Moment of inertia (experiment)

[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