ho4_rigid body rotational dynamics_09!30!11

8/3/2019 HO4_Rigid Body Rotational Dynamics_09!30!11

http://slidepdf.com/reader/full/ho4rigid-body-rotational-dynamics093011 1/26

This Print

Rigid Body Rotational Dynamics

John ChiassonBoise State University

September 30, 2011



ii



This Print

Contents

1 Rigid Body Rotational Dynamics 11.1 Moment of Inertia . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Newton’s Law of Rotational Motion . . . . . . . . . . . . . . . . . . . . . . . 21.3 Gears . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

1.3.1 Algebraic Relationships Between Two Gears . . . . . . . . . . . . . . 111.3.2 Dynamic Relationships Between Two Gears . . . . . . . . . . . . . . 12

1.4 Rolling Cylinder . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151.5 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22



iv



This Print

1

Rigid Body Rotational Dynamics

1.1 Moment of InertiaThe equations of motion of a rigid body that is constrained to rotate about a fixed axis arereviewed here briefly. Consider a cylinder which is constrained to rotate about a fixed axisas shown in Figure 1.1.

R

Axle

FIGURE 1.1. Cylinder constrained to rotate about a fixed axis.

The approach here is to obtain the equations of motion of the cylinder by first obtainingan expression for its kinetic energy. To do so, denote the angular speed of the cylinder by and the mass density of the material making up the cylinder by . Then consider the cylinderto be made up of a large number of small pieces of material ∆ where the piece hasmass

∆ = ∆∆∆

This is illustrated in Figure 1.2.

im

d r i

d

d

dr

ir

FIGURE 1.2. Cylinder is considered to be made up of small masses ∆ Drawn by Sharon Katz.

Each piece of mass ∆ is rotating at the same angular speed so that the linear speedof ∆ is = where is the distance of ∆ from the axis of rotation. The kineticenergy of ∆ is given by

=1

2∆

2

=1

2∆()2



2 1. Rigid Body Rotational Dynamics

The total kinetic energy is then

=X=1

( ) =X=1

1

2∆

2

=X=1

1

2∆ ()2 =

1

22

X=1

∆2

Dividing the cylinder into finer and finer pieces so that →∞ and ∆ → 0, the sum

X=1

∆2

becomes the integral

=

ZZZ

2

The quantity is called the moment of inertia . Using the kinetic energy of the cylindermay now be written as

=1

2

2

Assuming the axle radius is zero, the moment of inertia of the cylinder (assuming isconstant) is computed to be

=

Z 0

Z 0

Z 2

0

2 =1

2(2)2 =

1

22

where is the total mass of the cylinder.

1.2 Newton’s Law of Rotational Motion

The above expression for the kinetic energy is now used to derive a relationship betweentorque and angular acceleration. Recall from elementary mechanics that the work done on amass by an external force equals the change in its kinetic energy. In particular, consider anexternal force F acting on the cylinder as shown in Figure 1.3.

T F

N F

F

x

y y

x

z

FIGURE 1.3. Force F applied to the cylinder is resolved into a normal and tangential component.Drawn by Sharon Katz.



1. Rigid Body Rotational Dynamics 3

The cylinder is on an axle and therefore constrained to rotate about the axis. Figure1.3 shows the force F applied to the cylinder at the position ( ) (in polar coordinates)resolved into a tangential component (tangent to the rotational motion) and a normal

component . Using polar coordinates, we write this force as F = r + θ where r is aunit vector in the increasing direction and θ is a unit vector in the increasing direction.Similarly, x y and z are unit vectors in the increasing directions, respectively. The

torque about an axis is defined as the cross product τ , r × F where r is vector from theaxis to the point of application of the force and F is the applied force. We then have

τ , r× F = r×³

r + θ´

r× r |{z} 0

+ r× θ | {z } z

= z

= sin()z (1.1)

where is the angle between r and F. Recall from elementary mechanics that the magnitude of the cross product r× F is defined as sin() and the direction of r× F is perpendicular to

both r and F along the axis of rotation determined by the right hand rule1. As = sin()is the tangential component, the torque may be rewritten as

τ = z = z

or, in scalar form, as

=

The motivation for the definition of torque as given by (1.1) is that it is the cause of rotational

motion. Specifi

cally, the rotational motion about an axis is caused by the applied tangentialforce and the further away from the axis of rotation that the tangential force isapplied, the easier it is to get rotational motion. That is, the torque (cause of rotationalmotion) increases if either or increases which corresponds to one’s experience (e.g.,opening doors).

