# What is the "r" in the moment of inertia?

• vparam
In summary, my professor said that the moment of inertia of a system is equal to the sum of the moment of inertia of each of its masses. However, I don't see why we can use R here rather than the distance from the center of the pulley to the mass. Isn't the moment of inertia = mr^2, where r is the distance the center of the pulley to the position of m1? When we use ##\Sigma m_i r_i^2## to find an object’s moment of inertia, we are implicitly assuming that each individual particle has a circular path (radius ##r_i##) about the chosen axis of rotation (and also that the constituent particles have fixed
vparam
Homework Statement
Find the linear acceleration of an Atwoods machine with a pulley with moment of inertia I and radius R with masses m1 and m2 attached on either end.
Relevant Equations
∑τ = Iα
We solved this problem in class as follows:

Net torque about the center of the pulley taking counterclockwise rotation to be positive = m1gR - m2gR = I_tot α, where I_tot is the moment of inertia of the full system.

My professor said that I_tot = I + m1R^2 + m2R^2, where m1R^2 is the moment of inertia of mass 1 about the center of the pulley (assuming it is a point mass). However, I don't see why we can use R here rather than the distance from the center of the pulley to the mass. Isn't the moment of inertia = mr^2, where r is the distance the center of the pulley to the position of m1? This r ≥ R by the Pythagorean theorem, so I'm not sure where this comes from since the moment of inertia doesn't have a cross product in the formula. In a disk for example, when integrating I = ∫r^2 dm, we use the radial distance from the center, not this R which seems to be the horizontal component of r.

I resolved the problem using angular momentum (and verified with the textbook), and got the same answer as the professor, so it seems that his method does work though. So, where am I going wrong with my understanding of moment of inertia and this value of r?

Perhaps this could help, if I understand your question correctly:
http://hyperphysics.phy-astr.gsu.edu/hbase/mi2.html#rlin

The rotational inertia of the pulley adds up to the linear acceleration of both masses and should decrease the acceleration of the masses with respect to a similar machine with an ideal mass-less pulley.

There is no reason to consider the distances from the center of the pulley to each of the masses, which are constantly changing, as those have no influence on the natural rotational inertia of our massive pulley.

http://faculty.polytechnic.org/phys...undant_Atwood_machine_with_massive_pulley.pdf

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vparam said:
Net torque about the center of the pulley taking counterclockwise rotation to be positive = m1gR - m2gR = I_tot α, where I_tot is the moment of inertia of the full system.

My professor said that I_tot = I + m1R^2 + m2R^2, where m1R^2 is the moment of inertia of mass 1 about the center of the pulley (assuming it is a point mass). However, I don't see why we can use R here rather than the distance from the center of the pulley to the mass. Isn't the moment of inertia = mr^2, where r is the distance the center of the pulley to the position of m1?
When we use ##\Sigma m_i r_i^2## to find an object’s moment of inertia, we are implicitly assuming that each individual particle has a circular path (radius ##r_i##) about the chosen axis of rotation (and also
that the constituent particles have fixed relative positions).

This doesn't apply to an Atwood machine with moving masses. But (by considering angular momentum about the pulley’s axis) it is not too hard to show the effective total moment of inertia is ##I_{tot} = I + m_1R^2 + m_2R^2##. Ideally, your professor should have explained this!

Lnewqban
Steve4Physics said:
This doesn't apply to an Atwood machine with moving masses.
Why not? If the connecting string does not slip on the pulley, the (point) masses have a linear velocity and linear acceleration that matches the linear velocity and linear acceleration of a point on the rim of the pulley. Clearly, the length of the rope on each side does not matter and does not appear in the equations of motion. The FBD of the Atwood machine with a massive pulley leads to an identical angular acceleration (and hence linear acceleration of rope and masses) as the FBD of a massive pulley with additional point masses ##m_1## and ##m_2## attached on the rim at the two ends of a horizontal diameter.

