3-Dimensional Rotation: Gyroscopes

8.01

W14D1

Today’s Reading Assignment Young and Freedman: 10.7

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Problem Set 11 Due Thursday Dec 8 9 pm

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W014D2 Reading Assignment Young and Freedman: 10.7

2

Demo: Gimbaled Gyroscope (B140)

Rules to Live By: Angular Momentum and Torque

1) About any fixed point P

2) Independent of the CM motion, even if and

are not parallel

rL

P=

rL about cm+

rL of cm =

rL about cm+

rrP,cm×mtotal

rvcm

rτ about cm =

drL about cm

dt

about cmLr

ωr

Mini Demo: Pivoted Falling Stick

Magnitude of the angular momentum about pivot changes.

Direction of change of angular momentum about pivot is the same as direction of angular momentum about pivot

Demo: Bicycle Wheel Two Cases

Case 1: Magnitude of the angular momentum about pivot changes.

Direction of change of angular momentum about pivot is the same as direction of angular momentum about pivot

Case 2: Direction of angular momentum about pivot changes

Time Derivative of a Vector with Constant Magnitude that

Changes Direction

Concept Question: Time Derivative of Rotating Vector

A vector rA of fixed leng th A is rotat ing about the z -axis with the z-component of

angu lar ve loc ity ωz=ω. At t=0 it is pointed in the positive j-directi .on drA(t)dt is given by

1) −ωAsinωti−ωAcosωtj 2) ωAsinωti+ωAcosωtj 3) ωAcosωti+ωAsinωtj 4) −ωAcosωti−ωAsinωtj 5) −ωAsinωti+ωAcosωtj 6) −ωAcosωti+ωAsinωtj

Example: Time Derivative of Position Vector for Circular

MotionCircular Motion: position vector points radially outward, with constant magnitude but changes in direction. The velocity vector points in a tangential direction to the circle.

ˆd dr

dt dt

rv

ˆrr r

Generalization: Time Derivative of a Vector

Consider a vector

where

Vector can change both magnitude and direction.

Suppose it only changes direction then

drA

dt=Ar

ddt

=rA sinφ

ddt

=rA⊥

ddt

ˆ ˆz rA A A k r

A

r=

rA sinφ =

rA⊥

coszA A

Torque and Time Derivative of Angular Momentum

Torque about P is equal to the time derivative of the angular momentum about P

If the magnitude of the angular momentum is constant then the torque can cause the direction of the perpendicular component of the angular momentum to change

rτP

ext =d

rL P

dt

Introduction To Gyroscopic Motion

Gyroscopic Approximation

Flywheel is spinning with an angular velocity

Precessional angular velocity

Gyroscopic approximation: the angular velocity of precession is much less than the component of the spin angular velocity,

rω =ω r

rΩ =Ω k

Ω= ω

Strategy

1. Calculate torque about appropriate point P

2. Calculate angular momentum about P

3. Apply approximation that to decide which contribution to the angular momentum about P is changing in time. Calculate

4. Apply torque law

to determine direction and magnitude of angular precessional velocity

drL

P/ dt

rτP

rL

P

Ω= ω

rτP =

drL P

dt

rΩ

Table Problem: Gyroscope: Forces and Torque

Gravitational force acts at the center of the mass and points downward. Pivot force acts between the end of the axle and the pylon. What is the torque about the pivot point P due to gravitational force

Table Problem Gyroscope: Time Derivative of Angular Momentum

What is the time derivative of the angular momentum about the pivot point for the gyroscope?

Torque and Time Derivative of Angular Momentum

Torque about P is equal to the time derivative of the angular momentum about P

Therefore

Precession angular speed is

rτPS

ext =d

rL P

dt

d mg =I1ω Ω

Ω=dmg / I1ω

More Detailed Analysis of Angular Momentum for

Gyroscopic Motion

Angular Momentum About Pivot Point

The total angular momentum about the pivot point P of a horizontal gyroscope in steady state is the sum of the rotational angular momentum and the angular momentum of center of mass

rL

P=

rL cm+

rrP,cm×M

rVcm

.

Angular Momentum about Center of Mass

The disk is rotating about two orthogonal axes through center of mass. It is rotating about the axis of the shaft, with angular speed ω. The moment of inertia of a uniform disk about this axis is I1 = (1/2) MR2. The disk is also rotating about the z-axis with angular speed Ω. The moment of inertia of a uniform disk about a diameter is I2 = (1/4)MR2. The angular momentum about the center of mass is the sum of two contributions

rL

cm=I1ω r + I 2Ω k =(1 / 2)MR2ω r + (1 / 4)MR2Ω k

Angular Momentum Due to Motion of Center of Mass

The angular momentum about the pivot point P due to the center of mass motion is

where is a unit vector in the positive z-direction and is the angular speed about the z-axis

k Ω

rr

P,cm×M

rVcm =Md2Ω k

Angular Momentum of Flywheel about Pivot Point:

rL

P=

rL cm+

rrP,cm×M

rVcm

rL

P=I1ω r + I2Ω k + Md2Ω k

rL

P=I1ω r + ( I2 + Md2 )Ω k

Gyroscope: Time Derivative of Angular Momentum

If the angular speed (precession angular speed) about the z-axis is constant then only the direction of the spin angular momentum

along the axis of the gyroscope is changing in time hence

drL

P

dt=

rL P,⊥

ddt

=I1ω Ω

rL

P,⊥ ≡rL spin =I1ω r

Torque and Time Derivative of Angular Momentum

Torque about P is equal to the time derivative of the angular momentum about P

Therefore

Precession angular speed is

rτPS

ext =d

rL P

dt

d mg =I1ω Ω

Ω=dmg / I1ω

Concept Question: GyroscopeFor the simple gyroscope problem we just solved,if the mass of the disk is doubled how will the newprecession rate Ω be related to the original rate Ω0?

Ω = 4 Ω0

2) Ω = 2 Ω0

3) Ω = Ω0

4) Ω = (1/2) Ω0

5) Ω = (1/4) Ω0

.

Table Problem: Tilted Gyroscope

A wheel of mass M and moment of inert ia cmI about its centra l axis , through

its cent er of mass , is a t one end of an axle of length d. The a xle is pivoted a t an ang le β with respect to the vertical. The wheel is set into motion so that it executes uniform precessi .on Its sp inangular velocity has magnitude ω an dis directed as shown in the figure below. Find the magnitude and direction of th eprecessional angular velocity rΩ in terms of the other parameters. Assume ω>>rΩ.

Demo: Gyroscope in a Suitcase

A gyroscope inside a suitcase is spun up via a connection to the outside of the suitcase. The suitcase is carried across the lecture hall. When the lecturer turns while walking, the gyroscope causes the suitcase to rise about the handle.

Table Problem: Suspended Gyroscope

A gyroscope wheel is at one end of an axle of length d. The oth er end of the axle is suspend ed from a str ing of leng th s. The whee l is set into motion so that it executes un iform precess ion in the hor izonta l plane . The str ing makes an ang le β with the vertical. The wheel has mass M and moment of inertia about it scenter of mass cmI. Its spin angula rspeed is ω. Neglect the m ass of th eshaft and the mass of the string. Assume ω>>rΩ. What is the

direction and magnitude of the precessional angular velocity?