Rotating ring on a rough surface- but with a twist

[palaphys]

Homework Statement

A ring initially at rest begins to roll (pure rolling) on a rough horizontal floor. It has two masses, m and 2m fixed at diametric points, as shown in the figure. If the mass of the ring is 2m, find the:
1. Angular acceleration of the ring
2. Frictional force acting on the ring
3. Normal reaction on the ring

Relevant Equations

Tau= I alpha ,
v= wr

Honestly, I was very confused looking at the problem. With intuition, it is clear that the ring will roll towards the right. But what direction would friction be acting in? That was my first thought.
However I am unable to figure this out, leaving me stuck here.

My only attempt here was to find the com of the ring, which may be useful somehow.
$$\[
\begin{align}
X_{\text{com}} &= \frac{(2m \cdot 0) + (m \cdot (-R)) + (2m \cdot R)}{5m} \\
&= \frac{0 - mR + 2mR}{5m} \\
&= \frac{mR}{5m} \\
&= \frac{R}{5}
\end{align}
$$

I humbly request someone to aid me in this problem and explain the mechanics here, thanks in advance. I'm only used to seeing rings roll where com coincides with geometric center.

 

[kuruman]

Homework Statement: A ring initially at rest begins to roll (pure rolling) on a rough horizontal floor. It has two masses, m and 2m fixed at diametric points, as shown in the figure. If the mass of the ring is 2m, find the:
1. Angular acceleration of the ring
2. Frictional force acting on the ring
3. Normal reaction on the ring
Relevant Equations: Tau= I alpha v= wr

Honestly, I was very confused looking at the problem. With intuition, it is clear that the ring will roll towards the right. But what direction would friction be acting in? That was my first thought.
However I am unable to figure this out, leaving me stuck here.

My only attempt here was to find the com of the ring, which may be useful somehow.
$$\[
\begin{align}
X_{\text{com}} &= \frac{(2m \cdot 0) + (m \cdot (-R)) + (2m \cdot R)}{5m} \\
&= \frac{0 - mR + 2mR}{5m} \\
&= \frac{mR}{5m} \\
&= \frac{R}{5}
\end{align}
$$

I humbly request someone to aid me in this problem and explain the mechanics here, thanks in advance. I'm only used to seeing rings roll where com coincides with geometric center.

You have listed relevant equations. How do they apply in this case?

 

[kuruman]

I'm not sure if I can add on by adding a similar question, if I am not allowed someone please tell me.

You cannot. Think of the confusion that might ensue as to who is giving what reply to what question. I have reported this post to the mentors' attention so that it can be moved to a separate thread.

 

[palaphys]

You cannot. Think of the confusion that might ensue as to who is giving what reply to what question. I have reported this post to the mentors' attention so that it can be moved to a separate thread.

Should i remove it manually or will the moderators move it to a separate thread?

 

[palaphys]

You have listed relevant equations. How do they apply in this case?

No idea, first off I'm not sure which point I would take torques about, but I say the bottom point would be more sensible as it is eliminates torque of friction

 

[kuruman]

Should i remove it manually or will the moderators move it to a separate thread?

Let the moderators take care of it. [Mentors' note: We have. It's in a new thread]

No idea, first off I'm not sure which point I would take torques about, but I say the bottom point would be more sensible as it is eliminates torque of friction

Hint: Where is the axis of rotation? What point is instantaneously at rest while the rest of the assempbly rotates about it?

 

[palaphys]

Let the moderators take care of it.

Hint: Where is the axis of rotation? What point is instantaneously at rest while the rest of the assempbly rotates about it

Bottom point?

 

[palaphys]

Bottom point?

Okay I see ui gotta use parallel axes thowreom

 

[kuruman]

Bottom point?

Yup.

 

[palaphys]

Yup.

What torques to consider? That is bothering me

 

[kuruman]

What torques to consider? That is bothering me

All of them. What is (are) the force(s) that can conceivably generate torque(s) about the bottom point?

 

[palaphys]

All of them. What is (are) the force(s) that can conceivably generate torque(s) about the bottom point?

Not sure, maybe the weight of the particle? But that doesn't make sense to me. Weight is exerted by the earth on those particles. How is that force acting on the ring?

 

[kuruman]

Weight is exerted by the earth on those particles.

Correct.