To summarize, τ is a vector pointing along the axis of rotation and the magnitude is givenby

| τ | = | | = | |

(Recall that the angular velocity vector ω = z also points along the axis of rotation where is the angular speed.)

With s,

ˆθ =

ˆθ, the change in work done on the cylinder by the external force Fis

= F · s = =

where , . Dividing by , the power (rate of work) delivered to the cylinder is givenby

=

=

1 Using your right hand, curl your fingers in the direction from the first vector r to the second vecotor F. Then your thumbpoints in the direction of r× F.




As the rate of work done equals the rate of change of kinetic energy, it follows that

=

µ1

22

¶=

(1.2)

or

=

This gives the fundamental relationship between torque and angular acceleration:

=

That is, the applied torque equals the moment of inertia times the angular acceleration. Thisis the basic equation for rigid body rotational dynamics.

Viscous Rotational FrictionAlmost always there are frictional forces, and therefore, frictional torques acting between theaxle and the bearings.2 This is illustrated in Figure 1.4.

y

x

Frictional force f

F

Axle

FIGURE 1.4. Viscous friction torque.

Often the frictional force is proportional to the angular speed and this model of friction iscalled viscous friction which is expressed mathematically as

τ = − ω = − z

or, in scalar form,

= −

where 0 is the coe ffi cient of viscous friction .

2 An interesting exception are magnetic bearings where the axel is levitated by magnetic fields so that there is no mechanicalcontact.




Sign Convention for Torque

Suppose the axis of rotation is along the axis. Recall that the definition of torque is

τ , r× F = sin()z = z

where is the angle between r and F. The magnitude of the cross product r× F is sin()

and the direction of r× F is perpendicular to both r and F along the axis. In engineeringapplications, the systems are designed so that the applied force is tangential to the rotationalmotion, i.e., = 2, = and

τ , z = z

If = 0 then the torque will cause the cylinder to rotate around the axis inthe direction indicated by the curved arrow. On the other hand, if = 0 then thetorque will cause the cylinder to rotate around the axis in the direction opposite to thatindicated by the curved arrow. Typically in engineering texts, the sign convention for torqueis indicated by a curved arrow as shown in Figure 1.5. (Physics texts prefer to write τ , z)

r

F

z

FIGURE 1.5. Sign convention for torque.

Example 1 Rack and Pinion System A rack and pinion system is illustrated in Figure 1.6 which is used to convert rotary motion

to linear motion and vice-versa. In Figure 1.6 the axis of the pinion (gear wheel) is consideredto be fixed in space. A torque applied to the shaft causes the pinion to rotate and to movethe rack in the direction through the contact of their teeth.

FIGURE 1.6. Rack and Pinion system (From System Dynamics by W. J. Palm III [1])




We take the input to be the torque (produced by a motor) and the output to be theposition . Because the teeth on the pinion wheel and rack are meshed together, there is analgebraic relationship between the angle and position given by = is the mass of the rack and is the inertia of the pinion gear wheel. Let be the force of the pinion toothon the rack tooth in the direction so that − is the reaction force of the rack tooth on thepinion tooth. Applying Newton’s law to the mass of the rack we have

=

Applying Newton’s law for rotational motion to the pinion gear we have

= −

where − is the reaction torque on the pinion gear produced by the rack. Multiply thefirst equation by and add to the second to obtain

+ =

In this equation we eliminate using the algebraic constraint = and rearrange it toobtain

( + 2) =

Computing the Laplace transform, the transfer function is then

()

()=

+ 2

1

2

Conservation of Energy

Let’s redo the above using conservation of energy. The kinetic energy of the rack andpinion system is

=1

2

2

+1

22 =

1

2 22 +

1

22

=1

2

¡2 +

¢2

Recalling that rate of work done equals the rate of change of kinetic energy [see equation(1.2)], that is,

=

( )

we have

1

=¡

2 + ¢

or

= ( + 2)

and thus ()

()=

+ 2

1

2

as before.




Example 2 Rack and Pinion System Connected to a Spring

Figure 1.7 shows a rack and pinion system with the rack connected to the wall through aspring and also some viscous friction damping between the rack and a table top. We considerthe torque (produced by a motor) to be the input and we take the position to be theoutput. is the radius of the pinion (gear), is the moment of inertia of the pinion about itscenter, is the moment of inertia of the motor, is the viscous friction coefficient between

the rack and the surface it is on, is the spring constant and is the mass of the rack.

