Moment of inertia (experiment)

[VVS2000]

Newton's first law:
"An object at rest stays at rest and an object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced force."

What does the above law say about the motion of an object that begins at rest and is subject to no forces?

It's not a trick question. It is a simple question intended to make you stop and think. But you have to actually answer it.

It says that the object won't accelerate unless acted by an external force.
But I don't know where you are going with this

 

[VVS2000]

We can help you do that. What was the experiment you did and how did you do it?

Ok the experiment I did was finding the moment of inertia of flywheel. Instead of the black object in the above figure, I had a flywheel and the thread was wound around the axle.
In the case of flywheel, the whole thing rotates. But there, only the top platform rotates and with the presence of ball bearings

 

[kuruman]

Ok the experiment I did was finding the moment of inertia of flywheel. Instead of the black object in the above figure, I had a flywheel and the thread was wound around the axle.
In the case of flywheel, the whole thing rotates. But there, only the top platform rotates and with the presence of ball bearings

And what readings did you take? How did you relate these readings to the moment of inertia?

 

[VVS2000]

And what readings did you take? How did you relate these readings to the moment of inertia?

Is this question regarding the previous experiment or the present one?

 

[sophiecentaur]

Plz tell what and all confusions you have about this setup.

The confusions are there because there is not a full description of the apparatus and there is not a full description of the method. Each of the helpers who have answered you will have a different picture in their heads of what you are trying to achieve so some of them will have been wasting their time trying to answer a 'different' question.
Please describe in full what the experiment involves and what measurements you will make. Many of us have done similar lab experiments with similar equipment so I am sure the answer is there, if only you ask a specific question. There are dozens of alternative experiments possible with this sort of equipment.
I'm afraid that describing what you are doing as 'trying stuff and seeing what happens' won't get us (or you) very far. You may be finding it hard to put your problem formally but you really need to try.

 

[kuruman]

The following is equivalent to what I assumed above. It's for sure not a very accurate way to measure the moment of inertia, but in principle it's possible:

http://webpages.ursinus.edu/tcarroll/labmanual/node9.html

This method could be quite accurate if one uses LoggerPro as is done according to the link. The link does not specify the method of extracting the moment of inertia from the data. When I taught this, the students measured the time interval between successive interruptions of a photogate after rotation by $\pi$. They fitted that to a polynomial and extracted the "true" $t=0$ when motion starts as opposed to "photogate" $t=0$ which is the time the photogate is interrupted for the first time. They found "true" $t=0$ by taking the time derivative of the polynomial and setting it equal to zero. Then they took the second derivative of the polynomial and evaluated it at "true" $t=0$ to get the acceleration $\alpha_0$ at $t=0$. Why is that important? Because "true" $t=0$ is the only time when the tension is $T=mg$ and the moment of inertia is simply $I=mgR/\alpha_0$. It worked very well and taught to the students not only rotational dynamics, but also the value of fitting procedures to finding unknowable parameters.

To @VVS2000 : This is an example of method or procedure.

 

[jbriggs444]

It says that the object won't accelerate unless acted by an external force.
But I don't know where you are going with this

You have spoken about a concern that ball bearings cause a system to move on its own. That concern is misplaced. Ball bearings eliminate an external force. They do not create one.

If the system moves on its own when you install ball bearings, it is likely because you have neglected a different external force that was already there. The ball bearings are not the problem. They are a solution that allowed the real problem to manifest.

You can get a very reproducible measurement for moment of inertia by gluing the wheel into place. But that measurement will reflect the properties of the glue more than it reflects the properties of the wheel.

 

[kuruman]

Is this question regarding the previous experiment or the present one?

The previous one. Then we will see if and how it can be adapted to the present one.

 

[VVS2000]

You have spoken about a concern that ball bearings cause a system to move on its own. That concern is misplaced. Ball bearings eliminate an external force. They do not create one.

If the system moves on its own when you install ball bearings, it is likely because you have neglected a different external force that was already there. The ball bearings are not the problem. They are a solution that allowed the real problem to manifest.

You can get a very reproducible measurement for moment of inertia by gluing the wheel into place. But that measurement will reflect the properties of the glue more than it reflects the properties of the wheel.

Ok so you are telling that ball bearings have no affect on the force by the suspended mass?

