IILL

o

o

2.67 rad/s

Conservation of Angular Momentum Practice Problem 2 n  A merry-go-round (r =2, I

= 500 kg m/s2) is rotating about a frictionless pivot, making one revolution every 5 s. A child of mass 25 kg originally standing at the center walks out to the rim. Find the new angular speed of the merry-go-round.

n  Ans.- ωf=(5/6)ωi

Conservation of Angular Momentum Practice Problem 3 n  The same child as in the

previous problem runs with a speed of 2.5 m/s tangential to the rim of the merry go round, which is initially at rest. Find the final angular velocity of the child and merry go round together.

n  Ans. ω= 0.208 rad/s

Conservation of Angular Momentum Practice Problem 1

n  A child of mass m = 30 kg jumps on the edge of a small merry- go- round ( ω = 1rad /s,r=2.5m, mass= 100kg). If the child walks in towards the center of the disk and stops at 05m from the center, what will happen be the new angular velocity of the merry go round ?

Practice Problem 3

n  A child of mass m = 30 kg jumps on the edge of a small merry- go- round ( ω = 1rad /s,r=2.5m, mass= 100kg). If the child walks in towards the center of the disk and stops at 05m from the center, what will happen be the new angular velocity of the merry go round ?

r m

r m

ω = 1 rad/s ω ? Start by calculating the moment of inertia for each

€

I = IMGR + Ichild

I =12MR2 +mR2

I = (1/2M +m)R2

€

I = IMGR + Ichild

I =12MR2 +mrnew

2

r m

r m

ω = 1 rad/s ω ? Now let angular momentum be conserved

1.6 rad/s

€

I = IMGR + Ichild

I =12MR2 +mR2

I = (1/2M +m)R2

€

I = IMGR + Ichild

I =12MR2 +mrnew

2

€

L = L Iω i = Iω f

(12M +m)R2ω i = (1

2MR2 +mrf

2)ω f

Conservation of Angular Momentum Practice Problem 4a n  A particle of mass m

moves with speed v0 in a circle of radius r0 . The particle is attached to a string that passes through a hole in the table. The string is pulled downward so the mass moves in a circle of radius r. Find the final velocity.

n  Ans. v= (r0/r) v0

Conservation of Angular Momentum Practice Problem 4b n  Find the tension T in the

string in terms of m, r, r0 and vo.

n  Ans.

3

20

20

rvmrT=

Pg 130 PR

Momentum Formula for Kinetic Energy

mpK

mpmK

mpmmvK

vmp

mvp

2

2

2

2

21

2

212

21

=

=

⎟⎠

⎞⎜⎝

⎛==

=

=

n  Often it is useful to have the formulas for kinetic energy written in terms of momentum.

ILK

ILIK

ILIIK

IL

IL

2

2

2

2

21

2

212

21

=

=

⎟⎠

⎞⎜⎝

⎛==

=

=

ω

ω

ω

Some interesting Calculus relationships

RotationalKIW

dIWdtd

ddtdIW

dtd

dIdFrdW

dsdsFrFFdrW

o

Δ→=

==

==

→→=

=→=

=→•==

∫

∫

∫∫ ∫

∫∫

ωω

ω

ω

ω

ωωωθ

θωω

α

θαθτθ

θθ

0)

2(

rddsrs

length arc small,

2

tangent

More interesting calculus relationships

FvPP

tttWW

rFW

=

=

=ΔΔ

=→Δ•=

Δ•=

τω

ωθθ

τθτ ,