Transcript Lecture 10.2

Welcome back to Physics 215
Today’s agenda:
• Extended objects
• Torque
• Moment of inertia
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Lecture 10-2
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Current homework assignment
• HW8:
– Knight Textbook Ch.12: 8, 54, 58, 60, 62, 64
– Due Monday, Nov. 5th in recitation
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Motion of center of mass of a system:
åF
ext
= MaCM
The center of mass of a system of point objects
moves in the same way as a single object with
the same total mass would move under the
influence of the same net (external) force.
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Throwing an extended object
• Demo: odd-shaped object – 1 point has
simple motion = projectile motion
- center of mass
• Total external force Fext
aCM = Fext /M
• Translational motion of system looks
like all mass is concentrated at CM
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Equilibrium of extended object
• Clearly net force must be zero
• Also, if want object to behave as point at
center of mass  ALL forces acting on
object must pass through CM
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A few properties of the center of mass of
an extended object
• The weight of an entire object can be thought of as being
exerted at a single point, the center of mass.
• One can locate the center of mass of any object by
suspending it from two different points and drawing
vertical lines through the support points.
• Equilibrium can be ensured if all forces pass through CM
• An object at rest on a table does not tip over if the center
of mass is above the area where it is supported.
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Conditions for equilibrium
of an extended object
For an extended object that remains at rest
and does not rotate:
• The net force on the object has to be zero.
Fnet = å Fext = 0
• The net torque on the object has to be zero.
t net = åt = 0
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Preliminary definition of torque:
The torque on an object with respect to a given pivot
point and due to a given force is defined as the product
of the force exerted on the object and the moment
arm. The moment arm is the perpendicular distance
from the pivot point to the line of action of the force.
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Computing torque
F
q
d
O
Physics 215 – Fall 2014
|t| = |F|d
= |F||r|sinq
= (|F| sinq)|r|
r
component of
force at 900 to
position vector
times distance
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Definition of torque:
t =r ´F
where r is the vector from the reference point (generally
either the pivot point or the center of mass) to the point
of application of the force F.
| t |= r F sinq
where q is the angle between the vectors r and F.
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Vector (or “cross”) product of vectors
The vector product is a way to combine two vectors to
obtain a third vector that has some similarities with
multiplying numbers. It is indicated by a cross ()
between the two vectors.
The magnitude of the vector cross product is given by:
A ´ B = ABsinq
The direction of the vector AB is perpendicular to the
plane of vectors A and B and given by the right-hand rule.
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Interpretation of torque
• Measures tendency of any force to cause
rotation
• Torque is defined with respect to some
origin – must talk about “torque of force
about point X”, etc.
• Torques can cause clockwise (+) or
anticlockwise rotation (-) about pivot point
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Extended objects
need extended free-body diagrams
• Point free-body diagrams allow finding net
force since points of application do not matter.
Fnet = å Fext
• Extended free-body diagrams show point of
application for each force and allow finding
net torque.
t net = åt
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A T-shaped board is supported such that
its center of mass is to the right of and
below the pivot point.
Which way will it rotate once
the support is removed?
1.
2.
3.
4.
Clockwise.
Counter-clockwise.
Not at all.
Not sure what will happen.
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Examples of stable and
unstable rotational equilibrium
• Wire walker – no net torque when figure
vertical.
• Small deviations lead to a net restoring
torque  stable
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Restoring torque
Consider displacing
anticlockwise
e.g., wire walker
 tR increases
 tL decreases
net torque causes
clockwise rotation!
dR
dL
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Rotations about fixed axis
• Every particle in body undergoes
circular motion (not necessarily
constant speed) with same time
period
• v = (2pr)/T = w r. Quantity w is
called angular velocity
• Similarly can define angular
acceleration a = Dw/Dt
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Rotational Motion
* Particle i:
|vi| = ri w at 90º to ri
pivot
w
ri
Fi
mi
* Newton’s 2nd law:
miDvi/Dt = FiT  component at 90º to ri
* Substitute for vi and multiply by ri :
miri2Dw/Dt = FiT ri = ti
* Finally, sum over all masses:
Dw/Dt) S miri2 = Sti = tnet
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Discussion
Dw/Dt) S miri2 = tnet
a - angular
acceleration
Moment of inertia, I
Ia = tnet
compare this with Newton’s 2nd law
Ma = F
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Moment of Inertia
N
I = m r + m2 r2 + … + m N rN = å mi ri
2
11
2
2
i=1
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* I must be defined with
respect to a particular axis
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Moment of Inertia of
Continuous Body
Dm a 0
å
N
Þ
I = å mi ri
i=1
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ò
r
Dm
Þ I = ò r dm
2
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Tabulated Results for Moments of Inertia of
some rigid, uniform objects
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(from p.299 of University Physics, Young & Freedman)
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Parallel-Axis Theorem
D
I = ICM + M D
CM
2
D
*Smallest I will always be along axis passing through CM
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Practical Comments on Calculation of
Moment of Inertia for Complex Object
1.
To find I for a complex object, split it into simple
geometrical shapes that can be found in Table 9.2
2.
Use Table 9.2 to get ICM for each part about the axis
parallel to the axis of rotation and going through the
center-of-mass
3.
If needed use parallel-axis theorem to get I for each
part about the axis of rotation
4.
Add up moments of inertia of all parts
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Beam resting on pivot
N
r
CM of beam
r
rm
x
M=?
Vertical equilibrium?
Mb = 2m
SF =
Rotational equilibrium?
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m
St =
M=
N=
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Suppose M replaced by M/2 ?
• vertical equilibrium?
• rotational dynamics?
• net torque?
SF =
St =
• which way rotates?
• initial angular acceleration?
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Moment of Inertia?
I = Smiri2
* depends on pivot position!
I=
* Hence a = t/I =
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Rotational Kinetic Energy
K = Si(1/2)mivi2 = (1/2)w2Simiri2
• Hence
K = (1/2)Iw2
• This is the energy that a rigid body
possesses by virtue of rotation
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Spinning a cylinder
2R
F
Cable wrapped around
cylinder. Pull off with
constant force F. Suppose
unwind a distance d of cable
• What is final angular speed of cylinder?
• Use work-KE theorem
W = Fd = Kf = (1/2)Iw2
• Mom. of inertia of cyl.? -- from table: (1/2)mR2
– from table: (1/2)mR2
w = [2Fd/(mR2/2)]1/2 = [4Fd/(mR2)]1/2
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Reading assignment
• Rotations
• Chapter 12 in textbook
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