- #36

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##

\mu <= cot \alpha

##

that's the answer but I don't really know what to do if the term becomes negative what could it possibly represent and mean

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Oh you are right, thank you!The case for part (b) is really not all that different, in terms of how you go about the problem. Now, you just need to find the forces in a third direction; the direction along the length of the cylinder (often denoted by p). Step one is breaking up gravity into its r, θ and p components, in terms of m, g, θ and α. From there, it should be...f

- #36

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##

\mu <= cot \alpha

##

that's the answer but I don't really know what to do if the term becomes negative what could it possibly represent and mean

- #37

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The expression ##\cos \theta = \sqrt{\frac{1 - \mu^2 tan^2 \alpha }{\mu ^2 +1}}## arose from setting the derivative of some function ##f(\theta)## equal to zero . What if the function doesn't have any point where its derivative is zero?if the term becomes negative what could it possibly represent and mean

- #38

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so for that case the maximum is reached with cos being real is angle pi/2 is it? assuming so

so is it ##

\omega^2 >=\frac{g}{r}(\frac{sin \alpha}{\mu} + cos \alpha)

##

- #39

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Yes, I believe that's right. You might check the special case where ##\alpha = \pi/2##.

- #40

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oh you it reduces toYes, I believe that's right. You might check the special case where ##\alpha = \pi/2##.

##

mg = N \mu

##

just out of curiosity why is the answer same as the first part if ##

\mu <= cot \alpha

##

what does it physically represent

- #41

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I was surprised, too, when I first saw that the answer for ##\omega_{\rm min}## is independent of ##\alpha## as long as ##\mu \le \cot \alpha##. I don't see a nice physical argument for this.oh you it reduces to

##

mg = N \mu

##

just out of curiosity why is the answer same as the first part if ##

\mu <= cot \alpha

##

what does it physically represent

The mathematical reason becomes clearer if you first show that for any ##\alpha##, the friction force ##f## and the normal force ##N## may be written as

##f = mg\sqrt{1-\beta^2} \,\,\,\,## and ##\,\,\,\, N = m\omega^2 R - mg \beta \,\,\,## where ##\beta## is defined by ##\beta = \cos \alpha \sin \theta##.

Then, ##\mu N \ge f## leads to

##\large \frac{\mu R \omega^2}{g}## ##\ge \mu \beta + \sqrt{1-\beta^2}##.

Thus, the minimum value of ##\omega## is determined by the maximum value of the right hand side of this relation over the domain of allowed values of ##\beta##. The only parameter on the right hand side is ##\mu##. So, you would expect that the maximum value of the right hand side depends only on ##\mu## and not on ##\alpha##.

However, there is a complication due to the fact that ##\beta## is limited to a maximum value of ##\cos \alpha## (since ##\beta = \cos \alpha \sin \theta##). Thus, you have the possibility that the maximum value of the right hand side of the above relation occurs at this upper limit of ##\beta## where ##\beta = \cos \alpha##. In this case, ##\omega_{\rm min}## depends on both ##\mu## and ##\alpha##, as you stated in post #38. This occurs whenever ##\mu \ge \cot \alpha##.

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- #42

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once again as usual sorry for insanely late reply didn't mean to be rude and forgot to thank you

thanks for the help i get it now

- #43

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- #44

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It took me a while to grasp what you are saying here. I think I got it. This graphical construction is a nice way to get to the answer.Sorry about bumping the problem, but apparently there is a different approach to part a and b using geometry. In part a it is clear that we have A fixed plane containing the vector g and vector w^2.R. In the reference frame rotating with the block we can see that the resulting field of these two vectors points in a circle (since vector g rotates) with that idea and some ideas about the friction and normal forces we can do letter a.

Yes, the graphical construction of the force diagram in the frame rotating with the cylinder for part b is more complicated than part a. I spent a lot of time on this. But once constructed, the diagram can be used to solve part b without a lot of calculation.Now my question is about letter b. Since the plane is moving with time, I can't see geometrically how things happen between vectors g and w^2.R. Only thing I know is that, because in reference frame rotating block is at rest, the forces mg, mw^2.R and the resultant of normal and friction form a triangle. So indeed they are at the same plane at each instant.

I'm not sure if my diagram is similar to yours. I chose an x-y-z coordinate system rotating with the cylinder as shown:

In the rotating frame of the cylinder, the block and the x-y-z axes remain fixed with the y-axis passing through the block.

The forces acting on the block are shown added together in the force diagram below where I reoriented the diagram so that the x-y plane is oriented horizontally in the figure.

The normal force N and the centrifugal force C are always parallel to the y-axis. In this orientation of the figure, the axis of the cylinder (z-axis) is vertical. In this frame of reference, the x, y, and z axes remain fixed while the vector representing the weight W rotates around the z-axis with angular speed ##\omega##. Thus, W sweeps out a cone with the tip of the W vector tracing a circle. W maintains a constant angle ##\alpha## to the x-y plane, where ##\alpha## is the amount of "tipping up" of the cylinder in the first diagram above. The friction force

The time-dependent vector S is defined as the sum of the forces N and

##\theta## gives the instantaneous angle that the tip of W has rotated from the x-axis.

##\varphi## is the angle shown in the triangle formed by

Is this similar to the diagram that you had in mind?

Note that the block is on the verge of slipping if ##\tan \varphi = \mu##. For the block to not slip, we must have ##\varphi \leq \arctan \mu## for all positions of W as W sweeps out its cone. Using this idea along with the geometry of the figure, you can deduce the minimum value of ##\omega## for which the block will not slip.

- #45

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- #46

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Your vector R is the same as my vector S. (I didn't want to use R, since this symbol is also used for the radius of the cylinder.) At any instant of time, all five vectors N, C, W,

It all seems complicated, but the only thing that we are really interested in is the maximum value of angle ##\varphi##. If ## \tan \varphi## exceeds ##\mu##, the block will slip.

The magnitudes of C and W remain constant in time. With that in mind, if you wanted to maximize ##\varphi## in the triangle CWS, how would you want W and S to be oriented relative to each other?

- #47

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As zwierz did, I have ##N=m \omega^2 R-mg sin \theta ## and ## N= \frac{mg cos \theta}{\mu}## and then ##\mu \geq \frac{g cos \theta}{\omega ^2 R-g sin \theta}## and so ## \omega^2 R \geq g sin \theta ## because the denominator must be > 1 (otherwise the fraction would be > 1 and the block would slide)

Why Is the normal force in this case not just ##N=mg sin\theta## ?? This is already the component of mg which is perpendicular to the surface. And since you have written ## N= \frac{mg cos \theta}{\mu}##, then I guess the frictional force is ## mg cos \theta##. Why is the frictional force the component of mg which is parallel to the surface?

- #48

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The normal force is found by using Newton's 2nd law in the radial direction: ##\sum F_r = ma_r##. What is ##a_r## for circular motion?Why Is the normal force in this case not just ##N=mg sin\theta## ??

Apply Newton's second law in the tangential direction: ##\sum F_t = ma_t##. The block moves inWhy is the frictional force the component of mg which is parallel to the surface?

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