Hoop rolling on cylinder without slipping

In summary, the conversation is about a problem involving a small hoop rolling without slipping on a bigger stationary cylinder. The goal is to find the angle at which the small cylinder falls off the bigger one. The conversation discusses using Newton's Laws and Torques to solve the problem, with some confusion about the direction of the theta unit vector and the role of friction in causing torque. The small angle approximation is also mentioned, but ultimately, the suggestion is made to use conservation of energy to find the velocity as a function of theta.
  • #1
cacofolius
30
0
Hi everybody, I know this problem has been posted before, but it envolved Lagrangian methods which I haven't seen yet. I would appreciate any help.

1. Homework Statement

A small hoop is rolling without slipping on a bigger cylinder which is stationary. I need to write Newton's Laws and Torques, and find out at what angle the small cylinder falls off the bigger one.

Homework Equations



##ζ=Iα##

##F=ma##

##I_{cm}= ma^2##

The Attempt at a Solution



My restrain equation, since it rolls without slipping is that ##Φ=\frac{(R+a)}{a}θ## , because I'm taking θ=0 at the top.

##\hat{θ}) Fr- mg sinθ=m(R+a) \ddot{θ}##

##\hat{r}) N-mgcosθ=-m(R+a) \dot{θ}^2##

And around the center of mass of the small cylinder, the only force that makes a torque is friction, so

##ζ=I\ddot{Φ}##
##-aFr \hat{k} = -ma^2 \ddot{Φ} \hat{k} = -m a^2 \frac{(R+a)}{a} \ddotθ## So,

##Fr=m \frac{a^2}{a} \frac{(R+a)}{a} \ddotθ = m(R+a) \ddotθ##

And replacing this in my theeta equation I get:

##\hat{θ}) m(R+a) \ddot{θ} - mg sinθ=m(R+a) \ddot{θ}##

And this is where I get confused, becuse I get ## mgsinθ=0## I now I'm making a silly mistake somewhere, like the direction of a vector or something, but I cannot see it. Thanks in advance.
 

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  • #2
For starters, I suggest that you reverse the direction of the theta unit vector, so that it points down slope rather than up slope. The system remains right-handed, but this will make things work out easier.

Now, what do you mean by the angle phi? Is it the absolute rotation, or the rotating of the contact point on the small cylinder?

Also, it looks like you have said that the radial force is the cause of the torque driving the angular acceleration of the small cylinder. This is wrong.
 
  • #3
Hi OldEngr63, thanks for your reply, I tried your approach with the angles, so I got

##\hat{θ}) 2m(R+a) \ddot{θ} + mg sinθ=0 ##

which is nice for small angle approximation and describing the oscillatory movement.
And Fr meant friction force. Sorry it took long for me to answer back. Thanks for your help!
 
  • #4
What is theta-hat?
 
  • #5
cacofolius said:
My restrain equation, since it rolls without slipping is that ##Φ=\frac{(R+a)}{a}θ## , because I'm taking θ=0 at the top.
How are you defining Φ? I assume it's the rotation of the hoop about its own centre, and in the diagram the black arrow represents a radius to the point that was originally at the bottom, i.e. arrow pointing down, yes?
The easiest way with these problems is generally to start with conservation of energy, if appropriate. That will give you the velocity as a function of theta.
cacofolius said:
## \hat {\theta} \right) 2m(R+a) \ddot {\theta} + mg \sin \theta = 0##
Better, but still seems to have a sign wrong. Wouldn't that make ##\ddot{θ}## negative?
cacofolius said:
small angle approximation
You don't need such.
cacofolius said:
describing the oscillatory movement
What oscillation?
OldEngr63 said:
What is theta-hat?
It's indicating that the equation relates to tangential direction.
 

Related to Hoop rolling on cylinder without slipping

1. What is the concept of hoop rolling on a cylinder without slipping?

The concept of hoop rolling on a cylinder without slipping is based on the principle of conservation of energy. This means that the hoop must maintain a constant speed while rolling on the cylinder without any slipping, which requires a delicate balance between the gravitational potential energy and the kinetic energy of the hoop.

2. How does the radius of the hoop affect its rolling on the cylinder?

The radius of the hoop plays a crucial role in determining the speed at which it can roll on the cylinder without slipping. A larger hoop has a greater moment of inertia and requires more force to roll without slipping, while a smaller hoop has a lower moment of inertia and can roll with less force. Therefore, the radius of the hoop must be carefully considered when studying hoop rolling on a cylinder without slipping.

3. What is the role of friction in hoop rolling on a cylinder without slipping?

Friction plays a critical role in hoop rolling on a cylinder without slipping. It provides the necessary force to prevent the hoop from slipping while also causing the hoop to lose energy through heat dissipation. The amount of friction between the hoop and the cylinder greatly affects the speed and stability of the rolling motion.

4. How is the energy of the hoop conserved during rolling on a cylinder without slipping?

The energy of the hoop is conserved during rolling on a cylinder without slipping through a combination of potential and kinetic energy. As the hoop rolls, it converts its potential energy, gained from the initial height, into kinetic energy, which maintains the speed of the rolling motion. The energy is conserved as long as there is no slipping, which would result in energy loss due to friction.

5. Can the hoop ever reach a point of no slipping while rolling on the cylinder?

Yes, under ideal conditions, the hoop can reach a point of no slipping while rolling on the cylinder. This point is known as the critical velocity, where the hoop is rolling at a speed that perfectly balances its gravitational potential energy and kinetic energy. However, in real-world scenarios, there will always be some level of friction that prevents the hoop from reaching this point, resulting in a slight slipping motion.

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