hotcommodity

Thanks for the replies. Well, the mass given for the rim and tire is 1.3 kg, and it's distance from the axis of rotation is .34 meters. Using I = MR2, the moment of inertia is .15028 kg*m2. I assume that you're saying the spokes would contribute 10% of the total rotational inertia, so that's 10% of .15028 kg*m2 which gives .015028 kg*m2. Then using the equation I = 1/3 MR2 with .015028 kg*m2 for the moment of inertia due to the spokes, and with a distance .34 meters from the axis of rotation, I get that the total mass of the spokes is .39 kg, which is about a third of the mass of the tire and rim. Am I following this correctly?

I'm stuck on a homework problem. There's a bicycle wheel, and we're given the mass of the wheel and the rim as well as their distance from the axis of rotation, and then we're asked to calculate it's moment of inertia. I understand that part. Then we're asked why the contributions due to the hub and the spokes can be ignored. I'm pretty sure that the contribution to the moment of inertia from the hub can be ignored because the hub is at the center of rotation, and would effectively have a distance of zero from the center of rotation. I'm stuck on the part about the spokes however. Unless they were of negligible mass, I don't see how their contribution to the rotational inertia can be ignored. I don't want the answer, I'm just looking for a push in the right direction. Any help is appreciated.

Oddly enough, we do have fasces in Congress: http://www.reference.com/browse/wiki/Fasces http://www.bloggers.it/theMinority/itcommenti/fasci2.jpg We may, as a country, have some characteristics of a facist state, however facism involves uniting people for a common cause. The Bush administration tends to leech off of society, rather than unite it.

You're saying that America can't function the way it does today without oil, and you're right, to a degree. However, with all of the educational facilities, innovative companies, and money that Americans have at their fingertips, I'm confident that this country could engineer a new way of living. A way of living that allows us to function as we function now. Saying you're willing to die for the progression of America is one thing, but saying that you're willing to die for oil infers you're willing to die for convention.

I doubt anybody loves oil enough to die for it. Some of the troops may have misconceptions about the war, but the politicians who lead them into war are well aware of the situation, that's why we pay them. It's rather pathetic when greed overcomes common sense and moral standing.

I think most, if not all, of the troops who have been injured or died in the Iraqi War would disagree with that statement.

Honesty can be so refreshing.

Thank you so much, these look like really good sites

You don't suck, you're trying to learn just like everyone else By the equation you wrote, I assume it looks like : [math] 2 - (-5) + 3^{2} [/math] There are essentially two things you need to do to make this a simple adding problem. First, consider what happens when you subtract a negative number. If I have [math] 2 - (-2) [/math] , I really have [math] 2 + 2[/math]. Minus a minus is plus a plus, as they say. Second, you need to calculate three to the second power. Anytime you have some number or variable squared (squared meaning to the second power), it just means you have that number or variable multipled by itself. For example [math] 2^{2} = 2 * 2 = 4 [/math]. Or if you have [math] 2^{3} [/math], you really have 2 multiplied by itself 3 times, that is [math] 2 * 2 * 2 = 8 [/math]. Knowing this, you should be able to answer your problem, I hope that helps.

I see, I think I have a better grasp on it now, thank you I was reading through my lab book a little further, and I'm still having a bit of trouble with uncertainties. Sometimes a value is given with ± 0.1, but sometimes they give a value with ± 0.02 or ± 0.04 and I'm just not getting where those numbers come from. Oddly enough, I'm getting the part about error propagation, but the basic determination of the uncertainty of variables using lab tools is throwing me. I know it's a bit hard to explain over the internet, so I'm wondering if there are any tutorial sites that lay all of this stuff out. Any help is appreciated.

Hello again, I have another question but this time concerning my physics lab. In lab we took measurements of a pendulum including its period in seconds, the length of the string the ball was attached to, the diameter of the ball, and the mass and weight of the ball. We considered the uncertainty in these measurements, and it was said that we could only measure the value of mass, period, etc. to some degree. I didn't understand this part very well. How do I know what degree of certainty I can measure an object to with tools like the meter stick, stopwatch, caliper, and so forth? Any help is appreciated.

Thanks for the reply. About the bullet and pendelum problem, I understand that the problem must be broken into two parts: analyzing the system before the collision, and analyzing the new system (the bullet and pendelum combined) after the collision. After the collision, the book says that total ME is conserved, and I know that to be true if and only if the work done by non-conservative forces on the bullet and pendelum is zero. I'm curious why the accelerating bullet isn't considered to be a propulsive force acting on both the bullet inside the pendelum and the pendelum itself. I know I'm overlooking a key concept, but I can't quite put my finger on it. And in case I don't get to it later, I'd like to thank you swansont for the help and insight you've given me throughout the summer, I know I'll put it to good use in the coming semester.