FIGURE 1.7. Rack and Pinion system (From System Dynamics by W. J. Palm III [1])

The input is the torque and the position is the output. Let’s first find the transferfunction using Newton’s laws.

The equations modeling this system are then

( + ) = − = − −

Also, we have the algebraic constraint

=

Let be the force of the pinion wheel tooth on the rack gear tooth. Then the reaction forceof the rack gear tooth on the pinion wheel is − resulting in a reaction torque on the pinionwheel of − . is not known and typically not measurable. Consequently, we eliminate from these two equations to obtain

( + ) + = − ( + )

or( + ) + = −−

where we used the fact that = implies = and = . Rearranging this becomes

( + + 2) + 2 + 2 =

Taking the Laplace transform with zero initial conditions gives

( + + 2)2 () + 2 () + 2 () = ()




The transfer function is then

() =

( + + 2)2 + 2 + 2 ()

Equations of Motion Using Conservation of Energy

Let’s now find the equations of motion using conservation of energy. If the external torque

rotates the system by , the work done on the system is . By conservation of energy, this is equal to the change in the kinetic energy of + and plus the change inthe spring’s potential energy plus the heat dissipated due to the viscous friction. In terms of the input power = put into the system by , we have = = and

|{z} Input Mechanical Power

=

⎛⎜⎜⎝

1

2( + )

2

+1

22 | {z }

Kinetic Energy

+1

22 |{z}

Potential Energy

⎞⎟⎟⎠+ () | {z }

Power Dissipated as Heat

= ( + ) + + + 2

Using the algebraic constraint = this becomes

()

= ( + ) () () + + + 2

Multiplying through by 2 and canceling the common factor we have

= ( + + 2) + 2 + 2

which gives the same transfer function as before.

Example 3 Rotational Spring-Mass-Damper System

Figure 1.8 () shows a rotational fluid system used to damp out angular oscillations of apulley wheel due to a not completely rigid crankshaft.

FIGURE 1.8. Rotational Spring-Mass-Damper System (From System Dynamics by W. J. Palm III[1])

In this example, the crankshaft angle is the input and angular position of the pulleywheel is the output. The cylinder inside the pulley wheel has inertia is surrounded by a




fluid and has the eff ect of damping out vibrations of the crankshaft. A rotational spring-mass-damper system shown in 1.8 () is used to model the system. We want to find the transferfunction form to . Applying Newton’s law for rotational motion to the two inertias and , we obtain

2

2= (− )− ( − )

22

= ( − )

Taking the Laplace transform (zero initial conditions) gives¡ 2 + +

¢ () = () + ()¡

2 + ¢

() = ()

Eliminating () gives

¡ 2 + + ¢ () = () +

2

2

2 + () = () +

2

+ ()

Multiply through by + ¡ 2 + +

¢( + ) () = ( + ) () + 2 ()

and then solve for ()

()we have

()

()=

( + )

( 2 + + ) ( + )− 2

= ( + )

3 + ( + ) 2 + +




1.3 Gears

This presentation is from that given on page 296 of K. Ogata’s book [2]. Using the elementaryrigid body dynamics developed previously, the model of a two gear system illustrated inFigure 1.9 below is now developed.

1

1

,2

2

1

1r

2r

2r

xF ˆ11F

xF ˆ22

F

x

y

zGear 1

Gear 2

,2

2

1

zn ˆˆ

m

m

L

L

FIGURE 1.9. Two gear system. Drawn by Sharon Katz.

In Figure 1.9, 1 is the torque exerted on gear 1 by gear 2

1 is the force exerted on gear 1 by gear 2 2 is the torque exerted on gear 2 by gear 1 2 is the force exerted on gear 2 by gear 11 is the angle rotated by gear 12 is the angle rotated by gear 21 is the number of teeth on gear 12 is the number of teeth on gear 21 is the radius of gear 12 is the radius of gear 2

Let F1 , 1(−x) so that if 1 0, the force is in the −x direction as shown in Figure 1.9.

