This preview shows page 1. Sign up to view the full content.
Unformatted text preview: Laboratory 8
Rotational Motion II: The Rotating Bar
In Laboratory 6, we investigated moment of inertia and the ways in which it differs from mass.
We also looked at a situation (the inclined plane) in which both translational and rotational kinetic
energy come into play, and saw how rotational kinetic energy plays a role in two-dimensional
inelastic collisions between extended objects. In this laboratory, we will expand our knowledge of
rotational dynamics to include the concept of torque, and we will examine energy conservation
in another context: a rotating bar. We will also use the rotating bar to explore the conservation of
angular momentum in two situations, one of which will lead directly into our analysis of gyroscopic
motion in the following laboratory. Theoretical Background: Force vs. Torque
Simply put, force is something that can change the velocity of a body. Similarly, torque is something that can change the angular velocity of a body. Even though forces are required to generate
torques, it is important to remember that torque is not a force. Torques and forces are both vectors,
so we have to deal with them using the tools of vector algebra.
The magnitude of a torque experienced by a rigid object about a particular axis of rotation can be
τ = r F sin(θ ),
where τ is the magnitude of the torque, F is the magnitude of the applied force, and r is the
magnitude of the vector (also known as the lever arm) that points from the axis of rotation to the
point at which the force is applied. (For diagrams, see Figures 1 and 2.) In this equation, we also
have an angle θ , which is the angle subtended from to .
What does this equation tell us? First, it tells us that we can only calculate torque relative to a
particular axis of rotation - the torques an object experiences will differ depending on which axis
we choose, because the r in Equation 1 will change. Second, it tells us that a force is needed to
create a torque, but that not all forces create torques - a force can only create a torque if it is applied
at some nonzero distance from the axis of rotation. A similar force applied further from the axis
of rotation will generate more torque. An illustration of this idea is shown in Figure 1.
Problem 8.1 Why is it easier to open a door by pushing on the handle than
by pushing near the hinge?
65 Most Torque Less Torque r No Torque
90 90 F F F Figure 1: Forces applied at different distances from a pivot. Equation 1 also tells us that a force applied perpendicular to will generate more torque than the
same force applied parallel to . Hint: What is sin(0)? See Figure 2.
r Most Torque Less Torque r No Torque r 90 r
F 45 F F Figure 2: Forces applied at different angles.
Problem 8.2 Give a physical/geometric interpretation of F sin(θ ). What
does this term represent?
Another important aspect of torque that is not obvious from Equation 1 is that, because torque is a
vector, it also has a direction associated with it. So far we have discussed the factors that impact
torque’s magnitude; now we will consider those that can change its direction.
The direction of the torque vector is determined by the directions of the and vectors. It is
given by the right-hand rule. To use the right-hand rule, you must lay your right hand ﬂat in such
a manner that your ﬁngers can ﬁrst (i) point in the direction of , and then (ii) can curl to point in
. Doing so successfully, you will ﬁnd that your outstretched thumb points
the same direction as F
in the direction of the vector. A convenient short-hand way to denote this is to use a tool from
vector algebra called the cross product, which is written as
= × ,
where is the torque vector.
In general, the vector C that is the result of taking the cross product of two other vectors and
, where C = × , will have a direction given by the right-hand rule and a magnitude equal to
66 B A
ABsin(θ ). Note that if you switch the product around, such that C = × , C’s magnitude stays the
same, but its direction is exactly opposite from what it used to be. When using the right-hand rule
in combination with the cross product, in other words, your ﬁngers should start out pointing in the
direction of the ﬁrst vector that is listed. We will see more cross-product notation in Laboratory 9.
Problem 8.3 If points in the positive x-direction, and points in the
negative y-direction, in which direction does the torque vector point? What
if points in the positive y-direction and points in the negative x-direction?
Problem 8.4 Find the magnitude and direction of the torque for each of
the following situations, where the force, , has magnitude F , and the
radius vector, , has magnitude r.
r Torque is important in rotational motion because it is the rotational analogue of force (but it is not
a force). In translational motion, a force is needed to make an object accelerate, and the amount
of acceleration a given force can produce is inversely proportional to the object’s mass. Rotational
motion is similar. A torque is needed to create an angular acceleration, and the amount of angular
acceleration a given torque can produce is inversely proportional to the object’s moment of inertia.
