PHY 111L - Rotational Dynamics
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Aerospace Engineering
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Dec 6, 2023
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PHY 111L Activity 09
Rotational Dynamics
Name:
Kyra Pell
Date:
11/13/2023
Objectives
1.
Introduce rotational kinetic energy in terms of moments of inertia
2.
Use the Work-Energy Theorem to determine the final velocity of objects rolling down an inclined plane
3.
Experimentally determine the velocity of a ring and disk and compare to calculated values (lab report)
Materials & Resources
1.
Stopwatch (can use cell phone)
2.
Inclined plane
3.
Ring and Disk of equal radius
4.
Protractor, Calipers, Scale
Introduction
Recall that the translational kinetic energy (
TKE
) of a body is given by:
KE
t
= ½mv
2
Similarly, the rotational kinetic energy (
RKE
) is expressed as:
KE
r
= ½Iω
2
where
I
is the body’s moment of inertia, and
is the angular speed. The total kinetic energy of an object in general
consists of both translational and rotational energies. Recall that the work-energy theorem states that the change in
potential energy equals the change in kinetic energy for a conservative system, or one in which no external forces
are acting.
When a body is both translating and rotating, the Work-Energy Theorem becomes:
KE
t
KE
r
PE
What this means is that, for example, some of an object’s gravitational potential energy may end up as rotational kinetic
energy as well as translational kinetic energy.
Under these conditions, the work-energy theorem gives:
½Mv
2
+ ½I
2
= Mgh
Now, for an object that rolls without slipping,
v = R
, where
v
is the translational velocity of the object,
is its angular
velocity, and
R
is its radius.
Then,
=
v
/
R
.
Substituting and solving the energy equation for the expected value for
v
e
gives:
v
e
=
√
(
2
gh
/(
1
+
I
/(
MR
2
)))
where
M
i s the mass,
R
is the radius, and
I
is the moment of inertia of the object, while
h
is the vertical (in gravity’s
direction) distance traveled by the object.
Note that
I
is divided by a factor of
MR
2
in the formula to leave
1 +
f
, where
f
is a fraction associated with the object’s shape.
We will use a timer and use the acquired data to find the velocity
associated with the kinetic energy of various objects.
Recall that since gravity provides a constant acceleration over short
distances near the surface of the Earth, the average velocity
(
d
/
t
)
equals
(v
f
+v
i
)
2
.
In order to simplify this further, set
v
i
= 0
by starting the object at rest.
Then
v = 2d
t
.
We will measure this with the timer and compare it to
v
e
.
1
d
h
Figure 9-1. A solid disk & a ring roll from rest on an inclined plane.
1) Velocity of a Rolling Ring
Procedure
:
1)
Set up the experiment as shown above.
We recommend trying to use the smallest inclined ramp as possible so you
have more time to accurately stop your timer (
θ=1
works good
).
Measure the angle
θ
and the travel distance
d
using a ruler.
Calculate the vertical distance
h.
Record your results below with units.
Show all of your work for the
h
calculations in the space provided to the right.
θ
=
7
(degrees)
h
=
Δy,
Δy=y2−y1,
Δy=5.50 (cm)−0 (cm),
d
=
35.4
(cm)
Δy= 5.50 (cm) =
h
h
=
5.50
(cm)
3)
Measure the mass of the ring and disk; record with units below.
If your scale can’t make the measurement, use the
following values:
M
r
= 131.7 g
&
M
d
= 138.2 g
M
r
= _______131.7
________ (g)
M
d
= __138.2
____________ (g)
4)
Use the calipers to measure the inner diameter
D
i
and the outer diameter
D
o
of the ring, and then calculate the inner
radius
R
i
and the outer radius
R
o
of the ring, as well as its moment of inertia using the equation for a thick ring
I
r
= ½M
r
(R
i
2
+ R
o
2
)
and its expected velocity
v
er
at the bottom using the equation derived in the Introduction.
You may use the outer radius for the value of R in the inertia formula
.
Record your measured and calculated results
with units below. Show work on bottom of page or separate page.
D
i
=
108.00
(mm)
D
o
=
113.74
(mm)
R
i
=
54.000
(mm)
R
o
=
56.870
(mm)
I
R
=
0.00040499
(kg x m^2)
v
er
=
0.74374
(m/s)
2
5)
Roll the ring down the inclined plane at least 5 times; be sure that the initial velocity of the ring is zero so that the
calculated final velocity is accurate.
Write your 5 times on a scratch sheet of paper.
Also take care to ensure the track
does not move so that the angle stays the same for each run.
Find the average travel time for the ring
t
r
, then
calculate its velocity at the bottom using
v
r
= 2d / t
r
.
