Rolling Dynamics of a Rotating Cylinder on an Inclined Plane

In summary, the problem involves a solid homogeneous cylinder of radius r placed at the bottom of an inclined plane with an angle α with the horizontal plane. It starts to roll upwards with angular velocity ω0 and no initial forward speed. The equations of motion are given by mdv/dt = F - mgsinα and I(dω/dt) = Fr, where F represents the friction force, m is the mass of the cylinder, g is the acceleration due to gravity, and I is the moment of inertia of the cylinder. By integrating these equations and using the initial conditions, the time taken for the cylinder to reach the highest point on the inclined plane can be determined to be ω0r/2g*sinα. However
  • #1
sergiokapone
302
17

Homework Statement



Rotating with angular velocity ##ω_0## solid homogeneous cylinder of radius ##r## placed without starting forward speed at the bottom of the inclined plane, forming an angle ##\alpha## with the horizontal plane, and starts to roll in up. Determine the time during which the cylinder reaches the highest position on the inclined plane.

Homework Equations



##m\dfrac{dv}{dt}=F-mg\sin\alpha##

##I\dfrac{d\omega}{dt}=Fr##

The Attempt at a Solution


From integrating of the above eqns, I get:
##\dfrac{\omega r}{2}-v=g\sin\alpha \cdot t-const##
Where the constant we can determine from the initial conditions, e.g.
##const=-\dfrac{mr\omega_0}{2}##
So, I get eqn
##\dfrac{\omega r}{2}-v=g\sin\alpha \cdot t-\dfrac{\omega_0 r}{2}##

I think that at the start of pure rotation, the cylinder has to stop, ie, it reaches the maximum height. How is it possible to argue?

Then ##v=-\omega r##, and ##\omega=0##

##0=g\sin\alpha \cdot t-\dfrac{\omega_0 r}{2}##
##t=\dfrac{\omega_0 r}{2g\sin\alpha}##
 
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  • #2
I am missing some reason for the cylinder to roll upwards. Friction? If yes, how much? This will certainly influence the motion.

I think that at the start of pure rotation, the cylinder has to stop
The cylinder can keep rolling upwards afterwards.
 
  • #3
mfb said:
I am missing some reason for the cylinder to roll upwards. Friction?
Yes, friction ##F##.

mfb said:
If yes, how much? This will certainly influence the motion.
Interesting quastion.

I think the friction force, which raises the body, must be greater than the rolling friction, which is neglected in this problem. It turns out that an arbitrarily small sliding friction force can pull the body. But it also means that the acceleration is negative and the velocity of the center of mass must decrease with raising of body. [strike]When friction is included at the start, the body instantly appears nonzero velocity of the center of mass. I wonder what it is then?[/strike]

mfb said:
The cylinder can keep rolling upwards afterwards.
Yes, by inertia, I realized this already.
 
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  • #4
You need to express your second equation in terms of m, dv/dt, and r. What is the moment of inertia of a cylinder expressed in terms of m and r? What do you get if you multiply dω/dt by r? From the second equation, what is F expressed in terms of m, r, and dv/dt? Take this result, and substitute it into your first equation.
 
  • #5
Chestermiller said:
You need to express your second equation in terms of m, dv/dt, and r. What is the moment of inertia of a cylinder expressed in terms of m and r? What do you get if you multiply dω/dt by r? From the second equation, what is F expressed in terms of m, r, and dv/dt? Take this result, and substitute it into your first equation.

I'm it's all done in the first post. Is not it?
Strange moment in this problem, it is that the time does not depend on the coefficient of friction. But it must, I think.
 
  • #6
Oh, this is a nasty question. It defies intuition because it requires unlimited coefficient of friction. The instant the cylinder is placed on the slope there will be an impact. This means work is not conserved. You have to use conservation of angular momentum to figure out the rate of spin immediately after contact.
The easiest way is to take moments about the point of contact. That way, the impulse itself (being along the line of the ramp) has no moment.
 
  • #7
sergiokapone said:
I'm it's all done in the first post. Is not it?
Strange moment in this problem, it is that the time does not depend on the coefficient of friction. But it must, I think.
In the first equation, F has the wrong sign. F is pointing down the incline. It is harder to accelerate or decelerate a rotating cylinder than one that is sliding without rotation. The equivalent mass should be 3m/2, not m/2.
 
