Question

Figure 10.32 shows an apparatus used to measure rotational inertias of various objects, in this case spheres of varying masses $M$ and radii $R .$ The spheres are made of different materials, and some are hollow while others are solid. To perform the experiment, a sphere is mounted to a vertical axle held in a frame with essentially frictionless bearings. A spool of radius $b=2.50 \mathrm{cm}$ is also mounted to the axle, and a string is wrapped around the spool. The string runs horizontally over an essentially frictionless pulley and is tied to a mass $m=77.8 \mathrm{g} .$ As the mass falls, the string imparts a torque to the spool/axle/disk combination, resulting in angular acceleration. The mass of the string is negligible, but the combination of axle and spool has non-negligible rotational inertia $I_{0}$ whose value isn't known in advance. In each experimental run, the mass $m$ is suspended a height $h=1.00 \mathrm{m}$ above the floor and the rotating system is initially at rest. The mass is released, and experimenters measure the time to reach the floor. Results are given in the tables below. Determine an appropriate function of the time $t$ which, when plotted against other quantities including $M$ and $R,$ should yield two straight lines-one for the hollow spheres and one for the solid ones. Plot your data, establish best-fit lines, and use the resulting slopes to verify the numerical factors $2 / 5$ and $2 / 3$ in the expressions for the rotational inertias of spheres given in Table $10.2 .$ You should also find a value for the rotational inertia of the axle and drum together. (FIGURE CANNOT COPY) $$\begin{array}{|r|r|r|r|r|r|}\hline \text { Sphere mass } M(\mathrm{g}) & 783 & 432 & 286 & 677 & 347 \\\hline \text { Sphere radius } R(\mathrm{cm}) & 6.25 & 3.86 & 9.34 & 9.42 & 9.12 \\\hline \text { Fall time } t(\mathrm{s}) & 2.36 & 1.22 & 2.72 & 3.24 & 2.91 \\ \hline\end{array}$$ $$\begin{array}{|r|r|r|r|r|r|}\hline \text { Sphere mass } M(\mathrm{g}) & 947 & 189 & 821 & 544 & 417 \\\hline \text { Sphere radius } R(\mathrm{cm}) & 6.71 & 5.45 & 6.55 & 4.67 & 9.98 \\\hline \text { Fall time } t(\mathrm{s}) & 2.75 & 1.41 & 2.51 & 1.93 & 3.47 \\\hline \end{array}$$

Name: Problem 79, Chapter 10: Essential University Physics
Uploaded: 2020-05-11T20:26:25-08:00
Duration: 35 min 47 s
Channel: Donald Albin

   Figure 10.32 shows an apparatus used to measure rotational inertias of various objects, in this case spheres of varying masses $M$ and radii $R .$ The spheres are made of different materials, and some are hollow while others are solid. To perform the experiment, a sphere is mounted to a vertical axle held in a frame with essentially frictionless bearings. A spool of radius $b=2.50 \mathrm{cm}$ is also mounted to the axle, and a string is wrapped around the spool. The string runs horizontally over an essentially frictionless pulley and is tied to a mass $m=77.8 \mathrm{g} .$ As the mass falls, the string imparts a torque to the spool/axle/disk combination, resulting in angular acceleration. The mass of the string is negligible, but the combination of axle and spool has non-negligible rotational inertia $I_{0}$ whose value isn't known in advance. In each experimental run, the mass $m$ is suspended a height $h=1.00 \mathrm{m}$ above the floor and the rotating system is initially at rest. The mass is released, and experimenters measure the time to reach the floor. Results are given in the tables below. Determine an appropriate function of the time $t$ which, when plotted against other quantities including $M$ and $R,$ should yield two straight lines-one for the hollow spheres and one for the solid ones. Plot your data, establish best-fit lines, and use the resulting slopes to verify the numerical factors $2 / 5$ and $2 / 3$ in the expressions for the rotational inertias of spheres given in Table $10.2 .$ You should also find a value for the rotational inertia of the axle and drum together. (FIGURE CANNOT COPY)
$$\begin{array}{|r|r|r|r|r|r|}\hline \text { Sphere mass } M(\mathrm{g}) & 783 & 432 & 286 & 677 & 347 \\\hline \text { Sphere radius } R(\mathrm{cm}) & 6.25 & 3.86 & 9.34 & 9.42 & 9.12 \\\hline \text { Fall time } t(\mathrm{s}) & 2.36 & 1.22 & 2.72 & 3.24 & 2.91 \\
\hline\end{array}$$
$$\begin{array}{|r|r|r|r|r|r|}\hline \text { Sphere mass } M(\mathrm{g}) & 947 & 189 & 821 & 544 & 417 \\\hline \text { Sphere radius } R(\mathrm{cm}) & 6.71 & 5.45 & 6.55 & 4.67 & 9.98 \\\hline \text { Fall time } t(\mathrm{s}) & 2.75 & 1.41 & 2.51 & 1.93 & 3.47 \\\hline \end{array}$$