Calculate the percent difference between
v
r
and
v
er
and record
your results below.
t
r
= _______0.88
________ (s)
v
r
= _____0.80455
__________ (m/s)
% difference for
v
r
and
v
er
: _1.96
%
Questions
:
1)
Is your measured value for
v
r
approximately equal to the expected value
v
er
for the ring, yes or no (circle one)?
2)
Which value for the final velocity of the ring is higher,
v
r
or
v
er
(circle one)?
3)
Briefly justify this experimental result in terms of the measurements made in order to calculate
v
r
(
d
and
t
r
).
In this experiment, the distance traveled (d) was measured to be 35.4-centimeter, later converted to 0.354 meters by
dividing by 100, and the time taken (tr) to travel that distance was measured to be 0.88 seconds. The formula vr = 2d / tr
was used to calculate the velocity. According to the formula, the velocity (vr) is determined by multiplying the distance by
2 and then dividing it by the time taken. By substituting the given values into the formula and performing the calculations,
we obtained a velocity of approximately 0.80455 m/s. Therefore, the experimental result of 0.805 m/s can be justified by
the measurements of distance and time made to calculate vr.
4)
Briefly justify this result in terms of any of the measurements needed to calculate
v
er
(
d, θ, M
r
,
D
i
, and
D
o
).
To justify the result of the formula ver= √(2gh/(1+ I/(MR^2 ))), we need to calculate the values of the variables d, θ, Mr,
Di, and Do mentioned in the question. In this formula, the variable ver represents the velocity of an object. Let's break
down the formula and calculate each term:
1. g represents the acceleration due to gravity and is given as 9.81( m/s^2).
2. h represents the height or depth from which the ring is released and is given as 5.50 (cm). We need to convert it to
meters by dividing by 100: h = 5.50 / 100 = 0.055 (m).
3. I represents the moment of inertia, which is a measure of an object's resistance to changes in rotational motion. It is
given as 0.00040499 kg x m^2.
4. M represents the mass of the ring and is given as 131.7 g. We need to convert it to kilograms by dividing by 1000: M =
131.7 / 1000 = 0.1317 kg.
5. R represents the radius, in this case outer radius, of the ring and is given as 56.87 mm. We need to convert it to meters
by dividing by 1000: R = 56.87 / 1000 = 0.05687 m.
Now, let's substitute these values into the formula and calculate `ver`:
ver = √(2gh/(1 + I/(MR^2)))
ver = √(2 * 9.81 * 0.055 / (1 + 0.00040499 / (0.1317 * (0.05687^2))))
ver = √(1.0791 / (1 + 0.00040499 / (0.1317 * 0.0032342197)))
ver = √(1.0791 / (1 + 0.00040499 / 0.000425944))
3
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ver = √(1.0791 / (1 + 0.950806339))
ver = √(1.0791 / 1. 950806339)
ver = √0.553155881
ver ≈ 0.74374 (m/s) (rounded to 7 significant figures)
Therefore, the calculated result for ver is approximately 0.74374 m/s.
2)
Velocity of a Rolling Disk (
Lab Report)
Procedure:
1)
Use the calipers to measure the diameter
D
d
of the disk, calculate its radius
R
d
, and record your results below.
Calculate its moment of inertia using
I
d
= ½M
d
R
d
2
, as well as its expected velocity
v
ed
as in step 4 above.
Show your
calculations for
I
d
and
v
ed
in the space below.
D
d
=
112.80
(mm)
Id
=
0.5
×
0.1382
(
kg
)
×
0.056400
(
m
)
2
=
0.00021980
(
m
/
s
)
R
d
=
56.400
(mm)
I
d
=
0.00021980
(kg x m^2)
v
ed
=
0.8417
(m/s)
2)
Repeat the procedure outlined in step 5 above to find the average of the travel time of the disk, as well as its velocity
at the bottom using
v
d
= 2d / t
d
.
Calculate the percent difference between
v
d
and
v
ed
and record your results below.
Show work below.
Note: do not confuse the travel distance
d
with the subscripts that refer to the disk.
t
d
= ______0.84
________(s)
v
d
= _____0.84286
_______ (m/s)
% difference for
v
d
and
v
ed
: __0.16
___ %
4
V
ed =
2
×
9.81
(
m
s
2
)
×
0.055
(
m
)
¿
¿
1
+
0.00021980
(
kg x m
2
)
(
0.1382
(
kg
)
×
0.056400
(
m
)
2
)
¿
Questions
:
1)
For which object did you calculate the largest percent difference between the two velocities?
Provide a brief physical
or experimental justification for this result below.
