# A spring, disk and pulley system

• lorenz0
1. Wheel with radius ##R## on a horizontal surface: No friction
If the wheel sees an external torque it rotates, but does not translate. It slips.

2. Wheel on a horizontal surface: With Friction
If the wheel sees an external torque, it rotates and translates. If it sees a torque greater than ## R \mu_s N## it slips. slips means ##R \omega \neq v_{CM}##

3. Wheel on a horizontal surface with friction being held by an external force that stops translation. For some range of torque less than ## \mu_s N## the wheel remains static and does not slip. If the torque exceeds ## \mu_s N ## the wheel slips, but does not translate.

Are all these scenarios consistent with some understanding, or do I still not get something?
1. Yes
2. Not quite. The radius of inertia matters. If you consider moments and forces you get ##\tau\leq R\mu_sN(1+k^2)##, where the radius of inertia is kR.
3. This one is ## R \mu_s N##. ## \mu_s N ## is not a torque.

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So static friction is just applied while the wheel is accelerating AND when the wheel is completely stationary?
There being no other horizontal forces, there is a nonzero frictional force if and only if there is a nonzero acceleration.
When there is no friction (on a frictional surface) the wheel could be stationary or rolling at constant velocity with a matching angular velocity.

1. Yes
2. Not quite. The radius of inertia matters. If you consider moments and forces you get ##\tau \leq R \mu_s N (1+k^2)##, where the radius of inertia is kR.
3. This one is ## R \mu_s N##. ## \mu_s N ## is not a torque.

3 was a typo

Can you elaborate on 2. Is that otherwise known as the radius of gyration?

3 was a typo

Can you elaborate on 2. Is that otherwise known as the radius of gyration?
Yes, same as radius of gyration. For a uniform disk, ##\tau\leq R\mu_sN\frac 32##.

Yes, same as radius of gyration. For a uniform disk, ##\tau\leq R\mu_sN\frac 32##.
So that is making a torque from the IC of zero velocity. Why does that arise as the criterion for slip? I notice that for the case of a flat disk it just differs from the frictional torque by a constant.

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So that is making a torque from the IC of zero velocity. Why does that arise as the criterion for slip? I notice that for the case of a flat disk it just differs from the frictional torque by a constant.
If no slip, ##mr\alpha=ma=F_{sf}\leq N\mu_s##.
Moments about point of contact: ##\tau=\alpha (mr^2+I_{cm})=\alpha mr^2(1+k^2)##
The result follows.

• erobz