How is that force acting on the ring?

At what point does the external force of gravity act when you have an extended body in space?

 

[palaphys]

Correct.

At what point does the external force of gravity act when you have an extended body in space?

Center of mass

 

[kuruman]

Center of mass

There you go. Now draw a free body diagram and pull everything together.

 

[palaphys]

There you go. Now draw a free body diagram and pull everything together.

My only major query is whether the two masses on the ends exert a force on the ring

 

[kuruman]

My only major query is whether the two masses on the ends exert a force on the ring

Why is it "major"? When you draw the FBD, what is your system?

 

[palaphys]

Why is it "major"? When you draw the FBD, what is your system?

It's only the ring

 

[palaphys]

It's only the ring

Major because it would produce a torque right

 

[kuruman]

And what would that torque about the bottom point be?

 

[palaphys]

And what would that torque about the bottom point be?

Alright I think I got this.

 

[palaphys]

The friction and normal force part, I still have no idea though

 

[kuruman]

The friction and normal force part, I still have no idea though

Before you go there you need to fix part (a).
How do you figure $I_O=5mR^2+5mR^2$?
Why is the net torque $5mg\times\dfrac{R}{5}$?

I recommend that you consider as your system the entire assembly of the two particles and the ring together. Then draw the FBD of that and see what you get. That should also help you figure out parts (b) and (c).

 

[Lnewqban]

Homework Statement: A ring ... has two masses, m and 2m fixed at diametric points, as shown in the figure. If the mass of the ring is 2m, ...

Is the total mass of ring-masses assembly 5m?

 

[palaphys]

Is the total mass of ring-masses assembly 5m?

Yes

 

[palaphys]

Before you go there you need to fix part (a).
How do you figure $I_O=5mR^2+5mR^2$?
Why is the net torque $5mg\times\dfrac{R}{5}$?

I recommend that you consider as your system the entire assembly of the two particles and the ring together. Then draw the FBD of that and see what you get. That should also help you figure out parts (b) and (c).

Why is it wrong? I tried considering all the particles as my system here. Also I used parallel axes theorem

 

[kuruman]

Why is it wrong?

I did not say that it is wrong. I asked you to show how you got your answers in more detail. Specifically,

• where did each $5mR^2$ term come from in your calculation of $I_O$?
• why is the net torque about point O equal to $5mg\times\dfrac{R}{5}$?

This is to make sure you know what you're doing.

 

[kuruman]

From the frame of reference of the point of IAR would every point on the ring appear to fo in circular motion?

I don't know what you are asking here but, if you look at posts #6 and #7, we have both agreed that the bottom point is the instantaneous axis of rotation. It is the only point at rest with respect to the floor while the rest of the assembly rotates about it.

Your calculation of the net torque is correct. Your calculation of $I_O$ is not. To find the moment of inertia about the bottom point O, don't cut corners. Find separate expressions for

• the moment of inertia of $2m$ about that point
• the moment of inertia of $m$ about that point
• the moment of inertia of the ring about that point

and add the three terms together.

 

[palaphys]

I don't know what you are asking here but, if you look at posts #6 and #7, we have both agreed that the bottom point is the instantaneous axis of rotation. It is the only point at rest with respect to the floor while the rest of the assembly rotates about it.

Your calculation of the net torque is correct. Your calculation of $I_O$ is not. To find the moment of inertia about the bottom point O, don't cut corners. Find separate expressions for

• the moment of inertia of $2m$ about that point
• the moment of inertia of $m$ about that point
• the moment of inertia of the ring about that point

and add the three terms together.

My question was, if you say at the IAR on the ring, would u see the other particles on the ring around u going in a circular path?

 

[palaphys]

I don't know what you are asking here but, if you look at posts #6 and #7, we have both agreed that the bottom point is the instantaneous axis of rotation. It is the only point at rest with respect to the floor while the rest of the assembly rotates about it.

Your calculation of the net torque is correct. Your calculation of $I_O$ is not. To find the moment of inertia about the bottom point O, don't cut corners. Find separate expressions for

• the moment of inertia of $2m$ about that point
• the moment of inertia of $m$ about that point
• the moment of inertia of the ring about that point

and add the three terms together.

How is the third point sensible though. I thought Parallel axis theorem was icm+md^2= Io. But cm is not at the center point ont he ring,.