Download - Rigid Body Dynamics chapter 10 continues
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