Essential University Physics

Richard Wolfson 3rd Edition

Chapter 10, Problem 79

Instant Answer

Solved by Expert Donald Albin

05/11/2020

Step 1

$I_{\text{subzero}} + mB^2 = 0.978$ $I_{\text{subzero}} = 0.978 - mB^2$ $I_{\text{subzero}} = 0.978 - (0.778 \times 0.025)^2$ $I_{\text{subzero}} = 0.978 - 4.9 \times 10^{-5}$ $I_{\text{subzero}} = 0.978 - 0.000049$ $I_{\text{subzero}} = Show more…

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Figure 10.32 shows an apparatus used to measure rotational inertias of various objects, in this case spheres of varying masses $M$ and radii $R .$ The spheres are made of different materials, and some are hollow while others are solid. To perform the experiment, a sphere is mounted to a vertical axle held in a frame with essentially frictionless bearings. A spool of radius $b=2.50 \mathrm{cm}$ is also mounted to the axle, and a string is wrapped around the spool. The string runs horizontally over an essentially frictionless pulley and is tied to a mass $m=77.8 \mathrm{g} .$ As the mass falls, the string imparts a torque to the spool/axle/disk combination, resulting in angular acceleration. The mass of the string is negligible, but the combination of axle and spool has non-negligible rotational inertia $I_{0}$ whose value isn't known in advance. In each experimental run, the mass $m$ is suspended a height $h=1.00 \mathrm{m}$ above the floor and the rotating system is initially at rest. The mass is released, and experimenters measure the time to reach the floor. Results are given in the tables below. Determine an appropriate function of the time $t$ which, when plotted against other quantities including $M$ and $R,$ should yield two straight lines-one for the hollow spheres and one for the solid ones. Plot your data, establish best-fit lines, and use the resulting slopes to verify the numerical factors $2 / 5$ and $2 / 3$ in the expressions for the rotational inertias of spheres given in Table $10.2 .$ You should also find a value for the rotational inertia of the axle and drum together. (FIGURE CANNOT COPY) $$\begin{array}{|r|r|r|r|r|r|}\hline \text { Sphere mass } M(\mathrm{g}) & 783 & 432 & 286 & 677 & 347 \\\hline \text { Sphere radius } R(\mathrm{cm}) & 6.25 & 3.86 & 9.34 & 9.42 & 9.12 \\\hline \text { Fall time } t(\mathrm{s}) & 2.36 & 1.22 & 2.72 & 3.24 & 2.91 \\ \hline\end{array}$$ $$\begin{array}{|r|r|r|r|r|r|}\hline \text { Sphere mass } M(\mathrm{g}) & 947 & 189 & 821 & 544 & 417 \\\hline \text { Sphere radius } R(\mathrm{cm}) & 6.71 & 5.45 & 6.55 & 4.67 & 9.98 \\\hline \text { Fall time } t(\mathrm{s}) & 2.75 & 1.41 & 2.51 & 1.93 & 3.47 \\\hline \end{array}$$

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Key Concepts

Experimental Data Analysis in Rotational Dynamics

Extracting physical parameters from experimental data often involves plotting combinations of measured quantities to yield linear relationships. By appropriately choosing functions of variables such as fall time, mass, and radius, one can use linear regression to determine slopes and intercepts that correspond to theoretical expressions in rotational dynamics, thereby validating or refining theoretical models.

Relationship Between Linear and Angular Quantities

In systems where there is a conversion between translational and rotational motion, such as through a string wrapped around a spool, linear acceleration of the mass is directly related to the angular acceleration of the rotating system. This interconnection is crucial for analyzing experiments that involve both linear kinetics and rotational dynamics.

Moment of Inertia of Spheres

Different objects have characteristic formulas for their moments of inertia based on their geometry and mass distribution. For spheres, the moment of inertia takes on specific forms—such as (2/5)MR² for a solid sphere and (2/3)MR² for a hollow sphere—which are important for experimentally verifying theoretical predictions in rotational mechanics.