Thank you for the reply. I have a question about the impulse that acts upon an object. The impulse is defined as an object's change in momentum, and so initial and final velocities must be given values, in addition to the mass, to solve for the impulse. I know how initial and final velocities in the kinematic equations are dependent upon the displacement of the object in question, but what are they dependent upon when it comes to impulse? I would assume it's the objects initial velocity when the force begins to act, and its final velocity the moment the force stops acting, but I'm not quite sure. Additionally, I was looking at the bullet and pendelum example, where you're given the masses of the two objects and the height the pendelum swings, and asked to find the bullets initial velocity. When looking at the part where the two masses are combined and swing to some height, the book assumes that total mechanical energy is conserved, in other words, the work done on the object by non-conservative forces is zero. My question is why the bullets velocity is not considered a propulsive non-conservative force. And finally, I have a question that somewhat relates to the two questions above. I'm curious what determines the "initial velocity" of the bullet. I would assume it's the speed the bullet assumes after the force of the exploding gun powder stops acting on it, but I'm not certain. It wouldn't have any forces acting on it horizontally, so I wonder if the kinematic equations play a role since it has a constant velocity in both the x and y directions. Any help is appreciated.

I appreciate the replies above. Insane, I understand the object in question will move in the same direction as the net force, but I'm curious why it must "accelerate from rest." For instance, I would assume that the object could move at a constant velocity, and then accelerate over some distance, and the work-energy theorem would still be valid because 1) a net force occured, and 2) a constant acceleration occured over some distance. I don't see why the object would have to accelerate from rest only, if that what swansont was getting at, but maybe he was just giving it as an example. I do have another question however. I ran across a problem where the work-energy theorem was applied to an object in uniform circular motion, and they substituted the tangential velocity in for the velocity in the equation Work= 1/2*m*(vf^2-v0^2). I'm curious how this is possible since the work-energy theorem was derived using the kinematics equations, which apply to straight line motion with respect to x, y, and z components.

Thank you for the reply. I'm curious why the displacement must be in the direction of the net force if the mass starts at rest. Aside from that, lets say someone is pulling a sled, and lets say the sled is accelerated from rest, but the net force acting on the object was at angle (less than 90 degrees for instance). Could I still use the work-energy theorem if I only consider the force acting in the direction of the displacement, instead of the net force alone?

Thanks for the reply above I have two questions: The first involves uniform circular motion, and it's somewhat of an extension of what I've posted above. The Earth is a rotating mass, and would therefore contain different points in its mass that display particular tangential velocities, and centripetal accelerations. If I pick a point on the outer edge of the Earth, the radius being 3.38E6m, and the period being 8.64E4s, I get that its tangetial velocity is about 245 m/s. This doesn't seem correct to me. Are objects, or points of mass, on the outer edge of the Earth experiencing this kind of tangential velocity, or am I applying the concepts and equations incorrectly? Secondly, I was reading through the derivation of the work-energy theorem, and it shows that the kinematic equations were used, and that it was assumed that the net force was always in the same direction as the displacement such that the cosine of the angle between the displacment vector and the force vector does not come in to play. So my question is if the work-energy theorem, or more specifically the equation that the work done on an object is equal to the objects change in kinetic energy, can be applied to situations where the force is not in the same direction as the displacement. Any help is appreciated.

Thanks for the reply, you were right. I was integrating e^(2x) wrong. At least this way I got the same answer that the book had.

I appreciate the replies above. Well I worked out the integral using integration by parts only, and I got an answer much different from the books. I let u= sin(3x), du= 3cos(3x), dv= e^(2x), and v= e^(2x). This gives me: [math] \int e^{2x} sin(3x) dx= sin(3x)*e^{2x} - 3 \int cos(3x)*e^{2x} dx[/math] Then I use integration by parts a second time, with u= cos(3x), du= -3sin(3x), dv= e^(2x), and v= e^(2x). This gives me: [math] \int e^{2x} sin(3x) dx=sin(3x)*e^{2x} - 3cos(3x)*e^{2x}- 9 \int sin(3x)*e^{2x} dx [/math] I figured that since I have the same integral on both sides, I can just add 9 of that integral to each side, then divide the right side by 10, which gives: [math] \int e^{2x} sin(3x) dx= (1/10)*(sin(3x)*e^{2x} - 3cos(3x)*e^{2x} + C)[/math] So to check my answer I made the integral a definite integral, and checked each answer using a program on my calculator. Neither the books answer or my answer appears to be correct. Also, about the "dx" in integration expression, I'm still not grasping the concept that easily, but I'll come back when I have more time and question what I'm having trouble with in more depth.

