Rotating cylinder on inclined surface

[user5]

Homework Statement

A cylinder is rotating about its axis and is placed on an inclined surface without linear velocity, the coefficient of kinetic friction between the surface and the cylinder is μ_k . During Δt₁ it stays at the same height till the rotation stops. From that moment it takes Δt₂ for the cylinder to reach the bottom of the inclined surface. What is the angle of inclination of the surface?
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My question is why does at time Δt₁ there is a kinetic friction but not static- because as the problem states cylinder doesn't move?

 

[MostlyHarmless]

Because its rotating. The problem says it doesn't have any LINEAR velocity, but it does have angular velocity.

 

[user5]

Why does at Δt₂ friction do no work? How do I know that all the way down there is only a static friction?

 

[MostlyHarmless]

Because once the cylinder stops rotating, it begins to roll down the incline, so since it is no longer "slipping" it feels little or no friction.

 

[Chestermiller]

Because once the cylinder stops rotating, it begins to roll down the incline, so since it is no longer "slipping" it feels little or no friction.

I think this response needs to be modified just a little for clarity. Since it is no longer slipping, it feels no kinetic friction. However, static friction is now present to provide the torque necessary to increase its angular velocity as it rolls down the plane.

Chet

 

[MostlyHarmless]

Yes, thank you chet.

 

[user5]

Why mgsinθ would not provide the needed ability to slip?

 

[haruspex]

Because once the cylinder stops rotating, it begins to roll down the incline, so since it is no longer "slipping" it feels little or no friction.

There's a bit more to the answer to this.
First, you need to understand what's happening in the first time period. Clearly the cylinder is slipping, and since it is staying in the same spot on the incline there is a precise relationship between the kinetic friction force, the angle of the incline and the weight of the cylinder. Note that this relationship is independent of the rate of spin. it follows that the force of kinetic friction is sufficient to prevent the cylinder from moving down the incline, so such movement can only occur when there is no long kinetic friction acting up the incline. I.e., the cylinder is now rolling.

 

[Chestermiller]

Why mgsinθ would not provide the needed ability to slip?

Since they don't tell you what the coefficient of static friction is, I guess you have to assume that, during the second part, the coefficient is high enough to prevent slippage and to allow the cylinder to roll instead. If there were ice on the incline, the cylinder could possibly slide down the incline without rolling.

 

[TSny]

If there's enough kinetic friction to hold the cylinder in place as it slips, then I think you can show that there will be enough static friction for the cylinder to roll down the plane without slipping as long as $$\mu_s > \frac{\mu_k}{3}$$ For "ordinary" materials, $\mu_s > \mu_k$.

 

[haruspex]

Since they don't tell you what the coefficient of static friction is, I guess you have to assume that, during the second part, the coefficient is high enough to prevent slippage and to allow the cylinder to roll instead.

There's no need for an assumption here. The information about the first time period ensures this. See my post #8.

 

[Chestermiller]

There's no need for an assumption here. The information about the first time period ensures this. See my post #8.

I agree. But there seemed to be some questions about the second time period that I wanted to say something about.

Chet