DYNAMICAL TORQUE AND STATIC EQUILIBRIUM
Learning Objectives
After you complete the homework associated with this
lecture, you should be able to:
• Explain what torque is, and give examples that show it is
related to force but differs from it.
• Describe how torque changes rotational motion in a way
analogous to how force changes linear motion.
• Explain how the sign and magnitudes of torque
components are determined in rotational situations.
• Calculate changes in motion of objects due to torques.
• Determine the forces that act on a static structure if it
does NOT change its translational or rotational state.
Cause of Rotation
Consider two equal but opposite forces acting on a stick
as shown
Y 3P
F ext = 0
We see
C
3P
F ext = m P
a CM
C
but object will begin to rotate about its CM
C
This rotation is caused by TORQUE (symbol P
τ)
Y
Pa CM = 0 in this case
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Angular Momentum
Torque Causes Change in Rotational State
C
Linear momentum P
P is the fundamental descriptor of
translational motion. It is changed by forces.
C
Angular momentum P
L is the fundamental descriptor of
rotational motion. It is changed by torques.
Angular Momentum of a Rigid Body
The total angular momentum of a rigid body (all components
P ) rotating about a symmetry
have same angular velocity ω
Changes in rotational motion are
caused by the torque P
τ vector. It is
not a force, but is related to force.
Torque is a vector, and thus has a direction. This
comes from fundamental definition of torque as the
cross product of position vector P
r with force vector P
F:
Pτ = Pr × P
F
L
axis is:
PL = I
axis
ω
P
PL is in the same direction as angular velocity ωP .
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We get its direction from the right-hand-thumb rule,
and its magnitude from
τ=|P
r ×P
F | = r F sin(θ)
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TORQUE AND ANGULAR MOMENTUM
C The time rate of change of a system’s total angular
momentum VECTOR equals the net external torque
VECTOR acting on it.
P
dL
' Pτnet ext ' j Pτext
dt
Torques produce change in angular momentum in a
way analogous to how forces change linear momentum!
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LINE OF ACTION AND MOMENT ARM
The torque due to force P
F is a function of P
F and the
position vector P
r where is P
F applied about point O.
Important
Important:
rz is called the
moment arm
τ
Angular Acceleration of Rigid Bodies
Consider an rigid object that can rotate about an
axis of symmetry, which we will label the z-axis. It has
rotational inertia I about this axis. We have
Pτnet '
P
dL
dt
PL = I ωP
and
If Iz is constant, taking z-component of these gives us
τnet , z '
d (Lz )
dt
'
d (I ωz )
dt
' I
d ωz
dt
3τz = I αz
<==> 3 Fx = M ax
rotational
linear
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Choosing Positive Direction of Rotational Components
1) In your diagram, put a dot
at the point about which
you will calculate
rotational quantities. If the
object is rotating, pick the
axis of rotation.
2) Draw a curved arrow (labelled z) around this
point to show the sense of a positive rotation
= |P
τ | = magnitude of torque = r F sin(θ)
= r [ F sin(θ) ] = r Fz = F [ r sin(θ) ] = F rz
This defines direction of z-axis and you can use it to
determine signs of z-components.
DEMO : wrench turns bolt
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Survival Methods for Determining τz
•
•
If a force tends to produce a positive rotation (i.e., in the
direction of the curved z-axis arrow), then the
corresponding torque’s z-component τz is positive.
If a force tends to produce a negative rotation (i.e., in the
direction of the opposite the curved arrow), then the
corresponding torque’s z-component τz is negative.
Example: A disk of mass M and radius R rotates on a
frictionless axis through its center. A block of mass m is
attached to a string that is wound around the disk’s inner
core of radius r. What is the linear acceleration of the
block?
τ1z = + F1 r1z = + F1 d1
= + F1z r1 (positive)
τ2z = S F2 r2z = S F2 d2
= S F2z r2
(negative)
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3τz = I αz
Form free-body of block
where I = ½MR2
3Fx = (ST) + (+ mg) = m ax
τT,z + V
τ S,z + V
τ Fg,z = I αz
one equation with
2 unknowns (T and ax )
(+ r T) + 0 + 0 = I αz
Get torques from extended free-body diagram
of the disk
Because the disk does
not translate, we could get the
magnitude of the support force P
S
from 3P
F = MP
a = 0. But why bother?
F g are both zero,
Since the moment arms of P
S &P
they produce zero torques and thus do not affect
the rotational dynamics!
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But
ˆ
r αz = ax (rolling without slipping)
r T = Iz [ ax / r]
r 2 T = Iz a x
multiplying by r:
r2 @ [force-eq on block]:
(S r2 T) + r2 m g = r2 m ax
adding to get rid of T: r2 m g = [r2 m + I ] ax
Then solve:
ax =
m r2 g
[mr2 + 12 MR2 ]
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STATIC EQUILIBRIUM
An object will not change its rotation if
Y
IP
α = 3 Pτ o = 0
Y
Pα = 0
Y
Example of an odd-shaped beam
3P
τo = 0
L
no rotation change
T?
Thus, for motionless object to remain motionless:
3P
F ext = 0
(no translation)
odd
beam
and
3P
τo = 0
Y
3 (τo)z = 0
(no rotation)
M
CAUTION : Remember to EXPLICITLY indicate a ref
point and draw a labeled curved arrow around it
for the choice of positive rotation.
Demo
Demo:
m
θ
free
pivot
support
D
A ball of mass M is
suspended from an
oddly shaped beam
of mass m. The
beam can freely
rotate about a pivot
support.
What is tension T in the cable?
What (reactance) support force P
S does the pivot supply?
Static Equilibrium of a Simple Beam
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If we just want T, we could get it directly
from torque eq because P
S produces zero
torque about O. But let us not see it for
the moment, and do the "drill". Note the
tension force P
T lies along the strut, but
support force P
S has unknown direction.
Vx + V
w x + Sx = 0
3Fx = Tx + W
= (+T cos θ) + Sx = 0
Hint
Hint: If you choose an origin O that lies along the line
of action of a force, that force’s moment arm is
zero. Thus it will not contribute the sum of
torques, which simplifies the algebra.
3 Fy = Ty + Wy + wy + Sy = 0
= +T sin θ + (SMg) + (Smg) + Sy = 0
2 equations in 3 Unknowns: T , Sx , Sy Y look for another equation
τ Sz = (ST rz) + ( + MgD) + ( + mg[½L]) = 0
3τOz = τTz + τWz + τwz + V
= S T (L sin θ) + M g D + ½ m g L = 0
T = [ M g D + ½ m g L ] / [L sin θ ] , then go above to get Sx and Sy
note: τTz = ST rz = STz L = S (T sin θ) L (both the same)
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