Transcript 141S13-NotesCh8b-June17

8.4 Rotational Work and Kinetic Energy
In the last 3 lectures, we have seen that there are numerous
analogues between translational and rotational parameters. We
continue along this line of reasoning with the study of rotational
work. If a force acts in a tangential direction, moving an object
along an arc length s, then the work done is
𝑊 = 𝐹𝑠 = 𝐹 𝑟⫠ 𝜃 = 𝜏𝜃
(as usual, 𝜃 must be expressed in radians). If this work is done
over a time t, then the rotational power during this time is
𝑊 𝜏𝜃
𝑃=
=
= 𝜏𝜔
𝑡
𝑡
In class, we will take a close look at what actually constitutes a
“tangentially-directed force”.
PC141 Intersession 2013
Day 17 – June 17 – WBL 8.4-8.5
Slide 1
8.4 Rotational Work and Kinetic Energy
We know that any work done on a system will change its kinetic
energy. Using the rotational form of Newton’s 2nd law (𝜏 = 𝐼𝛼),
we have
𝑊 = 𝜏𝜃 = 𝐼𝛼𝜃
Assuming that torque (and therefore 𝛼) is constant, the
rotational kinematic equations from chapter 7 tell us that 𝜔2 =
𝜔02 + 2𝛼𝜃, from which we find
𝜔2 − 𝜔02
1 2 1 2
𝑊=𝐼
= 𝐼𝜔 − 𝐼𝜔0
2
2
2
But the work-KE theorem says that 𝑊 = ∆𝐾. This allows us to
write an expression for rotational kinetic energy
1 2
𝐾 = 𝐼𝜔
2
PC141 Intersession 2013
Day 17 – June 17 – WBL 8.4-8.5
Slide 2
8.4 Rotational Work and Kinetic Energy
All that remains to be discussed in this section is the kinetic
energy associated with an object that is rolling. Recall from the
last lecture that rolling is a combination of translation (of the
CM) and rotation (about the CM). The total kinetic energy
involved in rolling is simply the sum of the translational and
rotational kinetic energies; for an object of mass M,
1
1
2
𝐾 = 𝐾𝑡 + 𝐾𝑟 = 𝑀𝑣𝐶𝑀 + 𝐼𝐶𝑀 𝜔2
2
2
Note that this equation is valid for any object that rolls without
slipping (sphere, cylinder, etc.); they would simply have different
𝐼𝐶𝑀 . Often, you will encounter rolling problems in which the
radius of the object isn’t provided. Don’t worry! Remember, 𝐼𝐶𝑀
𝑣𝐶𝑀
2
is proportional to R , but 𝜔 =
. Therefore, the radius cancels
𝑅
out of the K equation entirely.
PC141 Intersession 2013
Day 17 – June 17 – WBL 8.4-8.5
Slide 3
Problem #1: Bowling Ball
WBL LP 8.15
A bowling ball rolls without slipping on a flat surface. The ball has…
A
…rotational kinetic energy
B
…translational kinetic energy
C
…both translational and rotational kinetic energy
D
…neither translational nor rotational kinetic energy
PC141 Intersession 2013
Day 17 – June 17 – WBL 8.4-8.5
Slide 4
Problem #2: Tipped Pencil
WBL EX 8.55
A pencil 18 cm long stands vertically on its point end on a horizontal table. If it
falls over without slipping, with what tangential speed does the eraser end strike
the table?
Solution: In class
PC141 Intersession 2013
Day 17 – June 17 – WBL 8.4-8.5
Slide 5
Problem #3: Loop-the-Loop Revisited
WBL EX 8.63
A steel ball rolls without slipping down an incline into a
loop-the-loop of radius R. On the Day 14 lecture, we
considered a sliding mass in the same situation, and
calculated that it required a minimum speed of 𝒗 = 𝒈𝑹 at
the top of the loop in order to stay on the track. We then
calculated (using energy principles) that the sliding mass
had to be released from a height h = 2.5R in order to achieve this minimum speed.
For the rolling ball, the required minimum speed is the same, but the required height
of release is not. Calculate the new h.
Solution: In class
PC141 Intersession 2013
Day 17 – June 17 – WBL 8.4-8.5
Slide 6
8.5 Angular Momentum
In chapter 6, we introduced linear momentum, p = 𝑚v. A
related quantity is angular momentum, L. For a single particle,
the magnitude of angular momentum is the product of the
moment arm and the linear momentum:
𝐿 = 𝑟⫠ 𝑝 = 𝑟⫠ 𝑚𝑣 = 𝑚𝑟⫠ 𝑣 = 𝑚𝑟⫠2 𝜔
It has SI units of kg·m2/s. For a rotating object, 𝑟⫠ is simply equal
to r, as the linear momentum of any particle in the object is
tangential. Since each particle in the object has the same ω, we
can write the total angular momentum of the object as
𝐿=
𝑚𝑖 𝑟𝑖2 𝜔 = 𝐼𝜔
If we wish to take the vector nature of L into consideration, then
L = 𝐼𝛚
PC141 Intersession 2013
Day 17 – June 17 – WBL 8.4-8.5
Slide 7
8.5 Angular Momentum
Again referring to chapter 6, we know that a net force will alter a
∆P
system’s total linear momentum: F𝑛𝑒𝑡 = (this was an
∆𝑡
alternate form of Newton’s 2nd law). Likewise, by applying the
rotational form of Newton’s 2nd law, we find that (considering
only the magnitude),
∆𝜔 ∆(𝐼𝜔) ∆𝐿
𝜏𝑛𝑒𝑡 = 𝐼𝛼 = 𝐼
=
=
∆𝑡
∆𝑡
∆𝑡
In other words, the net torque on a system is equal to the time
rate of change of the system’s angular momentum.