Also, let r1 , 1(−y) so that τ 1 = r1 × F1 = 1 1(−y) × (−x) = 1 1(−z) = 1 1n. That

is, τ 1 = 1n where 1 = 1 1 and n ,− z is a unit vector. Similarly, let F2 , 2x so thatif 2 0, the force is in the x direction. Writing r2 , 2y it follows that τ 2 = r2 × F2 =2 2(−z) = − 2z = 2n with 2 = 2 2. The reason that r1 τ 1 F1 are referred to the basis

vectors −x−y−z while r2 τ 2 F2 are referred to the basis vectors x y z is so that therewill no minus signs in the gear relationships to be derived below.




1.3.1 Algebraic Relationships Between Two Gears

There are three important algebraic relationships between the gears.

1. The gears have diff erent radii, but the teeth on each gear are the same size so that

they will mesh together properly. Consequently, the number of teeth on the surfaceof each gear is proportional to the radius of each gear. For example, if 2 = 21 then2 = 21. In general,

21

=2

1

.

2. By Newton’s third law, the forces F1 = 1(−x) F2 , 2x are equal in magnitude, butopposite in direction so that 2 = − (− 1) = 1. Thus, as 1 = 1 1 and 2 = 2 2 it

follows that 2 1

=21

3. As the teeth on each gear are meshed together at the point of contact, the distancetraveled along the surface of the gears is the same. In other words, 11 = 22 or

2

1=

1

2

The first two algebraic relationships can be summarized as

2 1

=21

=2

1

and these ratios are easily remembered by thinking of gear 2 as larger in radius than gear 1.

Then the number of teeth on gear 2 must also be larger (because its circumference is larger)and the torque on gear 2 is also larger (because its radius is larger).The last algebraic relationship is summarized as

12

=21

=2

1

but it is more easily remembered by writing 11 = 22 which just states the distancetraveled along the surface of each gear is the same as they are meshed together.




1.3.2 Dynamic Relationships Between Two Gears

Consider the two gear system shown in Figure 1.10 below. The motor torque acts on gear1 and the torque is a load torque acting on gear 2.

2n

L

z

,2

2

1 f

2 f

1

1, m

2 J

Gear 1

Gear 2

1n

1 J

zn ˆˆ

FIGURE 1.10. Dynamic equations for a two gear system. Drawn by Sharon Katz.

In Figure 1.10, 1 is the moment of inertia of the motor shaft. 2 is the moment of inertia of the output shaft. 1 is the viscous friction coefficient of the motor shaft. 2 is the viscous friction coefficient of the output shaft.1 is the angle rotated by gear 12 is the angle rotated by gear 21 is the angular speed of gear 12 is the angular speed of gear 2 1 is the torque exerted on gear 1 by gear 2

2 is the torque exerted on gear 2 by gear 1

The sign conventions for the torques 1 2 are indicated in Figure 1.10. In particu-lar, if 0 1 0 then they oppose each other and similarly, if 2 0 0 then thesetwo torques oppose each other. A load torque is illustrated in Figure 1.11 in which the loadtorque on gear 2 is = 2 with 2 the radius of the pick up reel (gear 2).

Motor mg

L

2r

1r

m

Gear 1

Gear 2

FIGURE 1.11. Illustration of load torque. Drawn by Sharon Katz.

The above development is now put together to write down diff erential equations that




characterize the dynamic behavior of the gears. Recall that the fundamental equation of rigid body dynamics is given by

=

where is the total torque on the rigid body, is the moment of inertia of the rigid body and is its angular acceleration about the fixed axis of rotation. Applying this relationship,

the equations of motion for the two gears are then

− 1 − 11 = 11

(1.3)

2 − − 22 = 22

Typically, the input (motor) torque is known and further the output position 2 andspeed 2 are measured. Consequently, the variables 1 2 1 need to be eliminated which isdone as follows:

2 =2

1 1 =

2

1

µ − 11 − 11

¶

=2

1

µ − 1

µ2

1

2

¶− 1

µ2

1

2

¶¶

=2

1

−

µ2

1

¶2

12 −

µ2

1

¶2

12

(1.4)

Substituting this expression for 2 into the second equation of (1.3) results in

2

1

−

µ2

1

¶2

12 −

µ2

1

¶2

12

− − 22 = 2

2

Rearranging, the desired result is2

1

=³

2 + (21)2 1

´ | {z }

2

+³

2 + (21)2 1

´ | {z }

2 + (1.5)

Let = 21 denote the gear ratio, , 2 + 2 1 denote the total inertia re fl ected to the

output shaft and , 2 + 2 1 denote the total viscous friction coefficient re fl ected to the

output shaft , equation (1.5) can be written succinctly as

= 2

+ 2 + (1.6)

The net eff ect of the gears is to increase the motor torque from on the motor shaft to on the output shaft, to add the quantity 2 1 to the inertia of the output shaft and toadd 2 1 to the viscous friction coefficient of the output shaft.Remark

Everything could have been referred to the motor shaft instead of the output (load) shaft.To do so, simply substitute 2 = (12)1 into (1.5) to obtain

2

1

=

Ã 2 +

µ2

1

¶2

1

!

