Ok, thanks for the answers.
That's kinda what was bothering me, because I was thinking of other kinds of exercises. One in particular had two "concentric overlapping discs of pulleys" (I'm sure there is a better term to indicate this, but I don't know it) with two blocks hanging on either side...
Yea, got it. I think.
So if I drew the diagram only for the block, it would show these three forces: Weight, Normal force, and a Tension pointing to the right.
For the pulley, there would be two forces: F pointing down, and T pointing to the left.
The T from the block and the T from the pulley...
Homework Statement
There is a block of known mass m on a horizontal surface. The block is connected to a pulley via a rope. The other end of the rope is pulled vertically, downwards, with a known force F. The pulley has a known moment of inertia I and radius r.
Calculate the acceleration a of...
I have a general understanding of how torque works, at least for "simple" objects that can be drawn as a single "bar" under the effect of various forces. In this problem there is a slightly more "complex" object though, and I'd like to know if there is a way to solve it without doing what I did...
lol
So if I'm reading this right you mean that after the object detaches it would start its parabolic motion, but it "hits its head" on the drum because the parabola trajectory "goes up" higher than the drum's circumference. Then, once the drum's circumference is "higher" than the parabola...
I think I get why μ must be infinite. With X centripetal axis, Y tangential:
X axis:
+N + Wx = m*a
Y axis:
+Friction - Wy = 0
=> N*μ = Wy
=> N = Wy/μ
Since N must be zero at the point of detachment, then N = Wy/μ => we need μ to be infinite to make that fraction -> 0.
Just to be clear, you...
Everyone says it should be 0.85 (and I'm sufficiently convinced that's the right one), though the textbook says 0.91. Could just be an error on the textbook part, at this point that's my guess.
I'd like to quote it exactly, but it's not in English. You can trust my translation though -- I've...
Yea, but I'm better at solving things when I understand them than when I have to accept an "insight" that falls out of the schemes I'm familiar with. I guess I try too hard to make things fit into a pre-made picture that I have of the problem, and thus find it hard to "adjust" to different...
Oh, so you mean W_x and W_y. Yea.
So "tangential" here means "perpendicular to the centripetal axis". Yea.
Ok, thanks (to you and everyone else).
Alright. I still am not entirely convinced though; it kinda feels like we are making the most out of what we have, but we are missing a necessary...
Uhm...no, sorry, I just don't get it. I've always thought that the weight force is the same, so W can't possibly have any other value than m*g. The thing that would change is, with different angles, how much W_x and W_y we have, but W is the same.
So how can gravity "provide more and more"...
Ehr...tangential component of gravity?
If R is not perpendicular, then where is it?
Centrifugal force is not enough anymore to counterbalance the Weight (in the second drawing I made, the W_x). Or, in other words, the W_x cannot provide a big-enough centripetal force.
Ok, but how? I attempted...
It doesn't mention anything about friction, which was my very first problem (as I'm about to type).
It is constant rotation, yes.
By "frequency" it means rounds per second [Hz], so as I mentioned earlier I presume it wants me to find the acceleration, and from the acceleration I can then find...
Yea, that's the idea.
The weight, I guess. Like, at some point the weight component is too strong, so the centripetal force "can't keep up" (or rather, the centrifugal force, but I'm trying to avoid using that one). But there should be some R influence there too, no?
So...this?
Putting the Y...
Homework Statement
We have an object rotating inside a circular container. The rotation is vertical (see picture below).
The radius r of the container is given. Find the frequency (which then ties back to finding the centripetal acceleration) necessary, so that the objects rotating inside the...
So yea, that also explains why Nx is "counterproductive" in the suitcase-scenario (it points away from the center of rotation), and why Friction force must point inwards (if friction didn't, then nobody else would, so it must be friction that makes up the centripetal force). And to determine...
Yea, that's exactly the conclusion I was coming to right about now, since I was thinking that in all of these problems it always felt weird that centripetal force and these "counterbalancing" forces were all in the same direction. That's because they ARE (or create) the centripetal force.
So...
Ok, but does it hang at rest or does it hang while rotating along the pole? Given the numbers of this particular problem (one solution is zero, the other solution is impossible because of cosine > +1), my conclusion is that the only way the object can rotate is if it remains "along the side" of...
So basically "someone", namely Nx and/or Friction, has to "counterbalance" the centripetal force to prevent objects from flying off. In the bucket, N is already pointing towards the middle, so it can do the job on its own. If Nx is not enough to counterbalance centripetal force anymore, then the...
Ok ok ok I think I got it this time. When I do these exercises I need to put the sum of all the forces on the "centripetal force axis" equal to the centripetal force. So in this case I can do this:
X axis:
+Tx = F
(the only x component here is Tx, aside from F)
Y axis I can keep what I did...
Mmh. Why not the minus sign though? I thought + and - depended on the chosen axis. So if I chose +Y upwards and +X leftwards, for Tension it should be positive & positive, for F it should be positive too, but Weight force goes down, so it should be -
If that - was a +, then the solution would...
Homework Statement
An object of mass m is rotating, hanging from a string of known length l. The string is attached to a pole, which rotates with a known angular velocity ω and forms a to-be-determined angle α with the string. Find α.
I think I have solved it (the numbers match the results of...
And I know that this is getting really confusing to read, so I'll try to cut to the chase for now, and go straight for the crucial question then: the friction force's direction. So, the question: does friction force in a "object outside the cone-surface" (like the suitcase on the conveyor belt)...
Didn't I put it in (albeit in the wrong direction) with the yellow Ffr vector? Probably a bit hard to see, I reckon lol
Correct me if I'm wrong, but I think the maximum ω is the one I would find putting the sum of the forces = 0 on the X and Y axis. What I mean is, when the sum of the forces is...
Sorry for double-posting, but now that I think about it I don't actually know why the marble in the bucket would go up. I just know from experience that it would do that, but no idea as to why that happens.
After thinking more about the previous post, I realized that negative acceleration...
Note about the post prior to this one:
If I put the friction force UPWARDS (in the drawing it would point up and to the right, instead of down and to the left), then I get this acceleration:
[sin(α)-μ*cos(α)] * g/[cos(α) + μ*sin(α)] = a
With the data from the problem, this acceleration is...
N = Normal force
Ff = friction force
Fc = centripetal force
Y axis:
+Ny - Ffy - W = 0
N * cos(α) - μ*N * sin(α) - m*g = 0
N * [cos(α) - μ*sin(α)] = m*g
N = m*g/[cos(α) - μ*sin(α)]
X axis:
+Nx +Ffx - Fc = 0
N*sin(α) + μ*N*cos(α) = Fc
(using the N found on the Y axis, and being Fc = m*a where a...
You mean a maximum speed that allows the suitcase to stay on the surface, i.e. not-fly-off? If so, how would that work? (it might tie-in to what I say later in the post)
Oh, that's right lol Still, if that's friction's only job (to prevent further slipping, and therefore prevent change of radius...