Transcript Lecture-15-10

Lecture 15
Rotational Dynamics
Moment of Inertia
The moment of inertia I:
The total kinetic energy of a
rolling object is the sum of
its linear and rotational
kinetic energies:
Torque
We know that the same force will be much more
effective at rotating an object such as a nut or a
door if our hand is not too close to the axis.
This is why we have
long-handled wrenches,
and why doorknobs are
not next to hinges.
The torque increases as the force increases,
and also as the distance increases.
Only the tangential component of force
causes a torque
A more general definition of torque:
Fsinθ
Fcosθ
You can think of this as either:
- the projection of force on to the tangential direction
OR
- the perpendicular distance from the axis of rotation to line of the force
Torque
If the torque causes a counterclockwise angular
acceleration, it is positive; if it causes a clockwise
angular acceleration, it is negative.
Using a Wrench
You are using a wrench to
tighten a rusty nut. Which
a
b
arrangement will be the
most effective in tightening
the nut?
c
d
e) all are equally effective
Using a Wrench
You are using a wrench to
tighten a rusty nut. Which
a
b
arrangement will be the
most effective in tightening
the nut?
Because the forces are all the
same, the only difference
is the lever arm. The
arrangement with the largest
lever arm (case b) will provide
the largest torque.
c
d
e) all are equally effective
The gardening tool shown is used to pull weeds.
If a 1.23 N-m torque is required to pull a given
weed, what force did the weed exert on the tool?
What force was used
on the tool?
Force and Angular Acceleration
Consider a mass m rotating around an axis a
distance r away.
Newton’s second law:
a=rα
Or equivalently,
Torque and Angular Acceleration
Once again, we have analogies between linear and
angular motion:
The L-shaped object shown below consists of
three masses connected by light rods. What
torque must be applied to this object to give it an
angular acceleration of 1.2 rad/s2 if it is rotated
about
(a) the x axis,
(b) the y axis
(c) the z axis (through the origin and
(a)
perpendicular to the page)
(b)
(c)
The L-shaped object shown below consists of three
masses connected by light rods. What torque must
be applied to this object to give it an angular
acceleration of 1.2 rad/s2 if it is rotated about
an axis parallel to the y axis, and through the 2.5kg
mass?
Dumbbell I
A force is applied to a dumbbell
for a certain period of time, first
as in (a) and then as in (b). In
which case does the dumbbell
acquire the greater
center-of-mass speed ?
a) case (a)
b) case (b)
c) no difference
d) it depends on the rotational
inertia of the dumbbell
Dumbbell I
A force is applied to a dumbbell
for a certain period of time, first
as in (a) and then as in (b). In
which case does the dumbbell
acquire the greater
center-of-mass speed ?
Because the same force acts for the
same time interval in both cases, the
change in momentum must be the
same, thus the CM velocity must be
the same.
a) case (a)
b) case (b)
c) no difference
d) it depends on the rotational
inertia of the dumbbell
Static Equilibrium
Static equilibrium describes an object at rest –
neither rotating nor translating.
If the net torque is zero, it
doesn’t matter which axis we
consider rotation to be around;
you choose the axis of rotation
This can greatly simplify a problem
X
Center of Mass and
Gravitational Force on
an Extended Object axis of
m1
...
mj
X
xj
rotation
Fj = m j g
center of mass
m1 ... mj
xj
X
So, forget about the weight of all the
individual pieces. The net torque will be
equivalent to the total weight of the
object, pulling from the center of mass of
the object
axis of
rotation
xcm
F = Mg
Balance
If an extended object is to be balanced, it must be
supported through its center of mass.
Center of Mass and Balance
This fact can be used to find the center of mass of
an object – suspend it from different axes and trace
a vertical line. The center of mass is where the lines
meet.
Balancing Rod
A 1-kg ball is hung at the end of a rod
1-m long. If the system balances at a
a) ¼ kg
b) ½ kg
point on the rod 0.25 m from the end
c) 1 kg
holding the mass, what is the mass of
d) 2 kg
the rod?
e) 4 kg
1m
1kg
Balancing Rod
A 1-kg ball is hung at the end of a rod
1-m long. If the system balances at a
a) ¼ kg
b) ½ kg
point on the rod 0.25 m from the end
c) 1 kg
holding the mass, what is the mass of
d) 2 kg
the rod?
e) 4 kg
The total torque about the pivot
must be zero !!
The CM of the
same distance
rod is at its center, 0.25 m to the
right of the pivot. Because this
must balance the ball, which is
the same distance to the left of
the pivot, the masses must be
the same !!
mROD = 1 kg
X
1 kg
CM of rod
When you arrive at Duke’s Dude Ranch, you are greeted by the large
wooden sign shown below. The left end of the sign is held in place by a bolt,
the right end is tied to a rope that makes an angle of 20.0° with the
horizontal. If the sign is uniform, 3.20 m long, and has a mass of 16.0 kg,
what is
(a) the tension in the rope, and
(b) the horizontal and vertical components of the force, exerted by the
bolt?
When you arrive at Duke’s Dude Ranch, you are greeted by the large
wooden sign shown below. The left end of the sign is held in place by a bolt,
the right end is tied to a rope that makes an angle of 20.0° with the
horizontal. If the sign is uniform, 3.20 m long, and has a mass of 16.0 kg,
what is
(a) the tension in the rope, and
(b) the horizontal and vertical components of the force exerted by the
Torque, vertical force, and horizontal force are all zero
bolt?
But I don’t know two of the forces!
I can get rid of one of them, by choosing my axis of
rotation where the force is applied.
Choose the bolt as the axis of rotation, then:
(b)
F = ma implies Newton’s first law:
without a force, there is no acceleration
Now we have
Linear momentum was the concept that tied together Newton’s Laws,
is there something similar for rotational motion?
Angular and linear acceleration
1 2
I  ML
3
MgL
 MgL 2
3g
   Mg   L 2  cos 
  

2
2
I
ML 3
2L
Wood:
1 2
2 0
4  L 0 
 2L 
t fall :   t fall   0   0   t fall  t fall  
 2 0    
 g 
2

3
g
3




Ball:
1 2
t fall : y  t fall   0  y0  gt fall and y0  L sin  L
2
1 2
 L 0 
 0  L 0  gt fall  t fall  2 
 t fall  Wood!

