Rotating ring on a rough surface- but with a twist

[kuruman]

Can you not calculate $I_O$ separately for each of the three parts of the system? Then, $$I_O^{\text{(system)}}=I_O^{\text{(2m)}}+I_O^{\text{(m)}}+I_O^{\text{(ring)}}.$$

 

[palaphys]

Can you not calculate $I_O$ separately for each of the three parts of the system? Then, $$I_O^{\text{(system)}}=I_O^{\text{(2m)}}+I_O^{\text{(m)}}+I_O^{\text{(ring)}}.$$

What is ioring here

 

[palaphys]

Sorry for asking but could u pls tell how to do the next part? Just say by words, I will find out the equatiosn

 

[kuruman]

What is ioring here

It is the moment of inertia of the ring about bottom point O. Use the parallel axes theorem to find it.

Sorry for asking but could u pls tell how to do the next part? Just say by words, I will find out the equatiosn

Draw a FBD of the entire system. Put in all the forces and then apply Newton's second law in the vertical and horizontal direction.

 

[palaphys]

It is the moment of inertia of the ring about bottom point O. Use the parallel axes theorem to find it.

Draw a FBD of the entire system. Put in all the forces and then apply Newton's second law in the vertical and horizontal direction.

Pls say direction of friction

 

[Lnewqban]

Pls say direction of friction

If the ring were horizontally floating on a frictionless surface it would only rotate around a fixed vertical axis.
Use the third law of Newton to figure the direction.

 

[palaphys]

If the ring were horizontally floating on a frictionless surface it would only rotate around a fixed vertical axis.
Use the third law of Newton to figure the direction.

Sorry don't get you, i thought friction opposed relative sliding and I gotta see which direction sliding would take place..

 

[palaphys]

It is the moment of inertia of the ring about bottom point O. Use the parallel axes theorem to find it.

Draw a FBD of the entire system. Put in all the forces and then apply Newton's second law in the vertical and horizontal direction.

You mean fbd of cm?

 

[Lnewqban]

Sorry don't get you, i thought friction opposed relative sliding and I gotta see which direction sliding would take place..

In order to move in a straight line, the ring needs to have a point of support on which exert a force to impulse its center of mass linearly.
Action and reaction at play.

 

[kuruman]

You mean fbd of cm?

The center of mass is a geometrical point on which only gravity acts. You need a FBD in which you can put the forces that you need to find, friction and the normal force. What do you think you should draw a FBD of?

 

[palaphys]

The center of mass is a geometrical point on which only gravity acts. You need a FBD in which you can put the forces that you need to find, friction and the normal force. What do you think you should draw a FBD of?

Center of mass the 5m thing?

 

[palaphys]

The center of mass is a geometrical point on which only gravity acts. You need a FBD in which you can put the forces that you need to find, friction and the normal force. What do you think you should draw a FBD of?

But isn't the center of mass accelerating tangentially as well here? Pls help. What else to draw fbd of?

 

[kuruman]

Sorry don't get you, i thought friction opposed relative sliding and I gotta see which direction sliding would take place..

Consider this related question. You are at rest standing up and at some point you start walking forward. Clearly, you accelerate because your velocity changes from zero to a non-zero value. What force provides this acceleration? In what direction is it?

 

[palaphys]

Consider this related question. You are at rest standing up and at some point you start walking forward. Clearly, you accelerate because your velocity changes from zero to a non-zero value. What force provides this acceleration? In what direction is it?

Frictional force forward?

 

[palaphys]

But the what force is providing angular acceleration for fring? If it's friction, isn't it retarding the rotation?

 

[kuruman]

But the what force is providing angular acceleration for fring? If it's friction, isn't it retarding the rotation?

Only torques provide angular acceleration.

 

[kuruman]

Frictional force forward?

Yes. Do you think that there is a frictional force here? Why or why not?

 

[palaphys]

Yes. Do you think that there is a frictional force here? Why or why not?

Friction here is static right? It is only ensuring the condition for pure rolling

 

[palaphys]

Only torques provide angular acceleration.

I meant the torque due to friction

 

[kuruman]

I meant the torque due to friction

About what point? About point O it's zero.

Friction here is static right? It is only ensuring the condition for pure rolling

Right. In what direction does it point?

 

[palaphys]

About what point? About point O it's zero.

Right. In what direction does it point?