µ1

2

1

¶

+

Ã 2 +

µ2

1

¶2

1

!µ1

2

1

¶+




Multiply both sides by 12 results in

=

Ã 2 +

µ2

1

¶2

1

!µ1

2

¶2 1

+

Ã 2 +

µ2

1

¶2

1

!µ2

1

¶2

1 +1

2

or, finally, the desired form is

=

Ãµ1

2

¶2

2 + 1

!1

+

Ãµ1

2

¶2

2 + 1

!1 +

1

2

(1.7)

In this formulation, the load torque on the input shaft is reduced by 12 from that on theoutput shaft, and (12)2 2 has been added to the inertia of the motor shaft and (12)2 2has been added to the viscous friction coefficient of the motor shaft.




1.4 Rolling Cylinder

We now develop the equations of motion for a cylinder rolling on a surface without slipping.

Rolling on a Flat Surface Without Slipping

Consider the cylinder in Figure 1.12 which has mass and moment of inertia about itsaxis of rotation ( direction in the figure). The center of mass of the cylinder is on the axisof rotation in the midpoint of the cylinder. To say that the cylinder rolls without slippingmeans that

=

where is the angle the cylinder has rotated and is the distance the cylinder has rotatedalong the surface. The surface provides an upward normal force to cancel the downwardforce of gravity.

r

vcm

cmv

cmQ vv 2

0Pv

P

Q

r

r x

x

y

N

mg

FIGURE 1.12. Cylinder rolling on a flat surface with no slip.

As Figure 1.12 indicates, the velocity of the center of mass of the cylinder is denoted as

and

=

The cylinder rotates about the center of mass with angular rate

=

=

()

=

Note that at the point of contact of the cylinder with the surface, the speed of the partof the cylinder in contact with the surface is zero. There are no forces or torques on thecylinder.

=1

22 +

1

22 =

1

22 +

1

2

22 =

1

2

µ +

2

¶2

Cylinder Rolling Down an Inclined Plane

Figure 1.13 shows a cylinder rolling down an inclined plane. The cylinder has mass andmoment of inertia . The direction is taken to be positive going up the inclined planeand the direction is perpendicular to the inclined plane. There is a component of gravity sin() that is pushing the cylinder in the − direction down the inclined plane. Due tofriction (think of it as a rack and pinion) there is a force produced on the wheel at the




point of contact with the inclined plane (and by Newton’s 3rd law a force − produced bythe wheel on the the inclined plane). It is this force that produces the torque to turn thewheel. As illustrated in Figure 1.14, we can think of this force as the same as that in a rackand pinion system. Here the rack is the surface of the incline, the pinion is the cylinder andtheir meshed teeth model the surfaces interacting. This is not viscous friction and so there isno energy loss. (Think of viscous friction as two bodies rubbing/slipping against each other.)

r

)sin( mg

mgf F

x

y

)cos( mg

N Force on wheeldue to frictionbetween wheeland inclined plane.

FIGURE 1.13. Cylinder rolling down an incline under the influence of gravity.

f F

FIGURE 1.14. Modeling the interaction of the surface of the incline and the surface of the cylinderas a rack and pinion system.

We assume the cylinder rolls without slip so that

=

=

where in this case 0 as it is rolling down the inclined plane so that is decreasing.Equations of Motion using Newton’s Laws

Newton’s three laws of motion are valid with respect to non accelerating coordinate sys-tems. The rotational law = was derived from the three laws so it also valid in nonaccelerating coordinate systems. However, it also turns out that = still holds in anaccelerating coordinate system if the axis of rotation is through the center of mass .

The cylinder is not moving in the direction as there is a normal force = cos()in the + direction that cancels out the gravity component cos() in the − direction.




There is a friction force at the point of contact of the cylinder with the inclined plane.