  • #8
Chestermiller said:
In the first equation, F has the wrong sign. F is pointing down the incline. It is harder to accelerate or decelerate a rotating cylinder than one that is sliding without rotation. The equivalent mass should be 3m/2, not m/2.

But then what is the force begins drag up of the cylinder to the top? I think that only the friction force, so it must be pointing up the incline. I was wrong only in the sign in the second equation, since the friction slows the rotation.
 
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  • #9
All this discussion of F is irrelevant. Once rolling contact is established, and you know the speed at that point, you can use energy to calculate the height reached. Since deceleration will be constant, you can use the usual constant acceleration formulae to find the time.
But the first step is to find the initial speed up the ramp.
 
  • #10
sergiokapone said:
But then what is the force begins drag up of the cylinder to the top? I think that only the friction force, so it must be pointing up the incline. I was wrong only in the sign in the second equation, since the friction slows the rotation.

The incline is exerting a downward tangential force on the cylinder, also causing it to rotate. There is no force causing the cylinder to go up the incline. The cylinder goes up the incline as a result of its prior inertia. But it slows down as a result of the tangential gravitational component, and the tangential non-slip force component F.
 
  • #11
haruspex said:
All this discussion of F is irrelevant. Once rolling contact is established, and you know the speed at that point, you can use energy to calculate the height reached. Since deceleration will be constant, you can use the usual constant acceleration formulae to find the time.
But the first step is to find the initial speed up the ramp.

I am now so do the calculations, of course. But then where is the guarantee that the disc does not slip when climbing up? If it slips, the energy is not conserved, and the proposed method does not work you.
 
  • #12
Chestermiller said:
The cylinder goes up the incline as a result of its prior inertia. But it slows down as a result of the tangential gravitational component, and the tangential non-slip force component F.

The cylinder was no forward movement, but there were only rotating, so the center of mass can not move prior inertia. The friction force transferred energy of rotational motion into energy of translational motion, then it did the work.
 
  • #13
sergiokapone said:
I am now so do the calculations, of course. But then where is the guarantee that the disc does not slip when climbing up? If it slips, the energy is not conserved, and the proposed method does not work you.
Reread both my posts. Since you're not given the coefficient of friction, the problem cannot be solved unless you assume friction is adequate to prevent slipping at all times. (In the extreme, with no friction, it would spin forever at the bottom.) That splits the problem into two steps:
- what happens at initial contact? How can the rotation turn into a combination of rotation and linear movement 'instantly'? (see my first post)
- after that instant, there is rolling contact and you can use conservation of work (my 2nd post)
 
  • #14
haruspex said:
- what happens at initial contact? How can the rotation turn into a combination of rotation and linear movement 'instantly'? (see my first post)

haruspex said:
You have to use conservation of angular momentum to figure out the rate of spin immediately after contact.
The easiest way is to take moments about the point of contact.

From the conservation of angilar momentum about the point of contact looks like:
##\dfrac{3mr^2\omega_0}{2}=mv_0 r## It is Ok?
 
  • #15
haruspex said:
Since you're not given the coefficient of friction, the problem cannot be solved unless you assume friction is adequate to prevent slipping at all times.

Unless I'm overlooking something (very possible), I think you can allow slipping at the beginning. I believe the only requirement is to assume that the kinetic friction force is enough to initially make the cylinder climb the slope. Otherwise, the amount of friction is not important.

Using your idea of taking moments about a fixed point (in the Earth frame) that coincides with the initial point of contact makes the problem surprisingly simple.
 
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  • #16
sergiokapone said:
From the conservation of angilar momentum about the point of contact looks like:
##\dfrac{3mr^2\omega_0}{2}=mv_0 r## It is Ok?
No. The value on the LHS would be true if it were already rolling, i.e. if it had a linear component of motion rω0. If you subtract out the moment that would have about the point of contact then you'll get the right value.
Conversely, on the RHS, you only have the moment due to linear motion. You need to add in the rotational component.
 