I have a few questions, one concerning a basic integral that I'm having trouble working out, one about notation, and lastly about graphing a two dimensional function in R^3. This is the integral I'm having trouble with: [math] \int e^{2x} sin(3x) dx [/math] I'm pretty sure I have to start off by using u-substitution, and then use integration by parts. I think the u-substitution is what's giving me trouble. If I let [math] z= 3x [/math] then [math] dz= 3 dx [/math] and therefore [math] \frac{1}{3}dz = dx [/math]. If [math] z= 3x [/math] then [math] x= \frac{z}{3}[/math]. Then I would have [math] \frac{1}{3}\int e^{\frac{2z}{3}} sin(z) dz [/math] Is this correct? Secondly, I was wondering if someone could tell me what the "d" in dx means. I know that the derivative is [math] \frac{\Delta y }{\Delta x} [/math] as [math]\Delta x[/math] goes to zero. So does [math]\frac{dy}{dx}[/math] always imply that the change in x goes to zero? Why is "dx" in integration expressions? Lastly, i was playing around with my TI-89, and decided to graph y=sin(x) in 3D. It came up as a plane instead of a line. I guess I'm just having a hard time picturing why that is. Any help is appreciated.

I appreciate all of the suggestions above. I hope Glider is right about fleas being specific to what they like. My neighbor came home last night and offered to help me take care of the dog for the time being. We picked up some flee-shampoo and gave it a good bath. She said vinegar helps kill flees too, so we soaked it in some vinegar as well. It's still not potty trained, so it peed and pooed on my neighbors floor. Her reaction was classic, I had to go out on her balcony because I was laughing so hard. Anyways, we're still trying to find it a home. But for now, shes keeping the pup at her place while I sanitize mine.

Do you think running my sheets and clothes through the washer will get rid of what ever is on them?

So I've done something incredibly stupid. I picked up this stray dog about a day and a half ago, and have been trying to find it a home ever since. Ten minutes ago, I noticed it has fleas I read some stuff about fleas on wikipedia and I'm totally freaking out. It says they lay eggs and get everywhere. How can I sanitize myself, my apartment, and the dog inexpensively? I dont have the money to buy it a collar right now but I just need to know how to clean it all up. Any help is appreciated.

I really appreciate the help, thank you. I was working a problem on forces last night and got stumped, so I used my other physics text as a reference to help me out. It had the same problem I was working, only the exercise used arbitrary masses M and m to find the acceleration of two blocks attached by a massless rope, and the tension between them. This is the exercise: There are two things I dont understand about this example. First, on the left hand side it shows two axes, one is upright, but the other one was turned upside down to make "mg" a positive quantity, and I dont see why they did that. Secondly, I used the acceleration equation that they derived for the arbitrary masses, and it worked for the problem I was working on. But when I used the equation for the tension, it gave me the wrong answer. If I leave "g" out of the tension equation, it gives me the answer that the book has. Was the equation for the tension derived correctly, is my other book wrong, or is there some other concept that I'm missing?

I have another question about uniform circular motion. I was plugging in numbers to the equations that describe tangential velocity, radius, and period and found some differences between the gravitational force and other centripetal forces such as tension. For a centripetal force like tension, I found that: 1. As the radius increases, the tangetial velocity increases. 2. As the radius increases, the centripetal acceleration increases. 3. As the tangetial velocity increases, the centripetal acceleration increases. But for a centripetal force caused by gravitation: 1. As the radius increases, the tangential velocity decreases. 2. As the radius increases, the centripetal acceleration decreases. 3. As the tangetial velocity increases, the centripetal acceleration increases. Number 3 is the same in both cases, but 1 and 2 differ. Can this difference be attributed to the gravitational field?

Well, when I attended middle school and high school, I certainly noticed a moral decline among my peers. The kids hear rap/hip hop/violent music on the radio, they watch how these degenerates act on television, and they think it's alright to imitate them. The people that are in charge of what they listen to and watch on television tend to care less about the trends that they set in society, and more about making money. I see the same thing today that I saw 10 years ago, and its only getting worse. The unfortunate thing is that the juveniles you experienced on the bus will probably be raising kids of their own some day, and the question is if they'll have the sense to teach their kids to act any different. I'm not so sure this moral decline is a trend that will eventually die out. It's the type of thing that sends the respectibility of society spiraling downward.