# Rolling problem - disc on disc

1. Mar 27, 2012

### govindsuku

This is a typical case of friction wheels where one wheel drives the other...what I want to know is how can we model the problem when slipping occurs ..... also in case of no slipping how will the free body diagrams and forces be balanced on both disc?

2. Mar 27, 2012

### K^2

If no slipping occurs, you have this constraint.

$$\omega_1 R_1 = -\omega_2 R_2$$

If slipping does occur, you have a different constraint.

$$\frac{\tau_1}{R_1} = -\frac{\tau_2}{R_2} = ±F_f$$

Where Ff is the force of kinetic friction between the two disks, which you can find knowing the kinetic friction coefficient and the contact force between disks.

3. Mar 27, 2012

### govindsuku

thanks for the equation ..... but how can we ensure whether the disc are slipping on each other or not? (apart from visible slipping .... i.e. if the slip is not as significant to be perceived on sight but is significant enough when compared to the torque transferred?

4. Mar 27, 2012

### K^2

It's the same way as modeling a wheel rolling on the ground. First, you check if the angular velocities match, or are close enough to matching to say they do. If so, you apply the no-slipping constraint to compute the force required to maintain that constraint with applied external forces. If that force exceeds static friction, or if the angular speeds miss-match to begin with, you go with slipping constraint.

5. Mar 28, 2012

### govindsuku

ok.....just a doubt....in the above equation (case where slipping occurs)....suppose one disc is being driven by a constant power source.....then the torque acting on that disc will be Ff x R1 and the angular velocity will be constant power supplied divided by the torque.......and the torque transferred to disc 2 will be Ff x R2 ....when and how will equilibrium be established and how can i find the angular velocity of disc 2 when equilibrium is established?

6. Mar 28, 2012

### K^2

If your motor supplies constant torque regardless of angular velocity, you can't ever reach equilibrium. Fortunately, there are no such motors.

Consider, instead, a case where the wheel is driven by an electric motor, whose torque changes with angular velocity roughly like this. (I'm oversimplifying a bit, but it's not a bad model for simple DC motor.)

$$\tau = \tau_{max}(\omega_{max} - \omega)$$

Suppose, also, that the torque at ω=0 is such that it causes the wheels to slip. Now the driven wheel spins up quickly until the torque from motor balances torque from friction. This gives time for the second wheel, under friction torque, to catch up with the angular velocity of the former. At that point, the two wheels go back into a no-slipping mode, and keep accelerating together to level off at ωmax of the driven wheel.

If you also have load on the second wheel, then things get a little bit more interesting. But this is basically a greatly simplified model of a clutch. (For a real clutch, you also need to consider the flywheel.)

7. Mar 28, 2012

### govindsuku

well that is basically the problem i am trying to model.....(chassis dynamometer) i have the wheels of the car on rollers (disc on disc) ... the car is maintained at a particular throttle position (constant power) ....the wheels of the car cause the rollers to rotate......the shaft on which the rollers are mounted is loaded using a set of disc brakes.....the disc brake calipers which press on to the disc are all connected to an arm which is free to rotate about the shaft.....this arm is connected to a load measuring device......which reads the load applied by the arm and hence allows measurement of the torque experienced by the rollers......what i want to know is how the torque measured using this method will relate to the torque at the wheels of the car?

8. Mar 28, 2012

### K^2

In this setup, if the wheels slip, you aren't measuring anything useful. So just go ahead and assume no slipping. Let the system reach steady state with whatever torque you apply to dynamo wheels. Then, presumably, torque applied by your braking mechanism and torque applied by wheels balance each other out. Use Newton's 3rd at contact point to determine what torque the wheel is experiencing due to all that, and you should have the info you need.

Oh, and constant throttle won't give you constant power in this setup. Engine RPM will vary with amount of torque you apply to wheels, and power output depends on throttle position and RPM. The curve relating the two can be quite complex, but essentially, that's what you are measuring.

9. Mar 28, 2012

### govindsuku

ok....so the torque measured by the braking mechanism will be related to the torque at wheels by the ratio of roller radius to wheel radius...thats what i initally did....but the torque measured seems to be too low a value to deem it acceptable....that is why i wanted to investigate whether i was missing anything in these calculations......

10. Mar 28, 2012

### sophiecentaur

If there are no losses, then the Power (speed times torque) delivered to the brake will be the power supplied by the engine.
You seem to have been surprised by the torque involved. What actual power did this involve? Was the Engine working hard at the time (labouring with lots of noise and foot flat down)? Only when that's happening will you be getting the maximum power out of it.

It may be worth checking that you got your ratio of wheel diameters the right way up. (Sorry but it's just possible)

11. Mar 29, 2012

### govindsuku

Thanks for your help guys...there is no slipping in the setup , the load on the wheel is equal to the load due to friction and I did take the roller-wheel ratio correctly....the problem was with the load measuring device.....it was displaying a value off by a factor of 132 !!!! ... I was surprised when i calibrated it as well......so the torque comes well within the acceptable range!!! SO evrybody's happy !!!

Cheers!