  • #17
TSny said:
I think you can allow slipping at the beginning. I believe the only requirement is to assume that the kinetic friction force is enough to initially make the cylinder climb the slope.
Not sure what you mean by 'initially climb the slope'. If you mean rolling contact immediately then there is no slipping. If there's any slipping at all it will sap energy. You could assume some finite coefficient of friction and see what it leads, but I'd be astonished if the assumed unknown magically disappeared from the equations.

EDIT: I'm suitably astonished - see below.
 
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  • #18
haruspex said:
- after that instant, there is rolling contact and you can use conservation of work (my 2nd post)

##mgh=\dfrac{mr^2}{4}\dfrac{v_0^2}{r^2}## -concervation energy low - I'm not sure

##h=\dfrac{3r^2}{4g}\omega_0^2##

Distance along the incline ##x=\dfrac{h}{\sin\alpha}=\dfrac{v_0^2}{2a}##
##h=\dfrac{v_0^2\sin\alpha}{2a}##

from angular momentum conservation ##\dfrac{mr^2}{2}\omega_0=mv_0r## - I'm not sure

##v_0=\dfrac{\omega_0r}{2} ##
##a=2g\sin\alpha##

##0=\dfrac{\omega_0r}{2}-2g\sin\alpha \cdot t##

##t=\dfrac{\omega_0 r}{4g\sin\alpha}##
 
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  • #19
haruspex said:
Not sure what you mean by 'initially climb the slope'. If you mean rolling contact immediately then there is no slipping. If there's any slipping at all it will sap energy. You could assume some finite coefficient of friction and see what it leads, but I'd be astonished if the assumed unknown magically disappeared from the equations.

You can let the cylinder slip initially so that it's a force of kinetic friction that starts the cylinder moving up the slope. [EDIT: To be clear, Even though the cylinder slips when placed on the incline, it nevertheless starts accelerating up the incline immediately.] Eventually, that will change to rolling without slipping. You can go through a detailed calculation where you break it into parts and find the time for the slipping part and the additional time for the rolling w/o slipping. But, following your lead regarding taking moments about the initial point of contact, you don't need to go through all that.
 
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  • #20
TSny said:
following your lead regarding taking moments about the initial point of contact, you don't need to go through all that.
You're quite right - it can all be solved in terms of angular momentum. My thought experiment with no friction misled me. If the friction is very low, the initial movement will be down the ramp. If we allow the ramp to be as long as necessary, and the friction to be any nonzero value, the time taken to stop rotating is constant.
sergiokapone, your latest answer is almost right, but it's out by a factor of 2.
What do you think the angular momentum is about the point of contact before contact is made? What is the torque about the point of contact subsequently?

Correction: it's not a matter of whether friction is nonzero. As TSny said (took me a while to catch up), you have to assume it moves up the ramp at least a little bit. That will be true if F > mg sin(α). Otherwise the answer is zero time.
 
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  • #21
haruspex said:
sergiokapone, your latest answer is almost right, but it's out by a factor of 2.
What do you think the angular momentum is about the point of contact before contact is made? What is the torque about the point of contact subsequently?
I do not understand how can I calculate the angular momentum of a rotating body around a different axis, which rotates about its axis. Before contact, spin is ##\dfrac{mr^2}{2}\omega_0##, but it is the spin, i.e. angular momentum about cetnter of mass axis.

Again, if the body has already fallen. At the point of contact friction torque acts ##-F \cdot r##, which slows the rotation. And, of course, prior the torque the spin is not conserved.

Of course, if the law of conservation writing relative to the points of contact, the friction force does not create torque, and one can write the conservation of angular momentum, but I do not understand how?
 
  • #22
sergiokapone said:
I do not understand how can I calculate the angular momentum of a rotating body around a different axis, which rotates about its axis. Before contact, spin is ##\dfrac{mr^2}{2}\omega_0##, but it is the spin, i.e. angular momentum about cetnter of mass axis.
A body rotating about its mass centre has the same angular momentum about all points. (It's analogous to the fact that a "screw" - a pair of equal and opposite forces acting along different lines - has the same torque about every point.) Here's one way to see this: consider a body rotating about some point P distance r from its mass centre C. By the parallel axis theorem, the moment about P is (I+mr2)ω. But you can think of the body's movement at an instant as a combination of rotating about C plus a linear movement perpendicular to PC. The latter has angular momentum mr2ω about P, so rotation of the body about C must have angular momentum Iω about P.
Again, if the body has already fallen. At the point of contact friction torque acts ##-F \cdot r##, which slows the rotation. And, of course, prior the torque the spin is not conserved.

