I am having a disturbing problem with self-image. I'm 28- year old straight guy?

I am having a disturbing problem with self-image. I'm 28- year old straight guy? Topic: I am having a disturbing problem with self-image. I'm 28- year old straight guy?
July 18, 2019 / By Jerald
Question: and i'm thinking of getting married in the coming year but i am having disturbing thoughts about my self-image and it's shaking my self-confidence. My problem is that since i was young young my school peers used to describe me as looking pretty and they used to call me other names like soft, beautiful and having small penis coz i was skinny and petite back at school ! Even some girls used to describe some features in my face and body as being small, perfect or soft. For example one girl told me my hands are too small and that she preferred huge manly hands, another girl told me that girls hate guys whose facial features are close to perfect coz such features make men look soft. I once told a girl that i was thinking of marriage but she said that she couldnt imagine me getting married! I asked her why but she gave no answer, and that made me feel doubtful and in sever pain. All of these feed backs made me perceive myself as looking really soft although when i look in the mirror i find the opposite of this, even my attitudes shows that i'm everything but soft. People made me hate the way i look although i look really good and i admit i have some delicate facial features or let's say close to perfect features like a very nice looking nose,big eyes and my hands are somehow small. I think it's a silly thing to care about looks but it became an obsession and i'm always comparing my features to other guys and i started to believe that girls hate how i look, although many girls told me i look handsome. These thoughts made my confidence close to zero, and when i go to public places with a girl i start to feel that there's something wrong about me and my inner critical voice starts to tell me things like "u're soft, u're delicate, everybody here notice that, u're pretty boy" and i feel that some people are making fun of me, although i know i am exaggerating and i have no proof. Of course this leads to severe anxiety. I want to feel confident about myself and forget about these silly remarks made by mean guys and girls. I want to enjoy my life and marry the girl that i love and feel good about myself. Does looking so good mean that there must be something wrong with me as many people made me believe that? Does it mean i'm less manly? How can i become more confident and turn these thoughts into positive ones?
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Best Answers: I am having a disturbing problem with self-image. I'm 28- year old straight guy?

Gerrard Gerrard | 3 days ago
You dont mention your relationship with your fiance. I think this is a crucial point and the fact that you mention her only briefly suggests that you are not on a close enough plain with her. Do you really care what anyone else thinks? They are probably jealous and sense your insecurity. You need to throw your energy into your relationship with your loved one which if it is a healthy one should provide the answers you are looking for. Good luck
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Gerrard Originally Answered: Do most of the "Atheists" on here have some kind of self image problem?
Why does your avatar wear earrings? Do you have some kind of self image problem? Are you vain and shallow, and think you need shiny, twinkling baubles to attract people to your face? What's so bad about just having normal ears? Are you hell-bent on trying to improve how you look? (sigh) The poverty of critical thinking skills is going to be this nation's downfall. The great oak rots from the inside, and it starts with thinking like this.