2
 g 
Angular Momentum
Consider a particle moving in a circle
of radius r,
I = mr2
L = Iω = mr2ω = rm(rω)
= rmvt = rpt
Angular Momentum
For more general motion (not necessarily circular),
The tangential component of the
momentum, times the distance
Angular Momentum
For an object of constant moment of inertia,
consider the rate of change of angular momentum
analogous to 2nd Law
for Linear Motion
Conservation of Angular Momentum
If the net external torque on a system is zero, the
angular momentum is conserved.
As the moment of inertia decreases, the
angular speed increases, so the angular
momentum does not change.
Figure Skater
A figure skater spins with her arms
a) the same
extended. When she pulls in her arms,
b) larger because she’s rotating
she reduces her rotational inertia
faster
and spins faster so that her angular
momentum is conserved. Compared
c) smaller because her rotational
inertia is smaller
to her initial rotational kinetic energy,
her rotational kinetic energy after she
pulls in her arms must be:
Figure Skater
A figure skater spins with her arms
a) the same
extended. When she pulls in her arms,
b) larger because she’s rotating
she reduces her rotational inertia
faster
and spins faster so that her angular
momentum is conserved. Compared
c) smaller because her rotational
inertia is smaller
to her initial rotational kinetic energy,
her rotational kinetic energy after she
pulls in her arms must be:
KErot = I 2 = L  (used L = I ).
Because L is conserved, larger 
means larger KErot.
Where does the “extra” energy come from?
KErot = I 2 = L 2 (used L = I ).
Because L is conserved, larger 
means larger KErot.
Where does the “extra” energy come from?
As her hands come in, the velocity of
her arms is not only tangential... but
also radial.
So the arms are accelerated inward, and
the force required times the Δr does the
work to raise the kinetic energy
Conservation of Angular Momentum
Angular momentum is also conserved in rotational
collisions
larger I, same total angular
momentum, smaller angular velocity
Rotational Work
A torque acting through an angular
displacement does work, just as a force acting
through a distance does.
Consider a tangential force on
a mass in circular motion:
τ=rF
Work is force times the distance on the arc:
W=sF
s = r Δθ
W = (r Δθ) F = rF Δθ = τ Δθ
The work-energy theorem applies as usual.
Rotational Work and Power
Power is the rate at which work is done, for
rotational motion as well as for translational
motion.
Again, note the analogy to the linear form (for
constant force along motion):
Dumbbell II
A force is applied to a dumbbell
for a certain period of time, first
as in (a) and then as in (b). In
which case does the dumbbell
acquire the greater energy ?
a) case (a)
b) case (b)
c) no difference
d) it depends on the rotational
inertia of the dumbbell
Dumbbell II
A force is applied to a dumbbell
for a certain period of time, first
as in (a) and then as in (b). In
which case does the dumbbell
acquire the greater energy ?
If the CM velocities are the same, the
translational kinetic energies must be
the same. Because dumbbell (b) is
also rotating, it has rotational kinetic
energy in addition.
a) case (a)
b) case (b)
c) no difference
d) it depends on the rotational
inertia of the dumbbell
A 2.85-kg bucket is attached to a disk-shaped pulley of radius 0.121 m
and mass 0.742 kg. If the bucket is allowed to fall,
(a) what is its linear acceleration?
(b) What is the angular acceleration of the pulley?
(c) How far does the bucket drop in 1.50 s?
A 2.85-kg bucket is attached to a disk-shaped pulley of radius 0.121 m
and mass 0.742 kg. If the bucket is allowed to fall,
(a) What is its linear acceleration?
(b) What is the angular acceleration of the pulley?
(c) How far does the bucket drop in 1.50 s?
(a) Pulley spins as bucket falls
(b)
(c)
The Vector Nature of Rotational Motion
The direction of the angular velocity vector is along
the axis of rotation. A right-hand rule gives the sign.
Right-hand Rule:
your fingers should
follow the velocity
vector around the
circle
Optional material
Section 11.9
The Torque Vector
Similarly, the right-hand rule gives the direction
of the torque vector, which also lies along the
assumed axis or rotation
Right-hand Rule: point your
RtHand fingers along the force,
then follow it “around”. Thumb
points in direction of torque.
Optional material
Section 11.9
The linear momentum of components related to
the vector angular momentum of the system
Optional material
Section 11.9
Applied tangential force related to the torque vector
Optional material
Section 11.9
Applied torque over time related to change
in the vector angular momentum.
Optional material
Section 11.9
Spinning Bicycle Wheel
You are holding a spinning bicycle
wheel while standing on a
stationary turntable. If you
suddenly flip the wheel over so that
it is spinning in the opposite
direction, the turntable will:
a) remain stationary
b) start to spin in the same
direction as before flipping
c) start to spin in the same
direction as after flipping
What is the torque (from gravity) around the supporting point?
Which direction does it point?
Without the spinning wheel: does this make sense?
With the spinning wheel: how is L changing?
Why does the wheel not fall?
Does this violate Newton’s 2nd?