Torque and Angular Acceleration

Torque is the rotational equivalent of force and is responsible for causing angular acceleration in a rotating body. According to Newton’s second law for rotation, the net torque acting on an object is equal to the product of its rotational inertia and its angular acceleration, linking these key concepts in rotational dynamics.

Rotational Inertia

Rotational inertia, also known as moment of inertia, quantifies an object's resistance to changes in its rotational motion. It depends on the mass distribution relative to the axis of rotation and is a central concept in understanding and analyzing the dynamics of rotating systems.

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Transcript

00:01 Okay, so we have a very long problem here.

00:06 I took the liberty of drawing it, so i'm going to try to point to things in blue as necessary as i'm reading.

00:20 Also, i did use microsoft excel to solve this problem, and i can show this to you as well.

00:29 It says figure 10 .32, which i tried to draw on the screen, shows an apparatus used to measure rotational inertia of various objects.

00:43 In this case, spheres of varying masses, m, and radii are.

00:54 The spheres are made of different materials, and some are hollow, while others are solid.

01:05 Now, one thing that i know is that inertia for a hollow sphere is, let me look for that, hollow sphere is two -thirds mr squared.

01:27 And inertia for a solid sphere is two -fifths mr -squared.

01:41 So as it spoke, the question spoke about hollow spheres and solid spheres, it made me think of the different coefficients that come in front of mr squared.

01:57 It's either two -fifths or two -thirds.

01:59 Okay, let me continue reading.

02:02 To perform the experiment, a sphere is mounted to a vertical axle held in a frame within a frame.

02:11 Essentially frictionless bearings.

02:15 So up here is the sphere.

02:21 A spool of radius b, so i wrote radius b on there, equals 2 .5 centimeters.

02:31 So i should write that down, but i'm going to write 0 .0 to 5 meters, okay, is also mounted to the axle.

02:51 And a string is wrapped around the spool.

02:54 The string runs horizontally over an essentially frictionless pulley and is tied to a mass m.

03:03 So we can see that over here, tying to m, and m is 77 .8 grams, but i would like to write it as 0 .078, kilograms.

03:24 I generally like to write things in si base units.

03:29 As the mass falls, the string imparts a torque to the spool slash axle slash disk combination resulting in angular acceleration.

03:42 So that mass is going to accelerate downward.

03:46 And at the same time, the spool is going to spin.

03:50 It's going to accelerate.

03:52 It's spinning, and the mass is tied to the pull by that axle.

04:01 The mass of the string is negligible, but the combination of axle and spool has non -negligible rotational inertia i -subs -zero.

04:15 We're going to read that we need to figure out what that i -subs -zero is.

04:21 That value is not known in advance.

04:25 In each experimental run, the mass is suspended at a height of 1 .00 meters.

04:33 So i wrote that down here.

04:35 It looks like i should have had two zeros.

04:38 Okay, so that's the height.

04:40 Above the floor and the rotating system is initially at rest.

04:46 So our initial angular velocity and our initial speed are both zero.

05:00 The mass is released and experimenters measure the time to reach the floor.

05:07 Results are given in the tables below.

05:11 Determine an appropriate function of time, t, which, when plotted against other quantities, including m and r, should yield two straight lines, one for the hollow spheres and one for the solid ones.

05:29 Plot your data established best fit lines and use the resulting slopes to verify the numerical factors two -fifths and two -thirds.

05:41 And i've already written those down.

05:43 They're in the expressions for the rotational inertia of spheres given in table 10 .2.

05:51 You should also find the value of the rotational inertia of the axle and the drum together.

05:58 They don't mean drum there.

06:00 I think that's a typo.

06:02 They mean the spool, the axle and the spool.

06:06 So the axle and the spool are rotating together.

06:11 So they're going to have one total inertia.

06:18 Okay.

06:21 That seems like a lot.

06:23 Nevertheless, the first thing that i'm going to do is i'm going to write the sum of of the forces equals mass times acceleration.

06:38 And i'm going to draw that these are the forces on the mass element.

06:45 So on that mass element, there is weight, which is m times g, and there's also tension.

06:58 Now we know that it's going to accelerate downward.

07:03 So i'm going to set my action, so that a is downward.

07:11 So what that's going to give me is, when i write this down, the sum of the forces are t minus mg, or rather mg minus t, because i'm taking downward as the positive direction.

07:42 So let me redo that.

