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Pre-Lab 8
Torque balance & Rotational Dynamics
References (optional)
Physics 121:
Tipler & Mosca, Physics for Scientists and Engineers, 6th edition, Vol. 1 (Red book).
Chapter 9, sections 9-2, 9-3, 9-4, example 9-11. Chapter 10, section 10-2.
Physics 114:
Walker, Physics, 3rd edition, Vol. 1 (Green book). Chapter 10, section 10-5. Chapter 11,
sections 11-1, 11-2, 11-3, 11-4, and especially 11-5.
1. Review of concepts
Torque & Rotational Equilibrium
The rotational analog of force is torque. The magnitude
of the torque τ that a force F exerts about the axis is
defined to be
(1)
τ = rFt
where r is the (radial) distance from the axis to the
point of application of the force, and Ft is the
component of the force vector F that is tangential to
the circle that is defined by the radius r . It is not hard
to see that this component of the force causes the body
to rotate about the axis.
An object in rotational equilibrium has a balance of
torques acting on it, just like an object in translational
Figure 1: Torque on an extended body
equilibrium must have a balance of forces acting on it.
depends on both r and Ft .
The net torque, which is the sum of all clockwise
torques and counter-clockwise torques about any axis, must equal zero in order for the object to
be in rotational equilibrium:
τ net =
∑
τi = 0
(2)
all torques
In other words, the net torque, which is the sum of all clockwise torques and counter-clockwise
torques about any axis, must equal zero in order for the object to be in rotational equilibrium.
The center of mass & torque from gravity
The gravity force can exert a torque. An extended object, like a stick or ruler, has mass
distributed over various points. The force of gravity acts on each one of these points, and if the
torques due to gravity do not cancel each other out, the object will tip: rotate about an axis. The
balance point of an object is the axis about which gravity exerts no net torque. A simple
calculation shows that the net torque from gravity can be expressed as the total weight force on
an object Mg times the horizontal distance between the axis of rotation and the center of
mass of the object. Thus, when the center of mass lies directly above or below the axis of
rotation, the torque due to gravity is zero. This is the principle of the balance: the balance arm
©2009 Department of Physics, University of Washington
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Pre-Lab 8: Torque balance & Rotational Dynamics
does not tilt when the torque of the object being weighed equals the torque from an object of
known mass.
The center of mass is a weighted average of the positions of the mass elements of an extended
object. Mathematically, if the positions of the mass elements m1 , m2 ,… , mn are given by the set
of vectors r1 , r2 ,… , rn , the center of mass vector R CM is given by
R CM =
m1r1 + m2r2 + … + mnrn
,
M
(3)
where M is the sum of the individual masses, M = m1 + m2 + … + mn .
Question 1
Three balls are attached to a rigid rod, as shown in the figure. When measured from the end of the
rod, a fulcrum is located at x f = 11 cm. Ball A has mass mA = 50 g and is located at x A = 2 cm.
Ball B has mass mB = 10 g located at xB = 15 cm. Ball C has mass mC = 20 g located at
xC = 23 cm. Calculate the x-coordinate of the center of mass of the rod and balls.
We can use the principle of the balance to locate the center of mass of an extended solid object
in the following way. Suppose an object such as a stick of a non-uniform shape can be placed
on a fulcrum, as shown in Figure 2.
Figure 2. Balancing an extended object. In this picture, the net torque from gravity is zero, since the center
of mass is directly over the fulcrum.
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Pre-Lab 8: Torque balance & Rotational Dynamics
Imagine laying a grid over the extended object with each cell of the grid containing a mass
element mi ; such a grid is indicated by the dotted lines in the figure. Each mi of the object will
feel a torque from gravity about the fulcrum which is the product of the weight force mi g times
the horizontal distance xiF between the mass element and the fulcrum. If the fulcrum is located
at the position rF (with components xF , yF , z F ) then the horizontal distance is the difference
between the distance of mi to the reference point and the horizontal distance of the fulcrum to
the reference point: xiF = xi − xF , and the net torque on the object is the sum of each individual
torque from each mi :
τ net = (m1 g ) x1F + (m2 g ) x2 F + … + (mn g ) xnF
= g [ m1 ( x1 − xF ) + m2 ( x2 − xF ) + … + mn ( xn − xF )]
= g [ (m1 x1 + m2 x2 + … + mn xn ) − (m1 + m2 + … + mn ) xF ]
(4)
= g ( MX CM − MxF )
= Mg ( X CM − xF ).
