Big Mechanics Review Problem Solution
Measuring Rotational Inertia
This was a scenario where the barbell below was centered on top of a rotating platform. The platform had a string wrapped around its spool, and that string ran horizontally, and then was hung over another pulley of negligible mass, and the mass m descended vertically from that negligible mass pulley. This was diagrammed in class. The mass m took a measurable amount of time to descend the height h while the platform rotated and swept out an angle θ in that time. The mass descent time was measured twice: once with the platform connected to the string and NO barbell attached to it and then again with the platform connected to the string and the barbell centered on top of the platform. tempty and tloaded are the names for these two descent times. In any trial, everything is started from a state of rest. R is the radius of the spool attached horizontally to the platform around which the string wrapped. The data below were measured in 2015.
Objective: To determine the Moment of Inertia of the Symmetrical Object
First Solution: In class, the first method of solution was summarized. It was centered around Newton’s 2nd Law. The FBD of the platform yields the following version of Στ = Iα applied to the platform: TR = Iα, where T is the tension in the string. The FBD of the mass m yields the following version of ΣF = ma: mg – T = ma, where T is the tension in the string. Combining these together along with combining them with the constant kinematics ideal, h = ½ at2 AND with the Chapter 8 definition α = a/R, yields: I = mR2[((½gt2)/h) – 1]. The purpose of this paper is for you to see if you can get the same thing using an energy budget approach.
Solutions in bold italics
Diagrams, apparatus and symmetrical object by itself:
x1
Lrod
Measured Givens:
h = 0.5 m
tloaded = 3.21 s
tempty = 1.15 s
R = 0.02548 m
m = 20 g
For #9 Only:
Lrod = 38 cm
Mrod = 26.7 g
m1 = m2 = 82.1 g
x1 = x2 = 7 cm
Objective: To determine the Moment of Inertia of the Symmetrical Object
For 1-7, you are to answer only for the situation when the platform was empty:
1. Find the hanging mass’s acceleration.
a = 2h/t2
2. Find the final instantaneous velocity of the hanging mass.
v = at = (2h/t2)t = 2h/t (This is also double the average velocity during t.)
3. Find the final instantaneous angular velocity of the rotating stuff.
omega = v/R = 2h/(tR)
4. State the amount of TOTAL kinetic energy that the system has at the final instant.
Ktot = mgh (because the KLost = Ugained. This K is in TWO kinds of motion.)
5. Find the amount of translational kinetic energy that the system has at the final instant.
Ktrans = ½mv2 = m(2h/t)2/2 = 2mh2/t2
6. State the amount of rotational kinetic energy that the system has at the final instant.
Krot = Ktot – Ktrans = mgh – 2mh2/t2 (the total K is in TWO places.)
7. Solve for I of the empty platform. Do it symbolic because I say so:
½ Iω2 = Krot = mgh – 2mh2/t2 Because Krot = ½ Iω2
And now we just plug in all previous results of this page to isolate I. Notice that it comes out identically to how it came our when only using Newton’s Laws:
½ I(2h/tR)2 = mgh – 2mh2/t2
I = mR2[[(0.5)gt2/h] – 1] after some simplifying
Hopefully, you are appreciating how neat the symbolic solution turns out. Look at all those familiar groups of expressions in the symbolic answer. This must reveal patterns and laws that can be explored. For example,
(0.5)gt2 is in the answer and is the distance that an object would travel if it were in free-fall for the particular time t.
8. Now you may plug in data, and solve for the I of the platform when it was empty. You can plug in tempty, h, g, m, and R. (You may not use the data in the column to the right.) Using tempty yields the I of the platform with no barbell on it. Then use the same expression to calculate the I of the platform with a barbell on it. The difference between these two I’s should be the I of just the barbell.
Answer: Ibarbell = ______kgm2
9. Use the published I formulas of rods and point masses to determine the theoretical moment of inertia of the symmetrical object whose essential measurements are given above.
This would mean you are treating the rod with the theoretical ideal, Irod = (MrodLrod2)/12, and you are treating each point mass with the theoretical ideal, Ipoint = m1x12 = m2x22, and the measured values of all these quantities are in the right column of the data sheet. Add the moment of inertia of the rod and of each point mass all together, and that should be the theoretical I of this barbell.
Answer: 0.001117 kgm2
10. #8 and #9 are supposed to match up pretty well as a check of your understanding.
That’s a nice all-encompassing problem to end the year.
Shortcut: You can subtract mR2[[(0.5)gtempty2/h] – 1] from mR2[[(0.5)gtloaded2/h] – 1] BEFORE plugging any numbers in, and when you do, stuff cancels and you get
I = (R2mg/2h)(tloaded2 – tempty2) = [(0.025482 m2)(0.196 N)/(1 m)](3.212 – 1.152) s2
for the experimental solution for the I of the barbell.