 

[jbriggs444]

Ok so you are telling that ball bearings have no affect on the force by the suspended mass?

Yes. The fact that you are asking this question leads me to question whether you know what a force is.

 

[VVS2000]

Yes. The fact that you are asking this question leads me to question whether you know what a force is.

Well ok. You can ask that question coz now with the glue analogy you gave, won't the properties of the ball bearings come into picture here? I am really confused here...

 

[VVS2000]

The previous one. Then we will see if and how it can be adapted to the present one.

https://vlab.amrita.edu/?sub=1&brch=74&sim=571&cnt=1
I don't have the manual but yeah it was very similar to this above link

 

[jbriggs444]

Well ok. You can ask that question coz now with the glue analogy you gave, won't the properties of the ball bearings come into picture here? I am really confused here...

The slipperier the ball bearings are, the less they intrude. The stickier the glue is, the more that it intrudes. You want a wheel that rotates freely at the slightest touch.

Ball bearings good.
Glue bad.

Teflon good
Steel bad

Molasses bad
Motor oil good

Bronze bushings good
Ball bearings better

 

[kuruman]

Graphite powder good
Sand bad.

 

[sophiecentaur]

The slipperier the ball bearings are, the less they intrude. The stickier the glue is, the more that it intrudes. You want a wheel that rotates freely at the slightest touch.

Ball bearings good.
Glue bad.

Teflon good
Steel bad

Molasses bad
Motor oil good

Bronze bushings good
Ball bearings better

I wonder how many of those comparisons would be revealed by thet OP's experiment. I know that the everyday cataloguing [sorry datalogging] equipment is a lot better than the old 'ticker tape timers' used to work. Ahh those were the days; not a computer in sight.

 

[gmax137]

Ok I will try putting up a diagram

Please do.

 

[Mister T]

But it's here when I realized that my angular acceleration could be affected by this ball bearing system and my value might be less.

Less than what? There is a torque on the disk that depends on both its momentum of inertia and the torque applied by the bearing. What did you get for the ratio of the torque to the angular acceleration?

 

[Mister T]

Ahh those were the days; not a computer in sight.

You were the computer!

 

[sophiecentaur]

You were the computer!

Yep. We did the sums on paper but the kids I taught used their calculators. Also, a bar graph, using lengths of tape, five ticks long, gave a v/t graph. Plus the area underneath was the distance traveled (= the total length of tape)

 

[VVS2000]

The slipperier the ball bearings are, the less they intrude. The stickier the glue is, the more that it intrudes. You want a wheel that rotates freely at the slightest touch.

Ball bearings good.
Glue bad.

Teflon good
Steel bad

Molasses bad
Motor oil good

Bronze bushings good
Ball bearings better

Again so sry. For a moment I understood what you said. But let me just explain my present dilemma.
Let's say you're walking on a road and you want to calculate some physical quantity given that you're walking on a road. Now obviously due to friction you won't get the accurate result so you do something to reduce the friction. Now what if in the process of reducing the friction you make the road itself very very soft, almost like ice for example. This is actually how I see the influence of ball bearings.
Plz correct me if there's a mistake

 

[kuruman]

When you walk on the road friction is needed to provide the forward acceleration. If you remove friction, you will not be able to propel yourself forward. For example, if you put on roller blades (they got ball bearings) the soles of your feet are decoupled from the road surface and you will not be able to move forward the way you normally do when you walk. However you will be able to walk normally on roller blades if you inject Gorilla
-Glue into the ball bearings and wait a few minutes for it to set. Do not try this at home unless you want to ruin a perfectly good pair of roller blades.

Now what if in the process of reducing the friction you make the road itself very very soft, almost like ice for example.

Ice hard
Marshmallows soft.

 

[jbriggs444]

Again so sry. For a moment I understood what you said. But let me just explain my present dilemma.
Let's say you're walking on a road and you want to calculate some physical quantity given that you're walking on a road.

Let us make that specific. You want to calculate your mass. You want to do this by having someone pull on you with a string and measure how long it takes to move a certain distance. So you strap on a pair of tennis shoes and stand on the road while your friend pulls on you with a 10 Newton force.

You don't budge. Your feet stay firmly planted on the road.

You swap out for a pair of roller skates. Your friend pulls and you move on down the road.

Which choice of footwear provided a useful measurement?

 

[gmax137]

Ok I will try putting up a diagram

Please do.