Of course, if the law of conservation writing relative to the points of contact, the friction force does not create torque, and one can write the conservation of angular momentum,
Exactly. You can either fix on the point of initial contact, in which case you have to consider the moments of the normal force and gravity, or allow it to move as the point of contact move (so the only force with moment is gravity). It will lead to the same equation. So: what is the torque about the point of contact?
 
  • #23
haruspex said:
So: what is the torque about the point of contact?
##-mgr\sin\alpha##

Eqn of rotational movement:
##\dfrac{mr^2}{2}\dfrac{d\omega}{dt}=-mgr\sin\alpha##
 
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  • #24
sergiokapone said:
##mgr\sin\alpha##

Right. So you have the initial angular momentum and the torque which opposes it. How long does it take to stop the cylinder's rotation?
 
  • #25
haruspex said:
Right. So you have the initial angular momentum and the torque which opposes it. How long does it take to stop the cylinder's rotation?

From integrating of the eqn of rotational motion ##t=\dfrac{r\omega_0}{2g\sin\alpha}##
 
  • #26
haruspex said:
By the parallel axis theorem, the moment about P is (I+mr2)ω. But you can think of the body's movement at an instant as a combination of rotating about C plus a linear movement perpendicular to PC. The latter has angular momentum mr2ω about P, so rotation of the body about C must have angular momentum Iω about P.

I have to think about it.
 
  • #27
Combination of ##mr^2/2 \cdot \omega_0## and ##mr^2\omega_0## leads to ##3mr^2/2 \cdot \omega_0## and then only in the case of pure rotation. How you got ##mr^2/2 \cdot \omega_0##?
 
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  • #28
haruspex said:
You can either fix on the point of initial contact, in which case you have to consider the moments of the normal force and gravity, or allow it to move as the point of contact move (so the only force with moment is gravity). It will lead to the same equation.

Applying the ##\tau## = dL/dt equation to a point that is accelerating can get you in trouble. Think about the case of a cylinder that is given an initial velocity and no rotation at the bottom of a frictionless incline. It will slide up the incline without developing any rotation. You can see that the ##\tau## = dL/dt equation doesn't hold for a coordinate system with origin at the point of contact and moving with the point of contact.
 
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  • #29
TSny said:
Applying the ##\tau## = dL/dt equation to a point that is accelerating can get you in trouble.
True.. so how to justify this method? It gives the right answer (i.e. the same I got by doing it the hard way). Keeping the point of reference fixed doesn't help, does it? Unless it's rolling the torque as measured from a point on the ramp does not match the torque on the cylinder.
 
  • #30
haruspex said:
True.. so how to justify this method? It gives the right answer (i.e. the same I got by doing it the hard way). Keeping the point of reference fixed doesn't help, does it? Unless it's rolling the torque as measured from a point on the ramp does not match the torque on the cylinder.

The general theorem ##\vec{\tau}= d\vec{L}/dt ## is true relative to any fixed point in an inertial frame. This relation is not true, in general, relative to a point that is accelerating relative to an inertial frame. The exception is the center of mass point. It can be shown that ##\vec{\tau}_{cm} = d\vec{L}_{cm}/dt## is true even if the center of mass is accelerating.

In this problem, the theorem as applied to a fixed point on the plane gives the correct equation:

##mgsin\alpha= -(I_{cm}d\omega /dt +mR\;dv_{cm}/dt)## where R is the radius of the cylinder.

Here, the general relation ##\vec{L} = \vec{L}_{cm} + M\vec{r}_{cm}\times\vec{v}_{cm}## was used.

If you tried to use these relations relative to the point of contact P (which is accelerating) you would get the incorrect equation ##mgsin\alpha= -I_{cm}\;d\omega /dt##.