PC141 Intersession 2013
Day 17 – June 17 – WBL 8.4-8.5
Slide 8
8.5 Angular Momentum
Conservation of Angular Momentum
If the net torque on a system is zero, then its angular momentum
must be constant. In that case, we can write
𝐼𝑓 𝜔𝑓 = 𝐼𝑖 𝜔𝑖
In some systems, a reconfiguring of the mass distribution (which
changes the moment of inertia) will result in a change in the
angular velocity. This effect can be demonstrated as shown in
the figure below. A man spins slowly
in a chair while holding weights in
his outstretched arms. When he
brings the weights in toward his
body (lowering the moment of
inertia), his angular velocity
increases.
PC141 Intersession 2013
Day 17 – June 17 – WBL 8.4-8.5
Slide 9
8.5 Angular Momentum
Figure skaters also exploit the conservation of angular
momentum. They start to spin with their arms (and one leg)
stretched outward. Bringing these limbs together along the
rotation axis leads to a substantial increase in the rotational
speed (up to 7
revolutions per
second).
Videos are not
embedded into the PPT
file. You need an
internet connection to
view them.
PC141 Intersession 2013
Day 17 – June 17 – WBL 8.4-8.5
Slide 10
8.5 Angular Momentum
The orbit of planets about the sun (or the moon about the Earth,
etc.) provides a good example of a system where angular
momentum about the central body is conserved. We know that
L is conserved in this case because we can easily show that the
net torque on the orbiting body is zero. This is because the
gravitational force on the orbiting body is directed toward the
central body – thus, 𝐫 and F are anti-parallel, and their crossproduct (𝛕) is zero.
Conservation of angular momentum in orbital mechanics can be
used to prove Kepler’s 2nd law (the law of areas), although we
won’t attempt to show that here.
PC141 Intersession 2013
Day 17 – June 17 – WBL 8.4-8.5
Slide 11
8.5 Angular Momentum
As a final example of angular momentum, consider the
helicopter shown below. Before it leaves the ground, there is no
angular momentum (nothing is moving at all). If there was only
one rotor, conservation of L dictates that as it rotated, the body
of the helicopter would have to rotate in the opposite direction
(the torque required to turn the rotor is internal; there are no
external torques).
By using a second rotor with an oppositely-directed rotation, the
angular momentum can be kept at zero without requiring the
body to rotate.
PC141 Intersession 2013
Day 17 – June 17 – WBL 8.4-8.5
Slide 12
8.5 Angular Momentum
The table shown here illustrates the many similarities between
the translational and rotational dynamical equations.
PC141 Intersession 2013
Day 17 – June 17 – WBL 8.4-8.5
Slide 13
Problem #4: Angular Momentum Increase
WBL LP 8.19 (corrected)
The angular momentum may be increased by…
A
…decreasing the moment of inertia
B
…decreasing the angular velocity
C
…increasing the product of the angular speed and moment of
inertia
D
None of these
PC141 Intersession 2013
Day 17 – June 17 – WBL 8.4-8.5
Slide 14
Problem #5: Rotating Disk
WBL EX 8.65
A 10-kg rotating disk of radius 0.25 m has an angular momentum of 0.45 kg·m2/s.
What is the angular speed of the disk?
Solution: In class
PC141 Intersession 2013
Day 17 – June 17 – WBL 8.4-8.5
Slide 15
Problem #6: Figure Skater
WBL EX 8.71
An ice skater spinning with outstretched arms has an angular speed of 4.0 rad/s.
She tucks in her arms, decreasing her moment of inertia by 7.5%.
a) What is the resulting angular speed?
b) By what factor does the skater’s kinetic energy change?
c) Where does the extra kinetic energy come from?
Solution: In class
PC141 Intersession 2013
Day 17 – June 17 – WBL 8.4-8.5
Slide 16
Problem #7: Angular Momentum of an Airplane
A 1200 kg airplane is flying in a straight line at 80 m/s, 1.3 km above the ground.
What is the magnitude of its angular momentum with respect to a point on the
ground directly under the path of the plane?
Note: The purpose of this problem is to illustrate that the presence of a non-zero
angular momentum does NOT imply that an object is rotating or moving in a curved
path.
Solution: In class
PC141 Intersession 2013
Day 17 – June 17 – WBL 8.4-8.5
Slide 17