Physics
10 – 1
Experiment 10
Newton’s Second Law for Rotation
Questions .
How do we give something a constant angular acceleration? How can we measure this
angular acceleration? What are the directions of an object’s velocity and acceleration vectors when it is subject to non-uniform circular motion?
Concepts .
When an object moves in a circle and its angular speed (omega, 𝜔) changes, the object experiences an angular acceleration (alpha, 𝛼). In straight-line motion, a net force makes something accelerate. For rotation, a net torque gives an object an angular acceleration. In this experiment, you will apply a constant torque to a circular disk that can rotate on a low friction bearing. The disk will ‘spin up’ to a higher and higher angular speed. You will calculate the disk’s constant, angular acceleration and relate it to the applied torque.
Figure 1. Figure 2.
Figure 3.
!
v 1
!
v 2
Δ !
v / Δt Δ !
v
!
v 1
!
v 2
acen atan
!
a = Δ !
v / Δt( )
!
a
Florida Institute of Technology ©2020 by J. Gering
10 – 2
Figure 4.
In Figure 1, a particle moves clockwise in
non-uniform circular motion. Its angular and
linear speeds increase. The subscripts 1 and 2
refer to early and later times.
Figure 2 shows the change in the particle’s
velocity vector. The acceleration vector is
just a scaled version of the change in the
velocity vector. The scaling factor is the time
interval.
Figure 3 shows how the acceleration
vector has tangential and centripetal
components. The tangential component is
given by , and as usual .
The radius of the circle and the angular
acceleration are constant. So, the tangential
component of the acceleration remains
constant in time. However, due to the speed-
squared term, the centripetal component
increases with time.
Figure 4 shows both the velocity (blue)
and acceleration (red) vectors of the particle
at four successive times. The velocity vector
is always tangential and grows linearly in
length. The acceleration vector also grows in
length. It starts out tangential but becomes
more and more centripetal.
For a constant angular acceleration, the angular speed,
w, obeys a rotational kinematic
equation similar to straight-line motion: . If the disk starts from rest, = 0.
The linear and rotational versions of Newton’s Second Law are extremely similar:
and . Analogous to the mass m in the linear version, the rotational mass
of the object about a given axis of rotation is called the object’s rotational inertia or the moment
of inertia, I. Here the word moment is a statistical term that measures how spread out the mass is
distributed from the axis of rotation. A dumbbell shaped object is more difficult to rotate than a
solid sphere of the same mass. This is because more of the dumbbell’s mass is farther from the
middle of the dumbbell, which is typically where the torque is applied. Differently shaped
objects have different rotational inertias. Also, the same object has different moments of inertia
depending on the choice of the rotation axis. We will use a solid disk of mass M and radius R
with an axis perpendicular to the plane of the disk and through the disk’s center. In this case, the
moment of inertia is I = ½MR2 .
!
a
!
v
a tan = r
α acen = v 2
r
ω(t) =
ω 0 +
αt ω 0
!
Fnet = m !
a !
τ net = I !
α
Florida Institute of Technology ©2020 by J. Gering
10 – 3
A low-detail diagram of the apparatus is shown in Figure 5. A hanging mass falls and
accelerates the disk by unwinding a string wound around one of three small (stepped) pulleys mounted to the disk. A photogate is used to measure the angular speed of the vertical pulley.
From this you can use Logger Pro to calculate the linear speed of the string and its acceleration.
Let us take as the linear speed of the string; the sub-script V stands for the vertically mounted pulley and S for the stepped pulley. Then the angular speed of each pulley is written as follows.
Each r is the radius of each pulley.
and (4)
The tension in the string is the force that exerts the torque. However, this tension is not simply the weight of the hanging mass. We will ignore the mass of the vertical pulley and all friction. Then Newton’s Second Law gives for the falling weight:
Ignoring the mass of the vertical pulley allows us to write . (5)
Applying the definition of torque to the disk and using the rotational version of N2 gives:
Substituting for the tension gives: .
