Moment of Inertia
1. Rotational version of Newton’s second law
Newton’s second law says the net force on an object is directly proportional to its acceleration, and the
proportionality constant is the mass of the object: 𝛴𝐹 = 𝑚𝑎.
The rotational version of this law says the net torque acting on an object about the rotation axis is directly
proportional to the angular acceleration, and the proportionality constant is the moment of inertia:
𝛴𝜏 = 𝐼 𝛼
Units: For torque, the units are Newtons  meters, and for angular acceleration, they are rad/s2. So, the
moment of inertia units can be deduced: kg m2.
2. Examples of moment of inertia
For a single point object rotating at distance l from an axis, we have
𝐼 = 𝑚 𝑙2
If there are several point objects all rotating around the same axis, then
just add these up, once for each object:
𝐼 = 𝑚1 𝑙12 + 𝑚2 𝑙22 + …
We will use this to calculate the theoretical moment of inertia of the system of two or four orbiters.
3. Measuring moments of inertia with a graph
In this lab, we will measure the moment of inertia for three systems of objects:
a) the rotodyne wheel alone,
b) two orbiting masses (excluding the rotodyne wheel), and
c) four orbiting masses (excluding the rotodyne).
For each, we will make a plot and the variables will be the net torque 𝛴𝜏 and the angular acceleration α.
The first equation can be written as follows to emphasize the variables:
[𝛴𝜏] = 𝐼 [𝛼]
If we plot the net torque versus the angular
acceleration, do you get a straight line or not? What
is the theoretical expression for the slope? What is
the theoretical value of the intercept on the graph?
Make sure you understand the answers to these
questions by comparing the above equation with the
generic equation for a straight line where x and y
are the variables:
[𝑦] = 𝑎[𝑥] + 𝑏
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It’s essential to measure the net torque and the angular acceleration independently. The next two sections
explain how to obtain these quantities so they can be filled in on your data table.
4. Finding the net torque values (Στ):
i) Consider the free-body diagram for the hanging mass, and apply Newton’s 2nd
law. Find the tension T in terms of the hanging mass m, g, and the linear
acceleration a:
ii) Consider the rotational version of Newton’s 2nd law, applied to the
rotodyne.
Find the net torque in terms of the tension T in the string and R, the
distance from the center of the wheel to the point of contact with the
string:
Note that there is a lip at the edge of the wheel that must be excluded from the distance R.
Combining the results from i) and ii) above, we get
Σ τ = 𝑚 𝑅(𝑔 − 𝑎)
Where m = hanging mass, and a = linear acceleration of the hanging mass. Use the above equation to find
the values in the last column of the tables.
5. Finding the angular acceleration values (α):
The hanging mass accelerates uniformly from rest, and we will use a stopwatch to find how long it takes to
move a fixed distance. With this information, the equations of constant acceleration can be used to find the
acceleration. Taking the downwards direction as positive, we list the usual five variables:
𝑥 = height of fall = h;
𝑎 = quantity to find;
𝑣0 = 0 m/s;
𝑣 = not known or needed;
𝑡 = time found with stopwatch
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Use the space above to show from the relevant equation that
𝑎=
2ℎ
𝑡2
Since the string wraps around the rotodyne, this acceleration is also the tangential acceleration of points at
distance R from the center of the wheel. The angular acceleration α can be found from:
𝛼=
𝑎
𝑅
Use this equation to find the second-to-last column in each of the tables.
6. Finding the moments of inertia of the orbiters alone:
Moments of inertia about a fixed axis are additive quantities, which means that:
𝐼𝑟𝑜𝑡𝑜 𝑎𝑛𝑑 𝑜𝑟𝑏𝑖𝑡𝑒𝑟𝑠 = 𝐼𝑟𝑜𝑡𝑜 + 𝐼𝑜𝑟𝑏𝑖𝑡𝑒𝑟𝑠
So, we can solve for the moment of inertia of the orbiters alone:
𝐼𝑜𝑟𝑏𝑖𝑡𝑒𝑟𝑠 = 𝐼𝑟𝑜𝑡𝑜 𝑎𝑛𝑑 𝑜𝑟𝑏𝑖𝑡𝑒𝑟𝑠 − 𝐼𝑟𝑜𝑡𝑜
Since we found the moment of inertia of the rotodyne alone, we can do this subtraction to find the result
needed.
7. What to turn in:
i) The cover page with your hand-written original data for times, and the results in the boxes below
the tables.
ii) Excel data tables. Be sure to follow the guidelines – units, alignment, significant digits, headings
etc.
iii) The three graphs, with appropriate (different!) headings and all the usual requirements. They can be
put on the same set of axes if you wish.
iv) Any other things requested by your instructor.
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Moment of Inertia: Cover Page
Name:
Partner: _______________________
height (m):
Lab day: ________
time: _________
average orbiter mass (kg):
radius R (m):
1. Rotodyne Alone:
Mass (kg)
t 1 (s)
t 2 (s)
t 3 (s)
t ave (s)
a (m/s2)
a (rad/s2)
t (mN)
a (m/s2)
a (rad/s2)
t (mN)
a (rad/s2)
t (mN)
0.050
0.075
0.100
0.125
0.150
Moment of inertia I of
rotodyne (kg m2):
2. Rotodyne and 2 orbiting masses (at outermost position)
Mass (kg)
t 1 (s)
t 2 (s)
t 3 (s)
t ave (s)
0.050
0.075
0.100
0.125
0.150
I for rotodyne + 2
orbiters (kg m2):
I for the two orbiting
masses only (kg m2):
l = orbiter distance
(m)
calculated I for the two
orbiters (kg m2):
3. Rotodyne and 4 orbiting masses (at outermost position)
Mass (kg)
t 1 (s)
t 2 (s)
t 3 (s)
t ave (s)
a (m/s2)
0.050
0.075
0.100
0.125
0.150
I for rotodyne + 4
orbiters (kg m2):
I for the 4 orbiting
masses only (kg m2):
l = orbiter distance
(m)
calculated I for the 4
orbiters (kg m2):
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