Recent content by anon90

I'm sorry that the description was that confusing, I've looked for some pictures online that may describe my situation and I've found the attached one. Basically, instead of having a hanging rod, said rod is laying on a horizontal plane, and then we hit one extreme with a flying bullet. During...

Hello everyone, I have a doubt regarding the conservation of angular momentum. When dealing with collisions between two objects, if the net external force is zero we know that the linear momentum is conserved; even when the system is not isolated, for instance because of gravity acting on the...

I don't think I have big issues understanding the conservation of energy, it's more about the concept of negative work and what that implies. You are probably right, still I wanted to try understanding the physical meaning behind it. Like, I think I'm fine when I'm supposed to understand the...

Hello everyone, I'd like to share a doubt I am currently struggling with. So we know that ΔU=−W, where ΔU is the difference of potential energy and Wthe work done by the force to move the body from point A to point B. When analyzing this for the gravitational force, since we have U=−GmM/R, with...

I see, thank you for your help and merry Christmass.

So, in the end, the two conditions should be μ≥1 θ=45° So the trick with this kind of problems is to figure out which forces can be though as impulsive in that laps of time, and then you just draw the fbd and solve it. Am I forgetting about anything else?

Right. f≤μsK=μsJsinθ Together with θ= 45° and Jcosθ−f=0, I get μ \geq 1

So, by your logic, I have μ \geq sqrt(2) /(2K) But I know what (1) -> Jsinθ=K, hence μ \geq 1 /J (2) All the values of μ satisfying (2) are good, and there's only one possible value for θ.

The thing is, I already know μ's value, I don't see why I should be looking for its restriction.

From just the two of them? JRsin(θ)-Jcos(θ)R=0 -> J(sin(θ)-cos(θ))=0 So θ=\pi/4 +n\pi But, if I consider −Jsinθ+K=0 (forces on y-axis) together with the fact that the frictional impulse is linked to the impulsive normal through that inequality, I get a different outcome; what's wrong with my...

Let's say K is the impulsive normal, then f \leq μK Jcosθ−f=0 -> Jcosθ≤μK −Jsinθ+K=0 -> Jsinθ=K (1) Jcosθ≤μJsinθ This should be the condition for the CM not moving, I don't think you told me whether it's correct or not. No rotation: same as before JRsin(θ)−fR=IΔω=0 Since f \leq μK I get Jsin(θ)...

I'm glad everything is finally clear. Now, from what I read so far, this problem should be no different from your average one where you just solve the forces diagram; the only main difference is that we care for impulsive forces only. I thought I would've needed to use something else so I wasn't...

But if I'm looking for solutions for which the CM remains at rest and the ball doesn't slip too then I have to use the static friction, don't I? I think it means when the ball stays still, i.e. no slipping and no CM translation. If either one of them exists then the velocity isn't null.

But I am studying statics here. https://www.physicsforums.com/threads/billiard-ball-hit-by-a-cue.788645/#post-4953283 That's how I got that inequality by looking at the force diagram, I don't think I should be using the kinetic friction force in that case. Well, zero, but I wouldn't even use...

But weren't we talking about the rotational velocity? I thought the translational part was ruled by Jcosθ≤μJsinθ The linear impulse/momentum relation is J=Δp, but why should I use it now that I'm studying the rotational aspect of the problem? But, since slipping isn't allowed, if the...