Compare the equation
to Newton’s second law for translational motion, = m , and you will get the idea.
a Theoretical Background: Angular Momentum
So far we have investigated two quantities in rotational motion that have analogues in linear motion:
torque (force) and rotational kinetic energy (translational kinetic energy). Here we will investigate
another quantity, momentum, that takes a similar form in both translational and rotational motion.
Recall that Newton’s second law can also be written as
= m = ∆ ,
where is the linear momentum. It is natural, therefore, to deﬁne a similar quantity for rotational
motion called the angular momentum, deﬁned as
= Iω ,
67 (3) where is the angular momentum. We can also rewrite Newton’s second law for rotational motion
= I =
In addition, just as force and torque are related through the equation = × , linear and angular
momentum are related via
= × .
This means that a particle with a certain linear momentum also has an angular momentum about
any point it is not moving directly toward or away from. (If and are parallel, is zero.)
L Angular momentum, like linear momentum, obeys a conservation principle. The Law of Conservation of Angular Momentum states that, in the absence of a net external torque, the total angular
momentum of a system will remain constant. If there is a net external torque, the change in is
always in the direction of the torque.
Problem 8.5 In Laboratory 6, was the torque experienced by the rolling
objects constant? In the current laboratory, is the torque experienced by the
rotating bar during a single ball drop constant? Why or why not?
Problem 8.6 In the two-dimensional collisions lab, objects that were not
spinning before the collision suddenly started spinning after the collision,
even though no external torques were present. How is it possible that angular momentum was conserved in that experiment?
A comparison of all of the equations for linear and rotational motion is shown in Table 1.
Newton’s 2nd Law Translational/Linear Rotational/Angular
= m = m
= I = I ∆ω
∆t Table 1: A comparison of some of the equations for translational and rotational motion. Experimental Background: The Rotating Bar
The rotating bar experiment is intended to present you with another demonstration of energy conservation in rotating systems, as well as illustrate the concept of torque. A photograph of the
apparatus is shown in Figure 3. While the bar is horizontal, a steel ball is placed in one of eight
pockets which have distinct distances from the rotor shaft. One set of four pockets is designed to
68 hold 1-inch-diameter steel balls, while the other set is designed to hold 1.25-inch-diameter steel
balls. Once a ball is in place, the bar is leveled and then released. Figure 3: Picture of the rotating bar apparatus. Place a steel ball in one of the eight pockets of the
rotating bar. Use a level to set the horizontal position, and then release quickly.
When the bar is released, it is no longer balanced (due to the presence of the ball) so it begins to
rotate. The side with the ball will move downward, and when the bar is approximately vertical, the
ball will fall out onto the mat below. The speed with which the bar rotates after the ball falls out
can be calculated in a straightforward way using energy conservation, speciﬁcally accounting for
how the loss of potential energy of the ball is turned into the resultant kinetic energy of the ball/bar
The change in potential energy of the steel ball as it travels from the point at which the bar is
horizontal to the point at which it falls out of its pocket is given by
∆Uball = m g (R + r + 0.0127) ,
where R is the distance from the axis of rotation of the bar to the ball’s center of mass, m is the
mass of the ball, and r is the radius of the ball. The 0.0127 m correction factor comes from the
fact that the center of mass of the ball is initially sitting a distance of 0.5 in (0.0127 m) above the
midline of the bar. (See Figure 3.) The additional correction factor, r, comes from the fact that the
ball “pushes” the bar slightly as it rolls out of the pocket.
69 As the bar rotates, this potential energy is converted into the ball’s translational kinetic energy at
Kball = m v2 ,
and the rotational kinetic energy of the bar,
Kbar = 1
Ibar ωbar .
2 We can account for this energy transfer mathematically by writing
∆Uball = Kball + Kbar .
We can substitute in the expressions for these quantities and solve for ωbar in terms of the energy
changes the ball experiences, which results in
ωbar = 2
Ibar (∆Uball − Kball ) . (5) Now if we have a set of measurements for ∆Uball , Kball , and ω 2 , we can calculate the moment of
inertia of the bar. That is what we will do in our ﬁrst experiment today.