About center of mass
Static friction points forward? Yes? No? I think so because the center of mass is probably accelerating forward.. i cannot think about how the center of mass is moving here.

 

[kuruman]

Static friction points forward? Yes? No?

Are you asking me or are you telling me? Reason it out.

i cannot think about how the center of mass is moving here

Have you seen bicycles with reflectors on the spokes of their wheels? Imagine a reflector at distance $\frac{1}{5}R$ from the axis of the wheel. The path that it describes as the bicycle moves forward is a cycloid. That's the path of the CM here.

 

[Steve4Physics]

May I throw-in a couple of questions for @palaphys ?

a) You are in your new Porsche , stationary, on level ground. . The ground is super slippery – negligible friction between the ground and the wheels. What happens when you try to accelerate?

b) What happens if there is friction (e.g. if the ground is dry concrete)?

The centre of mass of an object will only accelerate when a net force acts on the object. What can you deduce about the type and direction of the force which makes your Porsche accelerate?

 

[Lnewqban]

About center of mass
Static friction points forward? Yes? No? I think so because the center of mass is probably accelerating forward.. i cannot think about how the center of mass is moving here.

Please, stop and think about the excellent reasoning led by Kuruman.

You always have a pair of forces, action and reaction.
When you ask about friction force and its direction, it is important to be clear about what body the force of interest is acting upon.

You have the force with which the ring pushes on the rough surface, and you have the force with which the rough surface pushes on the ring.
Since at that always changing point of contact neither ring nor surface moves respect to the other, something else yields and moves forward, which is the geometrical center of the ring.

That forward movement forces the points of contact to separate, while new points of contact will come to establish a new pivot of zero velocity and no relative movement.

Please, see:
https://www.astro.ucla.edu/~malkan/astro8/physics1a/rolling.htm

 

[kuruman]

To @palaphys :
You have not provided so far the final answer for the angular acceleration in part (a). Please do before worrying about the force of friction. That's because your plan of attack should be to find the magnitude of the linear acceleration of the CM and from it the horizontal component of the acceleration of the CM.

I have to sign off for a while now.

 

[palaphys]

Are you asking me or are you telling me? Reason it out.

Have you seen bicycles with reflectors on the spokes of their wheels? Imagine a reflector at distance $\frac{1}{5}R$ from the axis of the wheel. The path that it describes as the bicycle moves forward is a cycloid. That's the path of the CM here.

Ok. I assume cycloid is some kind of curvilinear motion. Now in GENERAL rolling cases, where the center of mass is the geometric center, it just goes in a straight line right?. So I feel that the center of mass here may have two accelerations, namely linear and tangential. Am I right?

 

[palaphys]

May I throw-in a couple of questions for @palaphys ?

a) You are in your new Porsche , stationary, on level ground. . The ground is super slippery – negligible friction between the ground and the wheels. What happens when you try to accelerate?

b) What happens if there is friction (e.g. if the ground is dry concrete)?

The centre of mass of an object will only accelerate when a net force acts on the object. What can you deduce about the type and direction of the force which makes your Porsche accelerate?

a)the weels will slide and not sure if I would move
B) pure rolling would take place and I would go forward
Oh so you are saying that here as center of mass moves forward, it is due to friction?

 

[erobz]

Oh so you are saying that here as center of mass moves forward, it is due to friction?

If you would draw a FBD you should be able to see that if the wheel is accelerating to the right, there is some force acting to the right. What force can act to the right? Which way is normal force. Which, direction are the weight forces, is anyone pushing the wheel etc... are questions you should be asking yourself. At the very least by process of elimination you should see that there must be a horizontal force, and where it must act.

 

[Lnewqban]

... as center of mass moves forward, it is due to friction?

... and to the imbalance of masses around the ring: that is the driving torque.

If you take your point of view as the center of the ring, you will see the ring pushing the surface rearward.
In order to do that, the ring needs "to be able to grab" onto the surface.

 

[palaphys]

If you would draw a FBD you should be able to see that if the wheel is accelerating to the right, there is some force acting to the right. What force can act to the right? Which way is normal force. Which, direction are the weight forces, is anyone pushing the wheel etc... are questions you should be asking yourself. At the very least by process of elimination you should see that there must be a horizontal force, and where it must act.

But will friction not cause angular retardation if it acts towards the right? Then how will the no slip condition be maintained?