3-Dimensional Rotational Motion and Gyroscopes

8.01

W14D2

Today’s Reading Assignment MIT 8.01 Course Notes

Chapter 22 Three Dimensional Rotations and Gyroscopes, Sections 22.4

Announcements

Problem Set 12 Due Week 14 Thursday 9 pm Math Review Week 15 Tuesday at 9 pm in 26-152 Sunday Tutoring in 26-152 from 1-5 pm Final Exam Dec 16 Johnson Athletic Center 9am – 12 noon

Problem Solving Strategy

1.  Calculate torque about appropriate point S,

2.  Calculate angular momentum about S,

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 precession angular velocity

5.  Apply Newton’s Second Law to center of mass motion

dLS / dt

τS

LS

Ωω

τP = d

LS / dt

Ω

Review Table Problem: Gyroscope Consider a flywheel that is spinning with angular speed as shown in the figure. (a) What is the precessional angular speed ? (b) Does the gyroscope rotate clockwise or counterclockwise about the vertical axis (as seen from above)?

ω s

Table Prob. Gyroscope on Rotating Platform A gyroscope consists of an axle of

negligible mass and a disk of mass M and radius R mounted on a platform that rotates with angular speed Ω as shown in the figure below. The gyroscope is spinning with angular speed ω . Forces aF and bF act on the gyroscopic mounts. The goal of this problem is to find the magnitudes of the forces aF and bF . You may assume that the moment of inertia of the gyroscope about an axis passing through the center of mass normal to the plane of the disk is given by nI .

a) Calculate the torque about the center of mass of the gyroscope. b) Calculate the angular momentum about the center of mass of the gyroscope. c) Use Newton’s Second Law find a relationship between aF and bF , the mass

M of the gyroscope, and the gravitational constant g . d) Use the torque equation and Newton’s Second Law to find expressions for aF

and bF .

Table Problem: Mill Stone In a mill, grain is ground by a massive wheel that rolls without slipping in a circle on a flat horizontal mill stone driven by a vertical shaft. The rolling wheel has mass M , radius b and is constrained to roll in a horizontal circle of radius R at angular speed Ω . The wheel pushes down on the lower mill stone with a force equal to twice its weight (normal force). The mass of the axle of the wheel can be neglected. Express your answers to the following questions in terms of R , b , M , Ω , and g as needed. The goal of this problem is to find Ω .

a) What is the relation between the angular speed ω of the wheel about its axle and

the angular speed Ω about the vertical axis?

b) Find the time derivative of the angular momentum about the joint (about the point P in the figure above) d

LP / dt .

c) What is the torque about the joint (about the point P in the figure above?

d) What is the value of Ω ?

Worked Example: Sopwith Camel The Sopwith Camel was a single-engine fighter plane flown by British pilots during WWI (and also by the character Snoopy in the Peanuts comic strip). It was powered by a radial engine, and the entire engine rotated with the propeller. The Camel had an unfortunate property: if the pilot turned to the right the plane Tended to go into dive, while a left turn caused the plane to climb steeply. These tendencies caused inexperienced pilots to crash or stall during takeoff.

Discussion: Sopwith Camel From the perspective of the pilot, who sat behind the engine, did the engine rotate clockwise or counter clockwise? Odd tables: Argue on the basis of torque (and third law pairs). Even tables: Argue on the basis of conservation of angular momentum in the horizontal plane.

Torque: Assume one of the two possible directions ofrotation and see if it gives the correct result.

L

if cw rightturn

dL/dt

! on engine

! on plane

results in nosepitching down,so clockwise iscorrect

Worked Example: Sopwith Camel

Angular momentum: The pilot begins turning right usingthe rudder on the tail section, applying an external torquechanging Lz. The large horizontal component of the engine’sL swings right. If the pilot does not use the elevators toapply another external torque, the horizontal componentof the plane’s motion must counter the engine, to ensureno net change in horizontal angular momentum.

dL engine

dL plane,obtained byrotating planenose downward

L

if cw rightturn

Worked Example: Sopwith Camel

Concept Question: Stabilizing a Turning Car

When making a turn every car has a tendency to roll over because its center of mass is above the plane where the wheels contact the road. Imagine a race car going counter clockwise on a circular track. It could mitigate this effect by mounting a gyroscope on the car. To be effective the angular velocity vector of the gyro should point .