The above sequence of equations simply states that the net torque from gravity for an extended
object is simply the torque due to the total mass as if it were all located at the center of mass.
Notice that if the fulcrum is placed so that τ net = 0 , then we can rewrite the last line of the
previous sequence of equations as
xF = X CM when τ net = 0 .
(5)
In other words, when the fulcrum is placed directly under (or over) the center of mass, there is
no torque from gravity, and the object balances on the fulcrum. This is the situation shown in
Figure 2.
Question 2
In Question 1, is the structure in balance? That is, is the position of the top string right over the
center of mass? If not, how far must the knot in the string be moved so that it becomes in balance?
Note that in Eq. (5) that the mass M of the object has dropped out. By adding a test mass mT
of known value to the object at a known x-position xT , we can weigh the object: find the value of
M . A recalculation of the torque would simply add a term mT g ( xT − xF ) to the last line in Eq.
(4). Re-balancing the object (i.e., setting τ net = 0 ) would give the following equation:
mT xT = ( M + mT ) xF − MX CM
(6)
There are a couple of things to notice about the above equation. First, there is no g . Even
though the balance works because of gravity, it does not matter what the acceleration of gravity
is. It is only important that it exists. A balance would act the same on the Moon as on Earth.
Second, the form of the equation should be familiar: it is a line. If we were to make
measurements of xF and xT for various positions of the test mass mT , and then plot mT xT
versus xF , we would obtain a line whose slope is the sum of the masses M + mT and whose
intercept is MX CM . This means that we could find both M and X CM by moving a test mass
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Pre-Lab 8: Torque balance & Rotational Dynamics
around on an object and locating the balance point. This procedure will be tried in the first part
of the experiment.
Question 3
If the object discussed has M =250 g, mT =20 g and X CM =3 cm, what is the slope and intercept of
the line described by Eq. (6)?
Rotational dynamics
If the net torque about a fixed axis is not zero, the object will experience a change in the rate of
rotation about that axis, that is, it will experience an angular acceleration α . The magnitude of
this acceleration depends on the torque, the mass, and the distance of the mass from the axis.
Consider a point mass m on the end of a light rigid rod of length r , with the other end of the
rod fixed on an axis, as shown in Figure 3.
Figure 3. Mass on a rod.
A force acting on the mass causes a tangential acceleration at . The tangential component of
the net force, Ft , is related to the tangential acceleration by Ft = mat . The torque τ exerted
by Ft about an axis through the center of the circle and perpendicular to the plane of the circle
(
)
is τ = rFt = rmat = r 2 mat r = mr 2 α . The angular motion form of Newton’s second law follows
from this illustration,
τ = Iα ,
(7)
where we have defined a quantity I called the moment of inertia. For a mass point m ,
I = mr 2 . This is easily extended to a solid body. If a solid is rotated about a fixed axis, every
element of the body moves in a circle about the axis and the torque required is the sum of the
torques on all the elements. The moment of inertia for the solid about the axis of rotation is the
sum of the moments of inertia for all the elements,
I = ∑ i I i = ∑ i mi ri 2 .
(8)
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Pre-Lab 8: Torque balance & Rotational Dynamics
The prescription for calculating the moment of
inertia of an extended body is to multiply each
mass element Δmi by the square of its distance
ri from the axis of rotation and sum over all the
elements that make up the solid. The moment of
inertia is not an intrinsic property of the solid like
the mass, but depends on the relative position of
the axis and the parts of the solid. As an
example, consider a thin hoop of radius R as
shown in Figure 4. The moment of inertia about
an axis #1 through the center of the hoop,
Figure 4. Hoop viewed in perspective.
perpendicular to the plane of the hoop, is quite
simple to calculate since every mass element Δmi lies a distance R from the axis:
I hoop = ∑ Δmi ri 2 = ∑ Δmi R 2 = ( ∑ Δmi ) R 2 = MR 2 where M is the total mass of the hoop.
Note that the definition of the moment of inertia, Eq. (8), allows one to add up the individual
moments of inertia in any manner. This means that we could find the moment of inertia of a
complex object by first working out the moments for the different parts of the object and then
adding the results together. Suppose we know that the solid consists of separate parts with
moments of inertia about the common axis of rotation of I1 , I 2 ,… , I N . Then, the total moment of
inertia is just the sum of the moments of inertia of the composite parts, that is,
I = I1 + I 2 + … + I N . We will make use of this property in the experiment.