 

[sophiecentaur]

Ok I will try putting up a diagram

And a proposed method with the input variables and the output variable. What do you actually want to do? This is all so up in the air that I cannot see how any 'learning' can emerge from the process.

 

[VVS2000]

Let us make that specific. You want to calculate your mass. You want to do this by having someone pull on you with a string and measure how long it takes to move a certain distance. So you strap on a pair of tennis shoes and stand on the road while your friend pulls on you with a 10 Newton force.

You don't budge. Your feet stay firmly planted on the road.

You swap out for a pair of roller skates. Your friend pulls and you move on down the road.

Which choice of footwear provided a useful measurement?

Ok, fine but what if the whole road is made extremely smooth so as to slip?
Can you explain whether the influence of ball bearings is that of the roller blades or like a super smooth road? Or are both cases similar?

 

[VVS2000]

And a proposed method with the input variables and the output variable. What do you actually want to do? This is all so up in the air that I cannot see how any 'learning' can emerge from the process.

So sorry, I will.

 

[jbriggs444]

Ok, fine but what if the whole road is made extremely smooth so as to slip?
Can you explain whether the influence of ball bearings is that of the roller blades or like a super smooth road? Or are both cases similar?

Similar.

Roller skates on ice, tennis shoes on ice or roller skates on asphalt all work to minimize friction and allow you to move freely over the road surface. This reduces the contribution of the frictional force to:
$$F_{\text{friction}}+F_{\text{string}}=ma$$
If you are trying to determine m, based on the known $F_\text{string}$ and the measured a then the unknown $F_\text{friction}$ impairs your ability to do so. You want to reduce the unknown frictional force as much as possible.

 

[VVS2000]

Similar.

Roller skates on ice, tennis shoes on ice or roller skates on asphalt all work to minimize friction and allow you to move freely over the road surface. This reduces the contribution of the frictional force to:
$$F_{\text{friction}}+F_{\text{string}}=ma$$
If you are trying to determine m, based on the known $F_\text{string}$ and the measured a then the unknown $F_\text{friction}$ impairs your ability to do so. You want to reduce the unknown frictional force as much as possible.

Ok, now I got it...!

 

[Fred Wright]

I have the following suggestions for your experiment:
First, I would make the two pulleys to be identically the same (this greatly simplifies the analysis because you won't to deal with two unknown moments of inertia). I have made a sketch of your experiment.

In the following I calculate the moment of inertia of the pulley in terms of the time it takes for the mass to descend from height h to the floor using the technique of Lagrangian dynamics. Lagrangian dynamics is usually introduced in the upper division physics sequence but you can (and should) do a force analysis to verify my result. The Lagrangian of a system is,$$\mathcal {L}=K.E. - P.E.$$where K.E. is the kinetic energy and P.E. is the potential energy. For this system,$$K.E.= I \dot{\theta}^2+\frac{1}{2}m \dot{y}^2\\
P.E.=mg(h-y)$$where $\theta$ is the angle of rotation of the pulley. The arc length for a circle is $s=R\theta$ and with a moment of thought you can see that $\theta = \frac{y}{R}$. The Lagrangian is therefore,$$ \mathcal {L}=I\frac{ \dot{y}^2}{R^2}+\frac{1}{2}m \dot{y}^2-(h-y)mg$$I suggest you google Lagrangian mechanics to understand the following sequence of calculations,$$
\frac{\partial {\mathcal{L}}}{\partial{\dot{y}}}=2\dot{y}(\frac{I}{R^2} + \frac{1}{2}m)\\
\frac {d(\frac{\partial {\mathcal{L}}}{\partial{\dot{y}}})}{dt}=2\ddot{y}(\frac{I}{R^2} + \frac{1}{2}m)\\
\frac{\partial {\mathcal{L}}}{\partial{y}}=mg\\
2\ddot{y}(\frac{I}{R^2} + \frac{1}{2}m)=mg\\
$$We now integrate the acceleration twice assuming the initial velocity is zero, the initial displacement is zero, and the final position is h.$$
\ddot{y}=\frac{1}{2}(\frac{mg}{\frac{I}{R^2}+ \frac{1}{2}m})\\
\dot{y}-\dot{y_0}=\frac{1}{2}(\frac{mgt}{\frac{I}{R^2}+ \frac{1}{2}m})\\
\dot{y_0}=0\\
y-y_0=\frac{1}{4}(\frac{mgt^2}{\frac{I}{R^2}+ \frac{1}{2}m})\\
y_0=0\\
I=\frac{mR^2}{2h}(\frac{gt^2}{2}-h)

$$

 

[hutchphd]

Roller skates on ice, tennis shoes on ice or roller skates on asphalt all work to minimize friction and allow you to move freely over the road surface.