Substituting from Eqn. (5) and solving for the linear acceleration gives
(6)
Method .
Figure 5 depicts the second part of this week’s apparatus: a heavy disk that can rotate in a horizontal plane on a low friction bearing. A two-claw clamp and a support rod for a pulley have been omitted for simplicity. The acceleration of the string is measured by a photogate mounted around the pulley and connected to the Lab Pro interface. A similar arrangement was used in the Newton’s Second Law experiment.
v
ωV = v
rV
ω S = v
rS
T − mg = −ma
a =
αrS
τ ≡ rS
T = I
α
τ = rS mg − m
αrS( ) = I
α
a = mgrS
2
I + mrS
2
Florida Institute of Technology ©2020 by J. Gering
10 – 4
Figure 5. The Pasco Introductory Rotation Apparatus.
Procedure .
1) Use a Vernier caliper to measure the diameter of whichever stepped pulley you want to use. Divide the diameter by two to find r S .
2) Measure the mass M of the plastic disk. The disk’s mass is greater than 610 grams,
which is the maximum mass the triple beam balance can measure. However, hanging a
tare weight on a small metal stud found at the end of the scale extends the balance’s
range. These are called tare weights. One tare weight adds 500 grams to the reading. In reality, this mass is 147.5 grams but its torque on the apparatus yields a 500 gram
equivalent mass. Using the same proportions, a 200-gram hooked mass can be used as a tare weight. This will add 678 grams to the scale reading.
3) Measure the radius, R, of the plastic disk. We will ignore the slight difference in the
moment of inertia formula that would result from including the shape of the stepped
pulley. Calculate the moment of inertia of the disk using I = ½MR2 .
4) Set-up the apparatus as in Fig. 5. Use white, kite string or fishing line as the string. Do
not tie the string to the screw at the top of the stepped pulley; just wrap it around the
screw a few times. This way you can easily remove the string. Then thread it through
the holes drilled in the pulley and then wrap the string around the chosen pulley. For
example, if you use the largest diameter pulley, the string should go through three holes.
5) Setup the software for using the photogate and the 10-spoke pulley by executing these menu commands: Experiment > Set Up Sensors > Show All Interfaces >
> Set Distance or Length… > Ultra Pulley 10
Spoke In Groove.
6) Also increase the sampling rate to 50 samples / sec. Execute these commands:
Experiment > Data Collection.
7) Adjust the length of the string so the weight hanger does not strike the floor when all of the string has unwound. Place a foam pad beneath the hanging weight to avoid damage if
stepped pulley
rotating disk
hanging
mass
vertically
mounted
pulley
Florida Institute of Technology ©2020 by J. Gering
10 – 5 the string breaks. Use slotted masses and the Beck mass hangers. The plastic 5-gram mass hangers are fragile and can break too easily.
8) Lower a mass before recording data. Examine the apparatus to make sure the disk
accelerates, and that friction is minimal. Use Logger Pro to measure the linear
acceleration of the 10-spoke pulley for five different falling masses. Do not use more
than 150 grams of hanging mass.
9) Use Logger Pro to acquire a velocity vs. time graph for when the hanging weight falls.
Use the slope of a tangent line feature to obtain five accelerations for one run.
Record your data in a spreadsheet. Later, calculate the average and sample standard
deviation for the five accelerations.
10) Change the amount of hanging mass and perform additional trials.
11) For each trial, calculate the linear acceleration from Eqn. (6). Compare it to the measured value. Enter these theoretical accelerations into the spreadsheet.
12) Calculate the differences between all experimental and theoretical accelerations.
Compare each difference to the standard deviation in each experimental value. If the
standard deviation is as large or larger than the difference, then the results agree to within experimental error. There may be no need to propagate errors through Eqn. 6.
13) Comment on sources and types of errors in your discussion.
ze