Question 3
It is important to remember that the moment of inertia depends on both the mass and the distance of
the mass to the rotation axis. Look again at Figure 4. How would you rank the moments of inertia
about axes #2 (parallel to axis #1, at the edge of the hoop) and #3 (perpendicular to axis #1, along a
diameter of the hoop)? Hint: think about how far each mass element Δmi is from the axis, and how
that distance compares to the distance from axis #1.
2. Apparatus
The apparatus for Part 1 of the
experiment is very simple: it is
comprised of a half meter stick, a few
brass weights and some tape. You will
first create a “complex object” by taping
one of the weights to the stick, and then
you will use another weight as the test
mass in a procedure that makes use of
Eq. (6) to determine the mass and
center of mass of the “complex object”.
In Part 2 you will study rotational
dynamics by using the rotational analog
of the cart being pulled by a falling
weight. Instead of linear acceleration,
Figure 5. Turntable and rotary motion sensor.
the falling weight will cause an angular
acceleration by exerting a torque on a turntable pulley. This constant torque will cause a
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Pre-Lab 8: Torque balance & Rotational Dynamics
constant rate-of-change of ω , that is, a constant α . These quantities will be measured directly
by a rotary motion sensor and DataStudio.
The rotary motion sensor and turntable is pictured at right. The turntable has holes drilled in it
that allow objects to be attached, such as metal rings or disks. In the figure the object is a thick
steel ring. Underneath the turntable is a pulley (#1) that you can wind a string around. The
string is wound around pulley #1, passed over pulley #2 and attached to a hanger similar to the
one used in the “One-Dimensional Dynamics” lab. Releasing the hanger with its weight exerts a
torque on the turntable and the object, causing the turntable and the two pulleys to be
accelerated.
The sensor unit is attached to inputs 1 (yellow plug) and 2 (black plug) of the Pasco Interface.
The sensor provides a sample-signal for each ¼ degree of rotation of the shaft (devices of this
sort are called “shaft encoders”) and DataStudio processes these measurements and
generates table and graphs of data.
Finding the moment of inertia by using this apparatus requires applying Newton’s 2nd law in both
its linear ( F = ma ) and rotational ( τ = Iα ) forms, and then combining the equations by linking
the rotational acceleration α of the turntable to the linear acceleration a of the falling weight.
Since you will do this in lab, the following exercise will give you practice by analyzing a
simplified form of the turntable apparatus.
Question 4
A hanger of mass m is attached to a light string
that is wrapped around the rim of a wheel of
radius r . The wheel is released allowing the
weight to fall and cause the wheel to accelerate
rotationally. In the experiment, the rotational
acceleration α is measured. The acceleration of
gravity is g and the moment of inertia of the
wheel is I . Work out a formula for I in terms
of the other variables by following these steps:
a. Write Newton’s 2nd law for the falling weight.
Remember, there are two forces acting on it:
the gravity force mg and the tension in the
string T .
b. Write Newton’s 2nd law in rotational form for
the wheel. Express the torque in terms of the
tension T and wheel radius r . Express the
angular acceleration α in terms of − a r , since a positive (counterclockwise ) torque implies a
negative (downward) linear acceleration.
c. Combine the two equations to eliminate T from them and obtain a formula for the acceleration
a in terms of I , m , g and r . Check your equation to make sure it has the correct behavior.
For example, when I = 0 , a should equal − g . When I gets infinitely large, a should go to 0.
d. Finally, substitute a = − rα in your formula and solve it for I to obtain the required formula in
terms of the variables m , g , r and α .
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Pre-Lab 8: Torque balance & Rotational Dynamics
3. Best estimates & Standard deviation of the mean
Let’s review for a moment the development of our formulas and our understanding of
experimental uncertainty. The motivating concern is to give a quantitative meaning to the idea
of experimental reproducibility: a way to state numerically how closely repeated measurements
(and calculations derived from those measurements) lie together. This led to the idea of a
distribution—a collection of numbers derived from repeating the same process—and the two
parameters we use to describe that distribution: the mean, which gives the approximate center
of the distribution, and the standard deviation, which gives the approximate spread or width of
the distribution in the sense that approximately 2/3 of the distribution will lie within one standard
deviation of the mean.
The error propagation formulas give us a way to estimate the properties of other distributions
that would be derived from mathematical operations on the original distributions (like
multiplication by a constant, taking the sum or difference of two measurements, or multiplying
two measurements together) without having to work the calculations through for each and every
member of each distribution involved.