The disk has the largest percent difference between the two velocities. In this experiment the percent difference
formula used in both parts was,
pre cent difference
=
|
a
−
b
|
(
a
+
b
)
2
x
100
. For the ring precent difference I substitute vr for
a and ver for b, giving us the formula of,
pre cent difference
=
|
0.80455
(
m
s
)−
0.74374
(
m
s
)
|
(
0.80455
(
m
s
)
+
0.74374
(
m
s
))
2
x
100
, leaving me
with the result of approximately 1.96%. For the disk precent difference I substitute vd for a and ved for b, giving us the
formula of,
percent difference
=
|
0.84286
(
m
s
)−
0.84817
(
m
s
)
|
(
0.84286
(
m
s
)
+
0.84817
(
m
s
))
2
x
100
, leaving me with the result of approximately 0.16%
2)
If a sphere, a disk, and a ring with the same mass and radii, all begin traveling down the same slope at the same time,
which would arrive at the bottom first, second, and third?
Why?
I
sphere
= (2/5)M
s
R
s
2
To determine which object would arrive at the bottom of the slope first, second, and third, we need to consider their
shapes and properties. Let's start by discussing the basic properties of the three objects:
1. Sphere: A sphere is a three-dimensional object in the shape of a ball. It has a curved surface, and all points on its surface
are equidistant from its center.
2. Disk: A disk, also known as a circular plate, is a two-dimensional object with a flat, circular shape. It has a constant
thickness and all points on its surface are equidistant from its center.
3. Ring: A ring is a three-dimensional object with a circular shape. It has a hollow center and is formed by a circular band
of material.
Since all three objects have the same mass and radii, let's only compare the inertia values, which are directly proportional
to the moment of inertia. The moment of inertia for each object is given by the formulas:
I
sphere
= (2/5)M
s
R
s
2
,
I
disk
= ½M
d
R
d
2
, and I
ring
= ½M
r
(R
i
2
+ R
o
2
).
Considering the formulas, we can conclude
that:
1. For the sphere, the moment of inertia is (2/5) times the mass times the square of the radius.
2. For the disk, the moment of inertia is half the mass times the square of the radius.
3. For the ring, the moment of inertia is half the mass times the difference between the square of the outer and inner radii.
Now, let's analyze the moments of inertia. The highest moment of inertia belongs to the sphere, followed by the disk,
and then the ring. This means that the sphere has the largest resistance to changes in rotational motion, followed by the
disk, and finally, the ring. When objects roll down an incline, their moments of inertia affect how they rotate and translate
their motion. Objects with larger moments of inertia will roll more slowly down the slope compared to objects with
5
smaller moments of inertia. In this case, the sphere has the largest moment of inertia, so it will roll down the slope the
slowest and reach the bottom last. The disk has the second-largest moment of inertia, so it will reach the bottom after the
sphere. Finally, the ring has the smallest moment of inertia, so it will reach the bottom first.
3)
Why is the moment of inertia different for each object, and how does this affect each object’s rotational energy?
The moment of inertia influences how effectively an object can convert translational kinetic energy into rotational
kinetic energy, in other words how easily an object can spin or change its rotational motion. Objects with larger moments
of inertia require more energy to rotate at a given angular velocity compared to objects with smaller moments of inertia,
and this is reflected in their rotational kinetic energy.
4)
Does the final velocity of each object depend on its mass?
If so, how?
If not, why?
The final velocity of each object at the bottom of the slope depends on various factors, including the object's mass, the
distribution of that mass (moment of inertia), and the initial conditions (e.g., the height of the slope and the initial
velocity). However, when comparing objects with the same mass and radii in the context of the 2cd question, the mass
cancels out when considering energy conservation. Do remember that the formula between gravitational potential energy
and kinetic energy is.
mgh
=
1
2
mv
2
. Here,
m
is the mass of the object,
g
is the acceleration due to gravity,
h
is the
height of the slope, and
v
is the final velocity. Notice that the mass (
m
) appears on both sides of the equation and cancels
out, giving us the formula
gh
=
1
2
v
2
. Therefore, the final velocity (
v
) does not depend on the mass of the object as long
as the masses are the same initially.
5)
How would the results be affected if the objects did have an initial velocity?
Justify your answer.
When objects have an initial velocity, the final velocity at the bottom of the slope is influenced by both the initial
conditions (including the initial velocity) and the properties of the objects (including their mass and moment of inertia).
The relationship is more complex than in the case where objects start from rest, and additional information about the
initial conditions would be needed to determine the final velocities accurately.
Think about the conservation of energy
equation, taking into account both kinetic and potential energy, the equation is,
mg
ℎ
+
1
2
mv
initial
2
=
1
2
mv
final
2
. It is
crystal clear that the initial conditions would be needed to determine the final velocities accurately.
6
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