Problem 8.7 Draw pictures of the rotating objects in the inclined plane
and rotating bar experiments, including the forces they experience, their
torque vectors, and their angular momentum vectors before and after they
start rotating. Which of these four vectors has a magnitude of zero in both
Problem 8.8 If the ball is initially sitting in a hole that is far from the
rotational axis, will the bar end up with more or less rotational kinetic
energy [when the ball falls off] than if the ball were initially sitting closer
to the rotational axis? Why?
Problem 8.9 Can you derive Equation 5 on your own? Make the substitution vball = R ωbar in the expression for Kball in your derivation. Do you see
why this substitution makes physical sense? Experimental Background: Calculating ω bar
We can simply measure m and R using a scale and a ruler, but we must estimate ωbar using more
indirect means. A photogate placed above the bar is used to detect the times at which an end of
the bar passes through the vertical. Since the bar has traveled a rotational distance π in the time
between successive photo gate blockages, measurements of this time delay can be used to estimate
70 the bar’s average angular speed, ω . This is what the LoggerPro ﬁle rotating bar.cmbl does: it
divides π radians by the time delay between successive gate blockages to ﬁnd ω . Further, we
assume that ω = ωhor , where ωhor is the angular velocity at the time the bar rotates back into the
horizontal position, effectively being at the “midpoint” of successive photogate blockages. Rather
than set up LoggerPro’s data collection on your own, you will save time by using this template.
If there were no friction in our rotating system, ω would always equal ωbar (the angular speed when
the ball falls out), and a single measurement of ω would be enough to allow us to calculate Ibar
using Equation 5. However, since friction does play a role, we must calculate ωbar more creatively.
If you take some practice data, you will notice that ω decreases linearly over time. This decrease
is due to friction in the ball bearings that allow the bar to rotate smoothly. A graph of ω vs. θ
(the total angle through which the bar has rotated after the ball drops out) might look like that in
Figure 4. Figure 4: A graph of some ωbar data. As you can see, the angular velocity of the bar decreases
over time due to friction.
LoggerPro will make four estimates of ω at four successive times. We actually want to obtain the
y-intercept of the graph in Figure 4, because this gives us an estimate of ωbar , the bar’s angular
speed right after the steel ball falls out of its pocket. We can do this by ﬁtting a line to the angular
velocity data in LoggerPro and ﬁnding its y-intercept. 71 Experiments
Experiment 1: Measuring the Bar’s Moment of Inertia
In the ﬁrst experiment, we will use the equation we derived in the theoretical section, Equation 5,
to measure the moment of inertia of our rotating bar. We will then compare our result to the
theoretical value of Ibar that we would obtain if the bar were uniform and free of cavities.
1. Weigh one steel sphere with a 1.00-inch diameter and another with a 1.25-inch diameter, and
record their masses on your template. The template will convert these masses to kg for you.
2. Enter the number written on the surface of the rotating bar in the appropriate cell on your
lab report. The masses of the bars have been obtained ahead of time, so you will not need to
take apart your setup to obtain the mass of your bar. The bar length has also been measured
for you ahead of time.
3. Open the LoggerPro template ﬁle rotate bar.cmbl.
4. Place the 1.00-inch steel sphere in the innermost pocket of the rotating bar that holds it
5. Hold the bar so that it is parallel to the table surface before letting go. A level has been
provided to help you get the bar perfectly level.
6. Begin taking data and let go of the bar. It will rotate for several revolutions before coming
to stop. During this time, LoggerPro will obtain four measurements of the bar’s rotational
velocity as outlined above in the theoretical section. Fit these to a line and use the y-intercept
as your measurement of ωbar .
7. Perform steps 4-6 for the rest of the 1.00-inch holes. Notice that your lab report has the
distances of these holes from the axis of revolution of the bar pre-entered. Make sure that
when you paste your ωbar data into the template, you are putting it in the cells that correspond
to the right value of R.
8. Repeat steps 4-7 using the 1.25-inch-diameter steel ball and the corresponding larger holes.
9. Calculate the change in potential energy of the ball, ∆Uball , for all eight drops. Be sure to
correct for the fact that the ball is initially sitting above the midline of the bar. Note that you
can save a lot of time by cloning formulas using the procedure in Appendix A Section 2.3.