The error propagation formulas are especially useful in the cases where the actual distribution
of a particular quantity is either lost or simply unknown. In these cases, we may not know a
standard deviation, but rather a more general “uncertainty.” For example, a motion sensor may
not be able to measure a change in position of less than 1 millimeter, and so we say that the
uncertainty in the position is ±0.5 mm. Even though such an uncertainty is not a measure of an
actual distribution, it does indicate the level of reproducibility associated with that quantity.
Although the error propagation formulas allow us to estimate the parameters of derived
distributions, and enlarge the kinds of uncertainty we feed into our calculations (like the
instrumental resolution mentioned above), there are some restrictions you must remember, lest
you wind up with nonsense. First, the error propagation formulas are derived under the
assumption that each quantity in a calculation is independent of any other quantity; if one
quantity does depend on another, then additional steps must be included in the propagation
calculations to account for the dependence. Second, most propagation formulas are derived
assuming that the fractional uncertainty of every quantity is small, typically 10% or less. If the
fractional uncertainty is not small, you cannot apply the formulas for products & quotients or
raising to a power*; in these cases you would need to work with the original distributions directly.
The “best estimate”
The purpose of making repeated measurement trials is not merely to determine the
reproducibility of that type of measurement, but to obtain a good estimate of the quantity in
question. This is naturally (and correctly) given by the mean of the distribution. It is not hard to
accept the assertion that the mean is a “better” value than any single trial, nor would be hard to
believe that the mean of 100 trials is better than the mean of just 5 trials. But in what sense is
the mean “better”?
To make sense of this question, consider the following scenario. Assume that you have
measured the acceleration of gravity g 100 times. Your data set is, in the language of
statistics, a “sample distribution”, which is a subset of the much larger (indeed infinite) “parent
distribution”. If you were to take more measurements, your distribution would have more
*
For sums and differences, or constant multipliers, this second restriction is relaxed. The error propagation
formulas for these cases are exact.
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Pre-Lab 8: Torque balance & Rotational Dynamics
elements, but its mean and standard deviation would not change very much. In fact, if you kept
on taking measurements indefinitely, the mean and standard deviation would become quite
stable: the mean and standard deviation for 1000 measurements would be very nearly the same
as for 5000 measurements. The parameters would eventually converge to the limiting values
that describe the parent distribution.
How close to these limiting values are ones obtained from a set of 5, 10, or 100 measurements?
Imagine that you divide your set of 100 values into 20 groups of 5 each, and then find the mean
for each group of 5. This produces another distribution: a distribution of “means-of-5.” It is not
hard to see that the mean of the means-of-5 distribution is the same as the mean calculated
from the full set of 100; in this case the values in the groups of 5 are added together and divided
by 5, and then to find the mean of this group, these results are added together and divided by
20, which is the same result as if you had simply added all 100 values together and divided by
100 (=5×20).
The standard deviation of the means-of-5 is not the same as the standard deviation of the set of
100, however. To work this out, we appeal to our error propagation formulas. Remember what
we are after: the spread of values in the collection of means-of-5, not the spread of values in the
original set of 100. Each element in a set of 5 comes from a distribution that has the standard
deviation σ . To form the mean, we first sum the elements together. The first formula is the
sum/difference formula: the standard deviation of this sum would be equal to the sum of the
squares of the standard deviation associated with each element. The next step in forming the
mean is to divide by 5. Thus we apply the formula for multiplication by a constant, and divide
our previous result by 5. Let σ 5 denote the standard deviation of means-of-5 distribution. The
whole calculation looks like this:
σ5 =
σ 2 +σ 2 +σ 2 +σ 2 +σ 2
5
=
5σ 2
σ
.
=
5
5
(9)
In other words, the standard deviation of the means-of-5 is smaller than the standard deviation
of the original distribution by a factor of 5 . This makes intuitive sense: when we select 5
elements at random from a distribution and then find their mean, the result is a number
somewhere near the middle of the group. The spread of these middle values should be less
than the spread of the original distribution.
It is important to remember what this standard deviation, which is called the standard deviation
of the mean or standard error represents: it gives the range of mean values about the mean of
the parent distribution.
Thus, because the uncertainty in the mean value is less than the uncertainty in the individual
measurements, this standard deviation of the mean is a better estimate of the uncertainty in our
experimental result than the standard deviation of the whole data set. When you collect data in
the lab, you should from here on use the standard deviation of the mean for your uncertainty
(when it is larger than the instrumental resolution—see the earlier comments).
You will be asked to calculate the standard deviation of the mean in the lab and post-lab.
DBP (02/22/2010)
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