Proof
The external forces acting on the pulley vertical and consist of the weights of the two masses and the reaction force at the center. Only the weights produce torques about the center. Then $$m_2gR-m_1gR=(I_p+m1_R^2+m_2R^2)\alpha \implies \alpha=\frac{(m_2-m_1)gR}{(I_p+m_1R^2+m_2R^2)}.$$The linear acceleration of the rim and masses is $$a=\alpha R=\frac{(m_2-m_1)g}{(I_p/R^2+m_1+m_2)}.$$ This is the same as the linear acceleration of the masses in the Atwood machine with pulley of moment of inertia ##I_p##. Of course, this is an instantaneous value because as soon as the diameter connecting the masses rotates away from the horizontal, the net torque is reduced. So if one wishes to maintain the acceleration constant at its initial maximum value for a finite time interval, one constructs an Atwood machine and connects the masses with an ideal rope of non-zero length. The longer the rope, the more time one has to make measurements.

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Lnewqban
kuruman said:
Why not? If the connecting string does not slip on the pulley, the (point) masses have a linear velocity and linear acceleration that matches the linear velocity and linear acceleration of a point on the rim of the pulley. Clearly, the length of the rope on each side does not matter and does not appear in the equations of motion.
I entirely agree with your reasoning. I didn't make myself clear enough.

When I said:
“This doesn't apply to an Atwood machine with moving masses.”
I should have said:
“This [the use of ##\Sigma m_i r_i^2## where ##r_i## is the distance from mass-i to the axis] doesn't apply to an Atwood machine.”

I believe the OP’s original confusion (i.e. the key issue to be addressed) was believing that the rope's length affected the total moment of inertia.

Suppose, for example, the Atwood machine has a very long rope so ##r_1## and ##r_2## (distances to centre of pulley) are both extremely large. The OP would argue that ##m_1 r_1^2 + m_2r_2^2## should be used as the contribution to the total moment of inertia. But this will give a false (excessively large) result. This approach is inapplicable to the Atwood machine. The total moment of inertia w.r.t. the pulley’s axis shouldn’t depends on the rope’s length.

Instead we must use ##m_1R^2 + m_2R^2## where R is the pulley’s radius - exactly as you do in your derivation and as the OP's professor did.

Lnewqban
vparam said:
My professor said that I_tot = I + m1R^2 + m2R^2
Well, certainly that gives you the correct answer, but I don't know how your professor justifies it. The correct way to do it, in my opinion, is outlined in the link at the bottom of Post #2. But your professor's way is a short cut that makes the solution much much easier.

Lnewqban
Steve4Physics said:
I entirely agree with your reasoning. I didn't make myself clear enough.

When I said:
“This doesn't apply to an Atwood machine with moving masses.”
I should have said:
“This [the use of ##\Sigma m_i r_i^2## where ##r_i## is the distance from mass-i to the axis] doesn't apply to an Atwood machine.”
The instantaneous angular momentum of a hanging mass in an Atwood machine is ##\vec L=m\vec r\times \vec v## where ##\vec r## is the position of the mass relative to the axis of the pulley.
The angular velocity of the mass relative to the center of the pulley is $$\vec \omega=\dfrac{\vec r\times \vec v}{r^2}=\frac{\vec L}{mr^2}\implies \vec L=mr^2\vec\omega.$$ This is, of course, true for any mass moving in a straight line about a point off the straight line.

Lnewqban
kuruman said:
The external forces acting on the pulley vertical and consist of the weights of the two masses and the reaction force at the center.
This is not correct. The tensions in the string are the two downward forces acting on the pulley, and they are not equal to the weights of the hanging objects. See the link at the bottom of Post #2.

Edit: Your statement would be correct if you changed the word "pulley" to "system".

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kuruman said:
The instantaneous angular momentum of a hanging mass in an Atwood machine is ##\vec L=m\vec r\times \vec v## where ##\vec r## is the position of the mass relative to the axis of the pulley.
The angular velocity of the mass relative to the center of the pulley is $$\vec \omega=\dfrac{\vec r\times \vec v}{r^2}=\frac{\vec L}{mr^2}\implies \vec L=mr^2\vec\omega.$$ This is, of course, true for any mass moving in a straight line about a point off the straight line.
Agreed.

But we must note that the mass’s angular velocity (##\vec {\omega}##) about the pulley’s axis is, in general, not the same as the pulley’s angular velocity (call it ##\Omega##).

(##\vec {\omega} =\vec {\Omega}## only if the mass comes into contact, and rotates with, the pulley.)

##\vec {\omega}## depends on the mass’s height. Even if the mass were moving with constant velocity, ##\vec {\omega}## would be continually changing due to the changing geometry.