07:49 There we go.

07:50 Let's try this again.

07:53 M .g minus t.

07:57 And that's that.

07:58 That is going to equal the mass times the acceleration.

08:02 We can solve this for acceleration, and we get a equals mg minus t over m, which again shows us that a equals simplifying it a little bit more, g minus t over m.

08:36 Are going to use this later.

08:40 Okay, now i'm going to draw some of the torques.

08:44 So if i draw the spool, i'm going to draw it from above, and i'm going to change my colors here, maybe red, draw the spool from above, the torque, which is the tension times the distance, is pulling from above, and then our acceleration is defined to be in the same direction.

09:20 So, some of the torque equals i alpha.

09:32 Torque.

09:33 Some of the torque is i alpha.

09:36 The torque is the tension times b, which is the radius of the spool, and that equals i alpha t b equals we know that i is some constant i'm going to call that constant k k is either going to be two -thirds or two -fifths times m r squared now um good however our inertia is not just k times m r squared there's also the small inertia that we have to add on.

10:34 So the k times m r squared is the inertia of the sphere.

10:38 That small inertia i sub -zero is the inertia of the spool and maybe of the rod.

10:48 Okay, so now we need to take that times alpha.

10:55 Okay, next thing that we need to know, and i'll change colors here, is that y equal, this is for movement of the mass.

11:05 Y equals y sub 0 plus vy sub 0 t plus 1 half a t squared.

11:20 However, i defined a to be downward.

11:27 And so i need to change that to minus at squared.

11:34 Oops, i wanted that to be green.

11:37 Okay.

11:38 V sub 0 is 0.

11:43 Y at the end is also 0.

11:48 So 0 equals y sub 0.

11:52 Well, y sub 0 was h, which is 1, minus 1 1 1 1⁄2 a t squared.

12:08 And we can solve that for a.

12:12 A equals 2h over t squared.

12:25 Really cutting it short on room here.

12:27 2h over t squared.

12:29 So we're going to need that later.

12:33 And we're going to need this later.

12:40 There's three things we're going to need later.

12:42 We're also going to need something else.

12:44 So let me go to another page.

12:53 A equals alpha r.

12:58 But we don't have an r in this problem.

13:01 As you look back at how we've drawn this problem, there's no lowercase r.

13:06 R is the radius of this sphere, but this r that what i'm talking about now is the radius of the spool, which is actually b.

13:13 So, a equals alpha b.

13:16 And so if i solve that for alpha, we see that alpha is a over b.

13:26 Okay, so now i'm going to go back to the equation, that said, a equals t over m minus g, or g minus t over m, my bad, right here.

13:42 A equals g minus t over m.

13:44 So let me write it over here.

14:08 And that should be a lowercase m.

14:14 It's the mass of the object.

14:21 Okay, but we know what t is.

14:28 So, a equals, let's go back and look what t is.

14:35 I think i'm going back here.

14:43 Okay, so from this equation right here, t is kmr squared plus i sub zero times alpha over b.

14:57 Kmr squared plus i sub zero.

15:07 I've already forgotten.

15:08 Kmr squared.

15:23 Okay, so we got that.

15:26 Kmr squared i sub zero times alpha over m.

15:39 Okay.

15:42 But we know what alpha is from over, oh no, i have that on this page.

15:48 Alpha is a over b.

15:51 So that's going to give us g minus kmr squared plus i subzero over m over b.

16:13 And let's look at what else we've got here.

16:17 We know down here that a is 2h over t squared.

16:30 Okay, i'm not going to use that quite yet.

16:33 What i'm going to do is i'm going to add this to both sides, and i'm going to get that g equals a plus kmr squared plus i.

17:01 Sub 0 over m b a k m r squared plus i sub 0 okay good can factor out the a so that's going to be times 1 plus k m r squared plus i sub 0 over m b okay but now we know a trying to go back to it a is 2 h over t squared g is 2h over t squared times 1 plus k m r squared plus i sub 0 over mb.

18:14 Now what i want to do is i want to factor out the mb, the 1 over mb from the inside.

18:20 So that's going to give me 2h over mb t squared times mb.

18:59 I'm just realizing way back, way back, when i wrote this equation right here, i forgot to divide by b on both sides.

19:14 So i should have written kmr squared plus i sub zero alpha over b.

19:22 So when you realize you have a mistake, just go back and fix it.

19:27 That's going to put, it was right in here, kmr squared.

19:37 So that's going to put a b down here...

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