2

2= − cos() + = 0

2

2= − sin() +

2

2= −

=

Note that the third equation is written from the point of view of being on the axis of rotationwhich is accelerating down as the cylinder rolls down the incline. However, as the axis goesthrough the center of mass of the cylinder, = is still valid in this case.

Multiplying the second equation by and adding to the third equation we obtain

2

2+

2

2= − sin()

Using = (no slip condition) we have

2

2+

2

2= − sin()

or2

2= −

2

2 + sin() | {z }

constant

Equations of Motion Derived from Conservation of Energy

The kinetic energy of the cylinder is given by

=1

22 +

1

22 =

1

22 +

1

2

22 =

1

2

µ +

2

¶2

If the cylinder is at the position along the inclined plane, then it is at a height sin()above the horizontal. Thus the potential energy is

= sin()

so that the total energy of the cylinder is then

+ = 12µ +

2¶ 2 + sin()

The total energy is constant so we have

( + ) =

µ +

2

¶

+ sin() = 0

Thus2

2=

= −2

2 + sin()




Motorized Cylinder Going Up an Inclined Plane

Suppose the cylinder has a motor inside to produce a torque . Due to friction (think of it as a rack and pinion) there is a force produced on the wheel at the point of contactwith the inclined plane (and by Newton’s 3rd law a force − produced by the wheel on theinclined plane). It is this force that propels the cylinder up the inclined plane.

)sin( mg

f F

m ,

Reaction force on

wheel due tofriction between

wheel and inclinedplane.

x

y

r

)cos( mg

N

mg

FIGURE 1.15. Cylinder going up an incline using a motor.

Equations of Motion from Newton’s Laws

2

2= − cos() + = 0

2

2= − sin() +

2

2= −

=

Multiplying the second equation by and adding to the third equation we obtain

2

2+

2

2= − sin()

Using the no slip condition we have

2

2+

2

2= − sin()

or2

2

=

2

+

−

2

+

sin()

Equations of Motion Derived from Conservation of Energy

The kinetic plus potential energy is

+ =1

2

µ +

2

¶2 + sin()

The rate of change of the cylinder’s energy is equal to the mechanical power put intothe cylinder by the motor, i.e.,

( + ) =

µ +

2

¶

+ sin() = =




Thus

2

2=

=

µ +

2

¶ −

+

2

sin()

=

2 + −

2

2 + sin()




Problem 1 Wind Up Cable

In Figure 1.16 below, a shaft with a steel drum of total inertia is used to wind up acable in order to raise a mass . The back end of the shaft is connected to a wall througha rotary damper with damping coefficient . The input to the shaft is the torque andalso the weight produces a disturbance (input) torque on the shaft as well. The outputis taken to be the angular position of the shaft which has the same sign convention as .

Ignore the mass of the cable.

FIGURE 1.16.

(a) Find the equations of motion of the shaft.

(b) Compute the transfer function.

Problem 2 Modeling a Gear System Figure 1.17 shows a gear system for an elevator car. The input shaft has a moment of

inertia is 1 and a torque (from a motor) 1 applied to it. The output shaft has moment of inertia 2 and has a pulley attached to it used to raise and lower an elevator car with mass and a counter weight of mass . The pulley has radius and the position of the elevatorcar is denoted by . The algebraic gear relationships are

1

2

=12

= =2

1

=21

= 2 1




pr pulley radius

z

1

2

1

2

1

2

2

1

n

n

r

r N

LT

M m

FIGURE 1.17. Gear system for an elevator [1].

(a) Write down Newton’s equation for rotational motion for the input (first) shaft.

(b) Write down Newton’s equation for rotational motion for the output (second) shaft. Note

that both the car weight and the counterweight produce a load torque on the outputshaft. Your answer should have this load torque written down explicitly in terms of and .

(c) Using your answers in parts () and (), give a single diff erential equation for 2 withinput 1. That is, eliminate 1 2 1 1 from your equations.

(d) Assume there is no slip between the pulley and the cable (i.e., = 2) and that = 0means the elevator car is on the ground floor. Using your answer in part (c), writedown the diff erential equation for the elevator car position with the motor torque 1as input.




1.5 References

[1] William J. Palm III, System Dynamics, Second Edition , McGraw-Hill, 2010.

[2] K. Ogata, Modern Control Engineering , Prentice-Hall, Englewood Cliff s, NJ, 2002.

ho4_rigid body rotational dynamics_09!30!11

Documents