1) ahead 2) behind 3) to the left 4) to the right 5) up 6) down

Table Problem: Stabilizing a Car When an automobile rounds a curve at high speed (in the figure below the car is turning left), the loading (weight distribution) on the wheels is markedly changed. For sufficiently high speeds the loading on the inside wheel goes to zero, at which point the car starts to roll over. The tendency can be avoided by mounting a large spinning flywheel on the car.

a)  What should be the sense of rotation of the flywheel to help equalize the loading? (Be sure that your method works for cars turning in either direction.)

b) Show that for a disk-shaped flywheel of mass mw and radius R, the requirement for equal loading is that the angular speed of the flywheel, ωs, is related to the speed of the car vcm by where mc is the mass of the car and flywheel, and h is the height of the center of mass of the car (including the flywheel) above the road. Assume the road is unbanked.

ω s = 2vcmmch / mwR2

Appendix:

Why a Gyroscope Precesses

Deflection of Particle by Small Impulse

If the impulse << the primary effect is to rotate about the x axis by a small angle .

p1p

θ

I

Note the symbol I denotes the magnitude of the impulse and is not the moment of inertia

Deflection of Particle by Small Impulse

ave t= Δ = ΔI p F

( )ave avet tΔ = Δ = × ΔL r F τ

ave tΔ = × ΔL r F

Δ = ×L r I

The application of causes a change in the angular momentum through the torque equation.

I

L

Deflection of Particle by Small Impulse

I

L

As a result, rotates about the x axis by a small angle θ. Note that although is in the z direction, is in the negative y direction.

ΔL

Effect of Small Impulse on Tethered Ball

The ball is attached to a string rotating about a fixed point. Neglect gravity.

The ball is given an impulse perpendicular to and to . pr

Effect of Small Impulse on Tethered Ball

As a result, rotates about the x axis by a small angle . Note that although is in the z direction, is in the negative y direction. ΔL

L

I

Effect of Small Impulse on Tethered Ball

IThe plane in which the ball moves also rotates about

the x axis by the same angle. Note that although is in the z direction, the plane rotates about the x axis.

Effect of Small Impulse on Tethered Ball

Concept Question: Effect of Large Impulse on Tethered Ball

I

What impulse must be given to the ball in order to rotate its orbit by 90 degrees as shown without changing its speed?

Concept Q. Ans. : Effect of Large Impulse on Tethered Ball

I

must halt the y motion and provide a momentum of equal magnitude along the z direction.

cancels the z component of and adds a component of the same magnitude in the negative y direction. ΔL

L

Effect of Large Impulse on Tethered Ball

Effect of Small Impulse Couple on Baton

Now we have two equal masses at the ends of a massless rod which spins about its center. We apply an impulse couple to insure no motion of the CM.

Again note that the impulse couple is applied in the z direction. The resulting torque lies along the negative y direction and the plane of rotation tilts about the x axis.

Effect of Small Impulse Couple on Baton

Effect of Small Impulse Couple on Massless Shaft of Baton

Instead of applying the impulse couple to the masses one could apply it to the shaft to achieve the same result.

Concept Q.: Effect of Small Impulse Couple on Massless Shaft of Baton

To make the top of the shaft move in the -y direction in which direction should one apply the top half of an impulse couple?

The impulse couple Ib applied to the shaft has the same effect as the Ia couple applied directly to the masses. Both produce a torque in the - y direction.

Concept Q. Ans.: Effect of Small Impulse Couple on Massless Shaft of

Baton

Effect of a Small Impulse Couple on Massless Shaft of Baton

Trying to twist the shaft around the y axis causes the shaft and the plane in which the baton moves to rotate about the x axis.

Effect of a Small Impulse Couple on Disk

The plane of a rotating disk and its shaft behave just like the plane of the rotating baton and its shaft when one attempts to twist the shaft about the y axis.

Effect of a Small Impulse Couple on Non-Rotating Disc

This unexpected result is due to the large pre-existing . If the disk is not rotating to begin with, is also the final . The shaft moves in the direction it is pushed.

L

L ΔL

Effect of a Small Impulse Couple on Disk

It does not matter where along the shaft the impulse couple is applied, as long as it creates the same torque.

Effect of a Force Couple on Rotating Disk

A series of small impulse couples, or equivalently a continuous force couple, causes the tip of the shaft to execute circular motion about the x axis.

Effect of a Force Couple on Rotating Disk

d dt= ΩL L

ddtL

=τ

ΩL

=τ

IωΩ=τ

IωΩ =τ

The precession rate of the shaft is the ratio of the magnitude of the torque to the angular momentum.

Precessing Gyroscope