Rigid Body Dynamicschapter 10 continues

around and around we go …

Rigid Body Rotation

Rotation Axis

222

2222

2

2

1

2

1)(

2

1

2

1)(

2

1)(

IrmKE

rmrmKE

rv

vmKE

iiitotal

iiiii

ii

iii

Moment of Inertia:

2

2

i

iirmI

CORRESPONDENCE:

Im

v

Imv

22

2

1

2

1

RotationalRotational Kinetic Energy

We there have an analogy between the kinetic energies associated with linear motion (K = ½ mv 2) and the kinetic energy associated with rotational motion (KR= ½ I2)

Rotational kinetic energy is not a new type of energy, the form is different because it is applied to a rotating object

The units of rotational kinetic energy are also Joules (J)

Important Concept:Moment of Inertia

The definition of moment of inertia is

The dimensions of moment of inertia are ML2 and its SI units are kg.m2

We can calculate the moment of inertia of an object more easily by assuming it is divided into many small volume elements, each of mass mi

2i i

i

I r m

Moment of Inertia, cont

We can rewrite the expression for I in terms of m

With the small volume segment assumption,

If is constant, the integral can be evaluated with known geometry, otherwise its variation with position must be known

lim 2 20im i ii

I r m r dm

2I r dV

Question – WHAT IS THE MOMENT OF INERTIA OF THIS OBJECT??

Let’s Look at the possibilities

c d

Two balls with masses M and m are connected by a rigid rod of length L and negligible mass as in Figure P10.22. For an axis perpendicular to the rod, show that the system has the minimum moment of inertia when the axis passes through the center of mass. Show that this moment of inertia is I = L2, where = mM/(m + M).

Remember the Various Densities

Volumetric Mass Density –> mass per unit volume: = m / V

Face Mass Density –> mass per unit thickness of a sheet of uniform thickness, t : t

Linear Mass Density –> mass per unit length of a rod of uniform cross-sectional area: = m / L =

Moment of Inertia of a Uniform Thin Hoop

Since this is a thin hoop, all mass elements are the same distance from the center

22

2

I r dm R dm

I MR

Moment of Inertia of a Uniform Rigid Rod

The shaded area has a mass dm = dx

Then the moment of inertia is

/ 22 2

/ 2

21

12

L

L

MI r dm x dx

L

I ML

Moment of Inertia of a Uniform Solid Cylinder

Divide the cylinder into concentric shells with radius r, thickness dr and length L

Then for I

2 2

2

2

1

2z

I r dm r Lr dr

I MR

dV=L(2rdr)

Moments of Inertia of Various Rigid Objects

Parallel-Axis Theorem

In the previous examples, the axis of rotation coincided with the axis of symmetry of the object

For an arbitrary axis, the parallel-axis theorem often simplifies calculations

The theorem states I = ICM + MD 2 I is about any axis parallel to the axis through the center of

mass of the object ICM is about the axis through the center of mass D is the distance from the center of mass axis to the

arbitrary axis

Howcome??

The new axis is parallel to the old axis of rotation.

Assume that the object rotates about an axis parallel to the z axis.

The new axis is parallel to the original axis.

mi

L

ri

Not insameplane

From the top: NEW OLD

mi

Lri

i

i

i

i

rL

rL

rLLL

aa

rL

iioldnew

ii

iinew

ii

inew

iinew

mIMLI

mrmMLI

rmI

anote

mI

2

2

)2(

:

)(

2

21

2

2

2

2

Center of Mass

Remember the Center of Mass?

0

1

CM

ii

iCM mM

r

rr

Since for our problem the sum is ABOUT the center ofmass, rCM must be zero

So:

irLi

ioldnew mIMLI 22

ZE

ROInew = ICM + ML2

Parallel-Axis Theorem Example

The axis of rotation goes through O

The axis through the center of mass is shown

The moment of inertia about the axis through O would be IO = ICM + MD 2

Moment of Inertia for a Rod Rotating Around One End

The moment of inertia of the rod about its center is

D is ½ L Therefore,

21

12CMI ML

2CM

22 21 1

12 2 3

I I MD

LI ML M ML

Many machines employ cams for various purposes, such as opening and closing valves. In Figure P10.29, the cam is a circular disk rotating on a shaft that does not pass through the center of the disk. In the manufacture of the cam, a uniform solid cylinder of radius R is first machined. Then an off-center hole of radius R/2 is drilled, parallel to the axis of the cylinder, and centered at a point a distance R/2 from the center of the cylinder. The cam, of mass M, is then slipped onto the circular shaft and welded into place. What is the kinetic energy of the cam when it is rotating with angular speed about the axis of the

Torque (Another Vector): F

Torque

Torque, , is the tendency of a force to rotate an object about some axis Torque is a vector = r F sin = Fd = rXF

F is the force is the angle the force makes with the horizontal d is the moment arm (or lever arm)

More Torqueing

The moment arm, d, is the perpendicular distance from the axis of rotation to a line drawn along the direction of the force d = r sin Φ

Torque

The horizontal component of F (F cos ) has no tendency to produce a rotation

Torque will have direction If the turning tendency of the force is

counterclockwise, the torque will be positive If the turning tendency is clockwise, the torque will

be negative

Right Hand Screw Rule

Net Torque

The force F1 will tend to cause a counterclockwise rotation about O

The force F2 will tend to cause a clockwise rotation about O

F1d1 – F2d2

Torque vs. Force

Forces can cause a change in linear motion Described by Newton’s Second Law

Forces can cause a change in rotational motion The effectiveness of this change depends on

the force and the moment arm The change in rotational motion depends on

the torque

Torque Units

The SI units of torque are N.m Although torque is a force multiplied by a

distance, it is very different from work and energy

The units for torque are reported in N.m and not changed to Joules

Torque and Angular Acceleration

Consider a particle of mass m rotating in a circle of radius r under the influence of tangential force Ft

The tangential force provides a tangential acceleration: Ft = mat

Ftr=matr=m(r)r=mr2

More Associations:

mI

a

F

maF

I

SO?