But if we use ##\vec L = mR^2 \vec{\Omega}## (with ##R## = pulley’s radius) we can deal in terms of the pulley’s angular velocity. This is far mnore tractable than delaing with ##\vec {\omega_1}## and ##\vec {\omega_2}## fo example.

The system’s moment of inertia can then be expressed as ##I_{pulley} + m_1R^2 + m_2R^2##.

The torques on the system are then provided by the weights. That’s the ‘trick’ that the OP’s professor uses.

Mister T said:
This is not correct. The tensions in the string are the two downward forces acting on the pulley, and they are not equal to the weights of the hanging objects. See the link at the bottom of Post #2.

Edit: Your statement would be correct if you changed the word "pulley" to "system".
Perhaps I was not clear. The system in post #4 under "Proof" is
kuruman said:
a massive pulley with additional point masses ##m_1## and ##m_2## attached on the rim at the two ends of a horizontal diameter.
There are no strings attached and no tensions. External to the pulley, we have the supporting structure and the Earth which exerts gravitational forces on the two masses and the mass of the CM of the pulley . Only the weights of ##m_1## and ##m_2## exert torques about the axis of the pulley.

kuruman said:
Only the weights of ##m_1## and ##m_2## exert torques about the axis of the pulley.
Yes, but the object of mass ##m_1##, for example, exerts a force on the string from which it hangs, and that force is not equal to ##m_1g##. Likewise for the object of mass ##m_2##. Thus the net torque on the pulley is not equal to ##m_1g-m_2g##. That is the only part of your post that I disagree with.

Mister T said:
Yes, but the object of mass ##m_1##, for example, exerts a force on the string from which it hangs, and that force is not equal to ##m_1g##. Likewise for the object of mass ##m_2##.
Which string? Please read my posts carefully. The system, as described, is "a massive pulley with additional point masses m1 and m2 attached on the rim at the two ends of a horizontal diameter". It is shown below.

kuruman said:
Which string? Please read my posts carefully. The system, as described, is "a massive pulley with additional point masses m1 and m2 attached on the rim at the two ends of a horizontal diameter". It is shown below.

View attachment 299505

Ahhh... I now see what you meant about perhaps being unclear. When you referred to "the pulley" I was thinking that you were referring to the pulley presented by the OP, but instead you are referring to your pulley, which is a modification of the original pulley. Sorry about that.

kuruman
Steve4Physics said:
(##\vec {\omega} =\vec {\Omega}## only if the mass comes into contact, and rotates with, the pulley.)
Agreed. In fact if ##y=## the length of the string from the rim of the pulley to the mass, we have
##\tan\!\theta=\dfrac{y}{R}##. Take the time derivative to get
$$\dot{\theta}\left[ 1+\left( \frac{y}{R}\right)^2\right]=\frac{\dot y}{R}.$$ With ##\dot{\theta}=\omega## and ##\dot y=v##, this yields the angular velocity of a point on the string as a function of its vertical position and linear velocity, $$\omega=\frac{vR}{R^2+y^2}.$$It reduces to the familiar expression when ##y=0##.

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Lnewqban

## 1. What does the "r" in the moment of inertia represent?

The "r" in the moment of inertia represents the distance from the axis of rotation to the point mass. It is also known as the radius of gyration.

## 2. How is the "r" in the moment of inertia calculated?

The "r" in the moment of inertia can be calculated using the formula I = mr², where I is the moment of inertia, m is the mass of the object, and r is the distance from the axis of rotation to the point mass.

## 3. What is the significance of the "r" in the moment of inertia?

The "r" in the moment of inertia is important because it determines how the mass is distributed around the axis of rotation. Objects with larger values of "r" have a greater resistance to changes in their rotational motion.

## 4. How does the "r" in the moment of inertia affect an object's rotational motion?

The "r" in the moment of inertia affects an object's rotational motion by determining its rotational inertia. Objects with larger values of "r" have a greater rotational inertia, meaning they require more torque to change their rotational motion.

## 5. Can the "r" in the moment of inertia change?

Yes, the "r" in the moment of inertia can change if the mass distribution of an object changes. For example, if an object's mass is moved closer to or further from the axis of rotation, the value of "r" will change and therefore affect the moment of inertia.

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