It should be at least noted that Ice skates and roller states will produce a slightly different result for the same total mass because of the rotational inertia of the wheels. It may be slightly less obvious that the same thing is true for a ball bearing vs a sleeve bearing for essentially the same reason.
Also if the frictional force is independent of the speed can it not be eliminated using several different accelerating rates and creating the appropriate intercept?
A good drawing would be a friendly addition... @VVS2000?.

 

[vanhees71]

I have the following suggestions for your experiment:
First, I would make the two pulleys to be identically the same (this greatly simplifies the analysis because you won't to deal with two unknown moments of inertia). I have made a sketch of your experiment.
View attachment 258601
In the following I calculate the moment of inertia of the pulley in terms of the time it takes for the mass to descend from height h to the floor using the technique of Lagrangian dynamics. Lagrangian dynamics is usually introduced in the upper division physics sequence but you can (and should) do a force analysis to verify my result. The Lagrangian of a system is,$$\mathcal {L}=K.E. - P.E.$$where K.E. is the kinetic energy and P.E. is the potential energy. For this system,$$K.E.= I \dot{\theta}^2+\frac{1}{2}m \dot{y}^2\\
P.E.=mg(h-y)$$where $\theta$ is the angle of rotation of the pulley. The arc length for a circle is $s=R\theta$ and with a moment of thought you can see that $\theta = \frac{y}{R}$. The Lagrangian is therefore,$$ \mathcal {L}=I\frac{ \dot{y}^2}{R^2}+\frac{1}{2}m \dot{y}^2-(h-y)mg$$I suggest you google Lagrangian mechanics to understand the following sequence of calculations,$$
\frac{\partial {\mathcal{L}}}{\partial{\dot{y}}}=2\dot{y}(\frac{I}{R^2} + \frac{1}{2}m)\\
\frac {d(\frac{\partial {\mathcal{L}}}{\partial{\dot{y}}})}{dt}=2\ddot{y}(\frac{I}{R^2} + \frac{1}{2}m)\\
\frac{\partial {\mathcal{L}}}{\partial{y}}=mg\\
2\ddot{y}(\frac{I}{R^2} + \frac{1}{2}m)=mg\\
$$We now integrate the acceleration twice assuming the initial velocity is zero, the initial displacement is zero, and the final position is h.$$
\ddot{y}=\frac{1}{2}(\frac{mg}{\frac{I}{R^2}+ \frac{1}{2}m})\\
\dot{y}-\dot{y_0}=\frac{1}{2}(\frac{mgt}{\frac{I}{R^2}+ \frac{1}{2}m})\\
\dot{y_0}=0\\
y-y_0=\frac{1}{4}(\frac{mgt^2}{\frac{I}{R^2}+ \frac{1}{2}m})\\
y_0=0\\
I=\frac{mR^2}{2h}(\frac{gt^2}{2}-h)

$$

That's precisely what I gave some days ago. Note that the rotational kinetic energy is $I/2 \dot{\theta}^2$ though.

 

[Fred Wright]

That's precisely what I gave some days ago. Note that the rotational kinetic energy is $I/2 \dot{\theta}^2$ though.

After I posted I suddenly realized that demanding the alteration of the apparatus to have identical pulleys might not sit well with the lab instructor. The solution, however, can accommodate two pulleys by making the substitution, $$I=\frac{mR^2}{2h}(\frac{gt^2}{2}-h)\\
I\to \frac{1}{2}(I_1+I_2)\\
\frac{1}{2}(I_1+I_2)=\frac{mR^2}{2h}(\frac{gt^2}{2}-h)$$
To solve for the two moments of inertia, run the experiment with two different masses. You now have two equations with two unknowns.

 

[VVS2000]

That's precisely what I gave some days ago. Note that the rotational kinetic energy is $I/2 \dot{\theta}^2$ though.

Thanks for the proof