Gyroscopes and statics

Welcome back from Spring Break !

CAPA due Friday at 10pm •  We will finish Chapter 11

in H+R on angular momentum and start Chapter 12 on stability.

•  Friday we will begin gravitation.

Web page: http://www.colorado.edu/physics/phys1110/phys1110_sp12/

Announcements:

Invented by Leon Foucault

Summary of rotation: Comparison between Rotation

and Linear Motion

Angular Linear θ = x / R

ω = v / R

α = a / R

x

v

a

Comparison Kinematics

Angular Linear constant=!

!

" ="0

+#t

! ! " #= + +0 0

21

2t t

constanta =

v = v0+ at

x = x0+ v

0t +1

2at

2

!

" 2#"

0

2

= 2$%

!

"AVE

=1

2(" +"

0)

!

v2" v

0

2

= 2ax

!

vAVE

=1

2(v + v

0)

Comparison: Dynamics

Angular Linear

!

K =1

2I" 2

!

K =1

2mv

2

I = Σi mi ri2 m

F = a m τ = r x F = α I

L = r x p = I ω p = mv

!

"EXT

=dL

dt

!

FEXT

=dp

dt

W = τ •Δθ W = F •Δx

ΔK = WNET ΔK = WNET

Angular Momentum Conservation

y x

d m

Example: Throwing ball from stool

•  A student sits on a stool which is free to rotate. The moment of inertia of the student plus the stool is I. She throws a heavy ball of mass M with speed v such that its velocity vector passes a distance d from the axis of rotation. – What is the angular speed ωF of the student-

stool system after she throws the ball ?

top view: before after

d v M

I I ωF

Example: Throwing ball from stool... •  Conserve angular momentum (since there are no

external torques acting on the student-stool system): –  LAFTER = 0 = Lstool - Lball, Lball = Iballωball = Md2 (v/d) = M d v

–  LBEFORE = 0 , Lstool = IωF

0 = IωF - M d v

top view: before after

d v M

I I ωF

ωF = M v d / I

Clicker question 1 Set frequency to BA

How do the magnitudes of the angular momentum of the planet Lplanet (with the origin at the Sun) at positions A and B compare?

A.  B.  C. 

A S B

For any central force like gravity, the force vector is parallel (or antiparallel) to the vector so no torque can be applied.

Thus, a central force cannot affect the angular momentum.

•  A mass m=0.1kg is attached to a cord passing through a small hole in a frictionless, horizontal surface as in the Figure. The mass is initially orbiting with speed ωi = 5rad/s in a circle of radius ri = 0.2m. The cord is then slowly pulled from below, and the radius decreases to r = 0.1m. How much work is done moving the mass from ri to r ?

(A) 0.15 J (B) 0 J (C) - 0.15 J (D) = 0.05 J

ri ωi

Clicker question 2 Set frequency to BA

Angular Momentum is conserved!

€

I1ω1 = I2ω2; ω2 =I1ω1

I2

=r1

2

r22 ω1 =

.2( )2

(.1)2 (5) = 20rad /s

ΔK = WNET

!

K =1

2I" 2

€

Ki =12mri

2( )ω i2 = (.5)(.1× .22)(5)2 = .05J

€

K f =12mrf

2( )ω f2 = (.5)(.1× .12)(20)2 = .20J €

ΔK = K f −Ki = 0.15J

Gyroscopes and precession Remember centripetal forces cause radial acceleration which changes the velocity direction but not the magnitude.

Similarly, some torques change the angular momentum vector direction rather than the magnitude.

Bicycle wheel gyroscope

pivot

When the wheel is not spinning the torque from the weight causes the wheel to rotate downward causing an angular velocity into the page. The torque points into the page as well.

When the wheel is spinning with the angular momentum vector perpendicular to the weight vector the torque causes a change in the angular momentum vector direction causing it to precess which is another form of rotation.

Can work out precession frequency

Gyroscopic Motion

 Suppose you have a spinning gyroscope in the configuration shown below:

 If the left support is removed, what will happen ??

ω pivot support

g

Gyroscopic Motion...

 Suppose you have a spinning gyroscope in the configuration shown below:

 If the left support is removed, what will happen ?   The gyroscope does not fall down !

ω pivot

g

Gyroscopic Motion...

ω pivot

Gyroscopic Motion...

  The magnitude of the torque about the pivot is τ = mgR.   The direction of this torque at the instant shown is out of

the page (using the right hand rule).   The change in angular momentum at the instant

shown must also be out of the page!

ω pivot

R

mg

! =d

dt

L

Gyroscopic Motion...

•  Consider a view looking down on the gyroscope. –  The magnitude of the change in angular

momentum in a time dt is dL = Ldφ.

–  So

where Ω is the “precession frequency”

top view

dφ ( pivot

dL

dt= L

d!

dt! L"

Gyroscopic Motion...

! =dL

dt= L!

ω pivot

R

mg

Ω

€

Ω =mgRIω

! =!

L

€

Ω

3 gyroscopes (to give direction) and 3 accelerometers (to give velocity) can be combined to form an inertial navigation system.

Gyroscope uses The artificial horizon in planes uses gyroscopes to indicate the pitch and roll of the plane in the absence of external views out the window due to clouds or darkness.

Gyroscopes can also be used as stabilizers. Some ships use them to reduce effects of waves The international space station uses four 500 pound gyroscopes spinning at 6500 RPM to maintain orientation

Clicker question 3 Set frequency to BA

If there is an object which is completely stationary, which answer is the most correct. A.  Net force on the object is 0 B.  Net torque on the object about the center of mass is 0. C.  Net torque on the object about any point in the universe is 0. D.  Two of A, B, C are true. E.  All three of A, B, C are true.

If there is a net force on the object it must be accelerating.

A net torque about the center of mass would start it rotating.

A net torque about any point in the universe indicates it is rotating about that point and therefore moving.

Statics The theory behind statics is very simple. Since nothing is moving, the acceleration and angular acceleration is 0.

Statics has just two equations:

These are vector equations so we actually have more equations:

€

Fz∑ = 0

A statics problem

1 kg

A 10 m long, 20 kg board has 2 supports, 1 m from the board center of mass. There is a 1 kg block on top of one support. How far can a 100 kg person walk out from that support without the board tilting?

First we draw an extended free body diagram!

A statics problem Next we apply the statics equations.

Do we know anything else from how the problem is setup?

The only forces are in the vertical direction:

To apply the torque equation we need to pick an axis (can be anywhere in the universe). Let us choose the first support. Remembering to keep the signs straight, the torque equation is:

At this point, we have 2 equations and 3 unknowns which means we need more information to solve this.

A statics problem The question asked for the maximum distance without tipping the plank. What does that tell us?

Put into torque equation: Solve the force equation for N2:

If we had recognized N1=0 and used the 2nd pivot as our axis, we immediately get our final equation from the torques and would not even need the force equation.