Worry about concepts.Don’t worry about too many new

“formulas”.

Torque and Angular Acceleration, Extended Consider the object consists

of an infinite number of mass elements dm of infinitesimal size

Each mass element rotates in a circle about the origin, O

Each mass element has a tangential acceleration

Torque and Angular Acceleration, Extended cont.

From Newton’s Second Law dFt = (dm) at

The torque associated with the force and using the angular acceleration gives d = r dFt = atr dm = r 2 dm

Finding the net torque This becomes

2 2r dm r dm

Torque and Angular Acceleration, Extended final

This is the same relationship that applied to a particle

The result also applies when the forces have radial components The line of action of the radial component must

pass through the axis of rotation These components will produce zero torque about

the axis

Torque and Angular Acceleration, Wheel Example

The wheel is rotating and so we apply The tension supplies the

tangential force The mass is moving in

a straight line, so apply Newton’s Second Law Fy = may = mg - T

Problem

Find the net torque on the wheel in Figure P10.31 about the axle through O if a = 10.0 cm and b = 25.0 cm.

Anudder oneAn electric motor turns a flywheel through a drive belt that joins a pulley on the motor and a pulley that is rigidly attached to the flywheel, as shown in Figure P10.39. The flywheel is a solid disk with a mass of 80.0 kg and a diameter of 1.25 m. It turns on a frictionless axle. Its pulley has much smaller mass and a radius of 0.230 m. If the tension in the upper (taut) segment of the belt is 135 N and the flywheel has a clockwise angular acceleration of 1.67 rad/s2, find the tension in the lower (slack) segment of the belt.

Torque and Angular Acceleration, Multi-body Ex., 1

Both masses move in linear directions, so apply Newton’s Second Law

Both pulleys rotate, so apply the torque equation

Torque and Angular Acceleration, Multi-body Ex., 2

The mg and n forces on each pulley act at the axis of rotation and so supply no torque

Apply the appropriate signs for clockwise and counterclockwise rotations in the torque equations

32. The tires of a 1 500-kg car are 0.600 m in diameter and the coefficients of friction with the road surface are s = 0.800 and k = 0.600. Assuming that the weight is evenly distributed on the four wheels, calculate the maximum torque that can be exerted by the engine on a driving wheel, without spinning the wheel. If you wish, you may assume the car is at rest.

Work in Rotational Motion Find the work done by F on the

object as it rotates through an infinitesimal distance ds = r d

dW = F . d s = (F sin ) r d

dW = d

The radial component of F does no work because it isperpendicular to thedisplacement

Power in Rotational Motion

The rate at which work is being done in a time interval dt is

This is analogous to P = Fv in a linear system

PowerdW d

dt dt

Work-Kinetic Energy Theorem in Rotational Motion

The work-kinetic energy theorem for rotational motion states that the net work done by external forces in rotating a symmetrical rigid object about a fixed axis equals the change in the object’s rotational kinetic energy

2 21 1

2 2

f

if iW I d I I

dIdt

dIdd

dt

dIdIddW

Work-Kinetic Energy Theorem, General

The rotational form can be combined with the linear form which indicates the net work done by external forces on an object is the change in its total kinetic energy, which is the sum of the translational and rotational kinetic energies

Energy in an Atwood Machine, Example

The blocks undergo changes in translational kinetic energy and gravitational potential energy

The pulley undergoes a change in rotational kinetic energy

Summary of Useful Equations

Rolling Object

The red curve shows the path moved by a point on the rim of the object This path is called a cycloid

The green line shows the path of the center of mass of the object

Pure Rolling Motion

In pure rolling motion, an object rolls without slipping

In such a case, there is a simple relationship between its rotational and translational motions

Rolling Object, Center of Mass

The velocity of the center of mass is

The acceleration of the center of mass is

CM

ds dv R R

dt dt

CMCMdv d

a R Rdt dt

Rolling Object, Other Points

A point on the rim, P, rotates to various positions such as Q and P ’

At any instant, the point on the rim located at point P is at rest relative to the surface since no slipping occurs

Rolling Motion Cont.

Rolling motion can be modeled as a combination of pure translational motion and pure rotational motion

Total Kinetic Energy of a Rolling Object

The total kinetic energy of a rolling object is the sum of the translational energy of its center of mass and the rotational kinetic energy about its center of mass

K = ½ ICM w2 + ½ MvCM2

Total Kinetic Energy, Example

Accelerated rolling motion is possible only if friction is present between the sphere and the incline The friction produces the

net torque required for rotation

Total Kinetic Energy, Example cont

Despite the friction, no loss of mechanical energy occurs because the contact point is at rest relative to the surface at any instant

Let U = 0 at the bottom of the plane Kf + U f = Ki + Ui

Kf = ½ (ICM / R 2) vCM2 + ½ MvCM

2

Ui = Mgh

Uf = Ki = 0