Disc on a quarter circle ramp with slip

[infini]

see diagram

a circular disc of radius r has an initial angular velocity \omega and rolls down a quarter circle ramp of radius R. The initial angular velocity will be very high (e.g. 10,000 RPM) causing the disc to slip. The normalised longitudinal force (i.e. longitudinal force / maximum longitudinal force) is at a maximum value of 1 at a slip percentage of 20%. The slip profile is shown in the above graph.

I am trying to derive an expression that will let me calculate the horizontal speed of the disc at a distance L (m) after the end of the quarter circle ramp, based on the friction co-efficient, mass of disc, radius of disc, radius of ramp and initial angular velocity of disc. Relevant equations

slip ratio is defined as SR = (omega*r/V)-1 where V is the longitudinal velocity of the wheel. based on this definition SR = 0 under pure rolling.The attempt at a solution

I am struggling massively with this problem. So far i have managed to calculate the gain in velocity from the ramp itself, but this is based on the assumption of no rolling of the disc (i.e. it effectively assumes the disc is a point mass with zero rotation). My solution for this is based on considering the disc at angle \theta (theta) from its starting point (i.e. draw horizontal line from mass starting position to ramp curve centre and then a line between point mass and ramp curve centre, with \theta being the angle between the two):

a = F/m

friction force Fr = \muN where N = mg.sin\theta

resolving forces gives F = mg.cos\theta - \mumg.sin\theta

a = g(cos\theta-\musin\theta)

for a small change in \theta, d\theta, tangential displacement dx = Rd\theta

a = v.dv/dx = g(cos\theta-\musin\theta)Rd\theta

integrating this with respect to \theta and simplifying i end up with

v = \sqrt{2Rg(1-\mu)}But as i say this is a massive simplification because it does not consider the rolling motion of the disc or the tractive force generated by the initial angular velocity of the disc.

The thing that is really throwing me is the slip of the wheel from the initial angular velocity. The initial slip will be very high, but will reduce over time, while at the same time the horizontal disc velocity increases due to the tractive force. I am assuming i may need to estimate the relationship between the velocity and slip and then perhaps integrate over time rather than angular position, but quite honestly I am not massively hot on calculus and feel extremely lost!

The more i look at this the more complex a problem it seems to become. This is my first post and i don't expect anyone to solve this outright for me, but if someone can spot the method/approach and help guide me along the right path i would appreciate it immensely.

 

[berkeman]

see diagram

a circular disc of radius r has an initial angular velocity \omega and rolls down a quarter circle ramp of radius R. The initial angular velocity will be very high (e.g. 10,000 RPM) causing the disc to slip. The normalised longitudinal force (i.e. longitudinal force / maximum longitudinal force) is at a maximum value of 1 at a slip percentage of 20%. The slip profile is shown in the above graph.

I am trying to derive an expression that will let me calculate the horizontal speed of the disc at a distance L (m) after the end of the quarter circle ramp, based on the friction co-efficient, mass of disc, radius of disc, radius of ramp and initial angular velocity of disc.


Relevant equations

slip ratio is defined as SR = (omega*r/V)-1 where V is the longitudinal velocity of the wheel. based on this definition SR = 0 under pure rolling.


The attempt at a solution

I am struggling massively with this problem. So far i have managed to calculate the gain in velocity from the ramp itself, but this is based on the assumption of no rolling of the disc (i.e. it effectively assumes the disc is a point mass with zero rotation). My solution for this is based on considering the disc at angle \theta (theta) from its starting point (i.e. draw horizontal line from mass starting position to ramp curve centre and then a line between point mass and ramp curve centre, with \theta being the angle between the two):

a = F/m

friction force Fr = \muN where N = mg.sin\theta

resolving forces gives F = mg.cos\theta - \mumg.sin\theta

a = g(cos\theta-\musin\theta)

for a small change in \theta, d\theta, tangential displacement dx = Rd\theta

a = v.dv/dx = g(cos\theta-\musin\theta)Rd\theta

integrating this with respect to \theta and simplifying i end up with

v = \sqrt{2Rg(1-\mu)}


But as i say this is a massive simplification because it does not consider the rolling motion of the disc or the tractive force generated by the initial angular velocity of the disc.

The thing that is really throwing me is the slip of the wheel from the initial angular velocity. The initial slip will be very high, but will reduce over time, while at the same time the horizontal disc velocity increases due to the tractive force. I am assuming i may need to estimate the relationship between the velocity and slip and then perhaps integrate over time rather than angular position, but quite honestly I am not massively hot on calculus and feel extremely lost!

The more i look at this the more complex a problem it seems to become. This is my first post and i don't expect anyone to solve this outright for me, but if someone can spot the method/approach and help guide me along the right path i would appreciate it immensely.

Welcome to the PF.

Are there more details on the initial angular velocity \omega ? It seems a bit odd that the disk is not just released from rest at that position. Is the \omega positive or negative, and do you have an initial value? Does the disk have any initial downward velocity to match the \omega ?

 

[infini]

Welcome to the PF.

Are there more details on the initial angular velocity \omega ? It seems a bit odd that the disk is not just released from rest at that position. Is the \omega positive or negative, and do you have an initial value? Does the disk have any initial downward velocity to match the \omega ?

sorry, realized my explanation of the angular velocity was a bit vague. Essentially the disc will be 'spun up' to its initial angular velocity (values will range between 5,000 and 10,000 rpm) and then dropped from its initial position. The disc will rotate clockwise in the diagram above, so as to produce a tractive force at the disc-ground contact patch which works to accelerate the disc along the track.

I have been trying to work out a solution that ignores the ramp and just considers the disc being dropped onto flat ground, and then attempting to come up for an expression for the horizontal disc velocity. The main thing that is confusing me is how to estimate the reduction in slip over time (or distance), the corresponding increase in speed over time (or distance) and the relationship between the two...

 

[infini]

so now I am thinking that my initial calculations were wrong and i should be coming up with expressions in terms of angular velocity.

the way i see it, having an initial high degree of slip will produce a forward facing frictional force that causes the positive linear acceleration of the disc.

Im recommending reducing the problem to a flat ramp at an angle rather than the quarter pipe, to simplify things.

Im still having issue with the effects of the slip though. I understand the slip must produce a force that works to accelerate the disc. But i have no clue how i would model the fact that the slip is going to 'decay' over time. PLUS the fact that slip itself is a function of both angular velocity and linear velocity. So you have a situation where slip causes linear velocity to increase and angular to decrease, which in turn changes the slip values, which again means the velocities change and so on and so on...

I am comfortable deriving the equations of motion for a disc in pure rolling down an inclined ramp, but the addition of slip just seems to take the whole problem way over my head