10. Calculate Kball using the fact that vball = R ωbar .
11. Calculate ∆Uball − Kball .
12. A graph will magically appear in your lab report that has ωbar on the y-axis and ∆Uball − Kball
on the x-axis. Using Trendline (see Appendix A Section 2.4) ﬁt a line to these points and use
it to calculate your bar’s moment of inertia. 13. Divide your value for Ibar by M L2 to obtain a measured value for its geometric factor.
Problem 8.10 How far off is the bar’s geometric factor from the theoretical
value of 1/12 for a uniform, cavity-free bar? Is it higher or lower than
expected? Give one source of error that might explain this deviation in
geometric factor you observe.
Experiment 2: Manually Changing the Bar’s Moment of Inertia
In this experiment, we will add weight at a ﬁxed distance from the axis of rotation of the bar, and
we will again measure Ibar and compare our results to theory.
1. Using the holes that have been pre-drilled into the bar, attach two 50 g solid brass disks (the
same disks we used for the collisions laboratories) to the ends of the bars, one on each side,
making sure you do not cover up any of the holes that hold the steel balls. The radial distance
from the axis of rotation to the screw at the center of a disk is 13.7 in or 0.348 m.
Problem 8.11 (a) What do we expect to happen to the geometric factor of
the rotating bar when we add the extra weights? Explain why. (b) Using
Equation 3 from lab 7 and the fact that moment of inertia is a scalar,
calculate the expected moment of inertia for the ‘bar + brass disks’ system
and record it in your lab report. You will need to use your calculated value
for the bar’s moment of inertia from Experiment 1, and that the moment of
inertia for disk rotating about an axis perpendicular to its axis of symmetry
is I = (1/4) mr2 , where r = 2.55 cm for the brass disks.
Problem 8.12 Do we expect ωbar to increase or decrease (relative to its
value in Experiment 1) when we add the extra weight? Why?
2. Repeat the eight ball drops (using the two types of balls, in each of the four holes) for this
new conﬁguration. Once again, remember you can clone formulas using the procedure in
Appendix A Section 2.3.
3. A plot of ωbar vs. ∆Uball − Kball will appear. Fit a line to this plot and specify that the line
intersect the origin. From the slope of this line, calculate the new moment of inertia of the
bar. 73 Problem 8.13 Did ωbar go up or down after adding the extra masses to
the bar? Why does this make sense in light of what we learned in the
theoretical section about energy transfer to the bar?
Problem 8.14 How close was your new measured moment of inertia to the
calculated value from Problem 8.12? What is one source of error that could
explain any deviation you observe from your theoretical calculation?
Experiment 3: Angular Momentum and the Spinning Chair
You will now conduct a tactile demonstration of angular momentum conservation by holding a
spinning gyroscope while sitting in a swivel chair (see Figures 5 and 6) and then ﬂipping the
gyroscope over so that the direction of is reversed.
L Figure 5: The swivel chair and stainless steel rotor used to demonstrate the conservation of angular
momentum. 1. Spin up a heavy gyroscope to above 3200 RPM using the Dremel tool, which your instructor
will demonstrate. You will be able to tell when it is going fast enough because you will start
to hear an ominous hum from the bearings.
2. Sit (or have your lab partner sit) on the swivel chair while holding the gyroscope shaft with
both hands, as shown in Figure 6. Note the initial direction of for the gyroscope.
3. Quickly ﬂip the gyroscope’s axis of rotation by 180◦ . Note the ﬁnal direction of for the
for the person sitting in the chair.
gyroscope, and the direction of the new L 74 Figure 6: Initial position for holding the gyroscope while sitting in the swivel chair. With the rotor
spinning rapidly, twist the axis 180◦ from up to down. Problem 8.15 If the direction of the angular momentum vector, is
initially upward, and you ﬂip the gyroscope while sitting in the chair so
that now points downward, does the chair spin clockwise or counterL
clockwise as viewed from above? Describe why this happens using terms
and concepts from angular momentum conservation.
Problem 8.16 If we were to increase the gyroscope rotor’s mass but spin
it up to the same initial speed, how would this affect the motion of the chair
when the gyroscope is ﬂipped? Why?
Problem 8.17 If we were to increase the mass of the person sitting in
the chair, how would this affect the chair’s motion when the gyroscope is
ﬂipped? Why? 75 ...
View Full Document
This note was uploaded on 09/13/2011 for the course ECONOMICS 101 taught by Professor Gerson during the Spring '11 term at University of Michigan.
- Spring '11