If you happen to have rolling without slipping, then you can show that the equation relative to the fixed point on the plane can be written as ##mgsin\alpha= -I_{P}\;d\omega /dt## where ##I_P## is the moment of inertia relative to P. So, for rolling without slipping you will get the correct equation relative to point P if you use the moment of inertia relative to P.

It seems odd at first that in this problem the incorrect equation relative to P gives the correct result for the time it takes the cylinder to lose all of its initial angular momentum ##L_0##. But I think this is due to the fact that the torque ##\tau## relative to P is the same as the torque relative to the fixed point. ##\Delta t = |L_{0}/\tau|## depends only on the torque and the initial angular momentum.
 
  • #31
TSny said:
In this problem, the theorem as applied to a fixed point on the plane gives the correct equation:

##mgsin\alpha= -(I_{cm}d\omega /dt +mR\;dv_{cm}/dt)## where R is the radius of the cylinder.

Here, the general relation ##\vec{L} = \vec{L}_{cm} + M\vec{r}_{cm}\times\vec{v}_{cm}## was used.

This is should be the same as I used a separately the eqns of center mass motion and rotational motion in my initial post, but it is not.
 
  • #32
sergiokapone said:
This is should be the same as I used a separately the eqns of center mass motion and rotational motion in my initial post, but it is not.

I'm not sure about your sign conventions when you wrote ##I\dfrac{d\omega}{dt}=Fr##.

If you take positive values of ##\omega## to correspond to the direction of the initial angular velocity of the cylinder, then the equation would be ##I\dfrac{d\omega}{dt}=-Fr##. Then I believe everything is ok.
 
  • #33
TSny said:
I'm not sure about your sign conventions when you wrote ##I\dfrac{d\omega}{dt}=Fr##.

If you take positive values of ##\omega## to correspond to the direction of the initial angular velocity of the cylinder, then the equation would be ##I\dfrac{d\omega}{dt}=-Fr##. Then I believe everything is ok.

I have already written

sergiokapone said:
Again, if the body has already fallen. At the point of contact friction torque acts ##-F \cdot r##, which slows the rotation. And, of course, prior the torque the spin is not conserved.

In the first post was a mistake in the signum.
 
  • #34
sergiokapone said:
This is should be the same as I used a separately the eqns of center mass motion and rotational motion in my initial post, but it is not.

Do you see how your equations are consistent with what I wrote?

##m\dfrac{dv}{dt}=F-mg\sin\alpha##

##I\dfrac{d\omega}{dt}=-Fr##

Multiply the first equation by r and then add it to the second equation. This will get rid of the friction force F and you will end up with the same equation that I wrote.
 

1. What is the concept of rolling dynamics in a rotating cylinder on an inclined plane?

The concept of rolling dynamics refers to the motion and behavior of a rotating cylinder on an inclined plane. This includes the forces acting on the cylinder, such as gravity and friction, and how they affect its motion.

2. How does the angle of the incline affect the rolling dynamics of the cylinder?

The angle of the incline plays a crucial role in the rolling dynamics of a rotating cylinder. As the angle increases, the force of gravity pulling the cylinder down the incline also increases, leading to a faster and more unstable rolling motion.

3. What is the difference between a cylinder rolling without slipping and a cylinder slipping on an inclined plane?

When a cylinder is rolling without slipping, its bottom point of contact with the inclined plane is stationary. However, if the cylinder is slipping, this point is moving, and the cylinder is not rotating at a constant rate. This leads to different rolling dynamics and behavior.

4. How does the mass and radius of the cylinder impact its rolling dynamics on an inclined plane?

The mass and radius of the cylinder play a role in determining the moment of inertia, which affects the rotational motion of the cylinder. A larger mass or radius will result in a larger moment of inertia, making it more difficult for the cylinder to rotate and increasing its stability on the inclined plane.

5. Can the rolling dynamics of a rotating cylinder on an inclined plane be applied to real-world situations?

Yes, the principles of rolling dynamics on an inclined plane can be applied to many real-world situations, such as the motion of a car or a ball rolling down a hill. Understanding these dynamics can help engineers and scientists design and improve various mechanical systems.

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