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November 17

Two problems involving angular motion

I do not know how to approach these problems. The first is finding the initial angular velocity/speed of a wheel that rotates 5.3 radians to a complete stop in 3.4 seconds. The second is finding the angular acceleration of a 26 centimeter thick 2.9-kilogram cylinder that has a force of 16 Newtons being applied to it. --Melab±1 ☎ 05:02, 17 November 2012 (UTC)[reply]

The first problem isn't well-defined as stated; what's missing is presumably an assumption that the wheel is subjected to a constant torque, or equivalently, that it undergoes a constant angular acceleration. This problem might be easier for you to solve with a well-chosen change in variables. Think of a movie of the decelerating wheel that's played at the same speed as it was recorded, but backwards. The movie being played backwards will still be 3.4 seconds long, and will still show the wheel rotating 5.3 radians, still at the same constant angular acceleration except for a change in sign. What constant angular accelereration will produce a rotation of 5.3 radians in 3.4 seconds? And then given that value for the constant acceleration, what will be the speed of the wheel at the end of the reversed movie? Red Act (talk) 05:59, 17 November 2012 (UTC)[reply]

The basic procedure for the second problem is to look up the cylinder's moment of inertia at List of moments of inertia, determine the torque from the equation at Torque#Moment arm formula, and then find the angular acceleration from the equation at Angular acceleration#Constant acceleration. Red Act (talk) 06:19, 17 November 2012 (UTC)[reply]

However, we are missing some geometry info on the cylinder. Is that a hollow cylinder ? If so, and it's rotating about it's axis, we need an inside radius or diameter and an outside radius or diameter. Given that we have the thickness, we could find either one, if we had the other. We also need to know if it's made of a uniform density material. The length of the cylinder is not important, since we are given the total mass (it would be, however, if we were instead given the density). And where is this force applied ? Tangent at the outer diameter ? StuRat (talk) 22:32, 17 November 2012 (UTC)[reply]

civil engineering related

why moment of inertia of a hollow circular cross-section is more than a solid circular cross-section? — Preceding unsigned comment added by 49.137.37.116 (talk) 13:14, 17 November 2012 (UTC)[reply]

The weighted average distance from centre of all small mass elements making up a thin ring is equal to its radius, but the wesighted average distance from centre of all small mass elements in a solid cylinder must obviously be less that its' radius, as elements exist from the centre outwards. However, this does not mean that the moment of inertia will be less for a solid cylinder as its' total mass will be considerably greater. See the formulae in the List of Moments of Inertia article. Floda 121.221.79.138 (talk) 13:53, 17 November 2012 (UTC)[reply]

The moment of inertia for a single point is mr^2. Turning that point into a ring, or a cylinder, doesn't matter - you still have the same mass, rotating around the same axis at the same length. But making that ring solid means you take the mass that was at distance r, and distribute it in a column from the center point (moment of inertia zero) to the full distance r. However - the distribution is not even, because there's only one point at the very center. So a small segment of the ring, when distributed toward the middle, forms a narrow triangle, with the average point in the triangle at a distance of r over the square root of 2 away from the center. Twisted any which way, the result is half the moment of inertia. Wnt (talk) 20:49, 17 November 2012 (UTC)[reply]

if the eye changes focal length to focus on an object, why isn't our visual perception like a zoom lens?

I don't notice my visual field at infinity (f=17 mm) being that much wider than my visual field at f=22mm, at the minimum focusing distance, at least not on the order that I would expect a 30% change in focal length to achieve. 71.207.151.227 (talk) 17:45, 17 November 2012 (UTC)[reply]

That's because your eyes are not cameras and your brain is not a .jpg file. Visual perception doesn't really work like a camera. Of course, the laws of optics still apply to the lenses in your eyes, but beyond that how you perceive the electrical signals that get sent from your retina to your visual cortex is very different from what any camera does, and you can't really reliably draw analogies between the two. --Jayron32 19:46, 17 November 2012 (UTC)[reply]

In the eye, the image plane (retina) never moves in relation to the aperature (iris). Only the shape of the lens changes. Since the light moves in a relatively straight line from part of an external feature to a spot on the retina, the external feature remains about the same size. In a camera with a zoom lens, the distance from a lens to the film changes, but the lenses themselves are of solid and unalterable glass. Wnt (talk) 20:33, 17 November 2012 (UTC)[reply]
A zoom lens is not "solid and unalterable glass," but a collection of glass lenses which changes the spacing between elements to alter the focal length. The distance of some elements to the film may change both to alter the focal length and to adjust the focus. Edison (talk) 02:11, 18 November 2012 (UTC)[reply]
But if the focal length changes, the angle of view should change shouldn't it? A focal length is a measure of how strongly it bends rays overall, which in turn affects angle of view. It shouldn't matter by what method the focal length is changed. 71.207.151.227 (talk) 09:51, 18 November 2012 (UTC)[reply]
- - - I think NorwegianBlue has put his finger on it, below. To rephrase, there are two variables involved — focal length and magnification. A zoom lens changes both; your eye mostly changes just the former. --Trovatore (talk) 02:32, 19 November 2012 (UTC)[reply]
A more relevant comparison with a camera would be using manual focusing, which does not change the size of the image, but allows you to choose whether the foreground or the background should be in focus. --NorwegianBlue^talk 22:36, 17 November 2012 (UTC)[reply]

You've only got decent vision in your fovea anyway, so width of field is sort of irrelevant. Gzuckier (talk) 02:29, 19 November 2012 (UTC)[reply]

But the question is, what solid angle of the world is represented in that image on the fovea? If you could increase the magnification of your eye, then you could reduce that solid angle, zooming in on smaller features in the world. But you can't, to any significant extent. --Trovatore (talk) 02:38, 19 November 2012 (UTC)[reply]

Doesn't a zoom lens move a heck of alot more in relation to the film than the focal point of the eye's lens in relation to the retina?

Another opportunity for me to make a pitch for the excellent book, Applied Photographic Optics. Camera optics provide a user-interface that isolates focus and zoom as distinct entities - in other words, it is possible to focus, without changing field of view; and zoom, without losing focus. Very nice, expensive camera lenses can be focused with negigible parasitic zoom because they are complex multi-element glass. In my twelve-element 70mm-300mm telephoto lens, the "zoom" is actually driving a gear system that slides several elements - to correct for some optical aberrations, and to help stay mostly focused while continuously extending the focal length over two and a half octaves. If you have a simple lens, say a single element convex magnifier-glass - and you move it relative to the image plane to provide focus - then you'll also see a very obvious zoom or magnification effect. Several multi-element lens designs are discussed in the book, portions of which are available to browse for free online. Chapter 30.2 discusses practical considerations for distortion-free telephoto zoom. Nimur (talk) 13:25, 19 November 2012 (UTC)[reply]