Dob Dob
Hey check this out. No one is perfect except one man. Jesus. So everyone will have some faults. Makes us unique. Now you said u r handsome. A lot of guys don't even look nice at all. I think that should be a plus. That shoukd outway the small hands and delicate features. Also all girls like different things. The one u marry will be happy with you and the only one you should worry about. Also use these worries to your advantge. Why marry someone who is looking for a manly man and the relationship not work because you guys aren't compatable. This way if there really are problems with your looks whoever you marry will want you for you. That's what makes 50 year marriges. I used to have long hair and be fat. People thought I was a girl.about 8 yrs ago I started working out. I cut the hair. As I got older the more manly my face got. Now I have a fan club at work lol. But I'm married now and the same ones that shot me down years ago are the same ones flirting. But feels nice being the one turning them down now. I suggest you hit the weights. Not only is it healthy and will prolong your life it will leave in no ones mind any doubt about your masculanity. I bulked up in under a year with the right routine and right food. Also helped my brother lose over 100 lbs in a year. So if u need advice on it email me. Hit the weights and forget what people say. They have their on probs and when talking about someone else it makes them feel better. Oh also get a counsler. They can help u overcome these thoughts faster than anyone here can. You would be thankful u got one. God Bless.
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Boniface Boniface
Come ON, man, you can't be serious. You're too pretty is your problem? Do you have any idea how many ugly men there are in the world who wished they looked as good as you? And you're not gay, either? That's a blessing, son. So your looks create attention...welcome to the life of a celebrity. Some folks are gonna love you and admire you, some are gonna be jealous and hate your guts and fire spitballs at you. You CANNOT please everybody, and you better not go and try to make yourself ugly just to "fit in". Just be successful in what you do. Be successful in your job, in your family life, just be good at being you and be happy with your achievements. Your looks may set you apart from some others, but you aren't the blame for the hatred of others. They are responsible for that, and that's what makes them ugly. Be as beautiful on the inside as you are on the outside, and reserve access to your heart for only those who can appreciate you. All others, "Access Denied". Comprendez-vous?
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Adalia Adalia
believe me... many ppl think that girls only care abt how guys look,, but mature girls,, or in other words.. when girls meet someone they could fall in love with him disregarding his look,,, and abt that girl who wouldnt imagine u getting married, honestly i think sh elikes you...so.. dont care abt what pplm tell u abt how u look, ..becoz every person will have a different opinion in you.. so come'on enjoy life dude!! plus.. many girls like guys with "perfect features" .. ;)
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Adalia Originally Answered: Please help me in solving ***RING **** problem,,*****please click on the image to make it large *****?
Pearlsawme, it is correct that each of the left and the right mass has a velocity component of u towards right and a vertical velocity component of u. But, as per that logic the answer is 6 m u^2, which is the answer I initially gave and then, for some reason, changed. w = u/R Distance of left mass from the point of contact = R sqrt2 Therefore speed of the 2m mass on the left = u/R * R sqrt2 = u sqrt2 KE of the 2m mass = 1/2 * 2m * (u sqrt2)^2 = 2 m u^2 Speed of the mass m on the right = u sqrt2 KE of this mass = 1/2 * m * (u sqrt2)^2 = m u^2 Distance of the mass m on the top from the point of contact = 2 R Its speed = u/R * 2R = 2u Its KE = 1/2 * m * (2u)^2 = 2 m u^2 KE of the ring = 1/2 m u^2 + 1/2 * I w^2 = 1/2 m u^2 + 1/2 m R^2 (u/R)^2 = m u^2 Total KE of the system = 2 m u^2 + m u^2 + 2 m u^2 + m u^2 = 6 m u^2 Another way: Find moments of inertia about a horizontal axis through the point of contact with the ground. I of the 2m mass = 2m * (R sqrt2)^2 = 4 m R^2 I of the mass m on the top = m * (2R)^2 = 4 m R^2 I of mass m on the right = m * (R sqrt2)^2 = 2 m R^2 Iring = Icm + m R^2 (by parallel axis theorem) = m R^2 + m R^2 = 2 m R^2 Itotal = (4 + 4 + 2 + 2) m R^2 = 12 m R^2 Ek = 1/2 * Itotal * w^2 = 1/2 * 12 m R^2 * (u/R)^2 = 6 m u^2 ________________. @Pearlsawme, 1. Even after finding the center of mass of the system we arrive at the same answer. Ans: No, the answer will not be the same. Note that you will need to find the rotaional inertia about the center of mass of the system, which will be different from the rotational inertia of the center of the ring. You can try this but actually no need because calculations are too complicated and they do not serve any purpose because much simpler method are available. 2. Do the mass m on the left and right do have translationoal kinetic energy alone or they do have both types of energies? Ans: It depends on the frame of reference. With respect to a fixed point on the ground, the speed of each of these masses is usqrt2. If you consider this speed, then translational KE itself covers the whole KE. So no need to calculate rotational KE separately. With respect to the point of contact with the ground, each of these masses is in a pure rotation with angular speed u/R. So in that frame, you need to consider only the rotational KE as that will cover the entire KE. So it depends on the frame of reference. 3. The velocity of masses m on the right and 2m on the left is u√2. This is obtained by the resultant of two perpendicular vectors. Considering the mass m on the right, The translational energy is 1/2 mu^2 due to the velocity in the horizontal direction. The rotationoal energy is again 1/2 mu^2 due to the perpendicular velocity u. Their sum is mu^2. If we have taken the resultant velocity as u√2 and find the energy we get 1/2 m (u√2.}^2 = mu^2. Ans: It is correct that each of the masses on the left and on the right has KE = m u^2 but do not separate translational KE and rotational KE for the particles. Note that speed = v sqrt2. So considering translational KE alone will give you the complete KE of the particle. And, with respect to the point of contact, the rotational KE alone will give the total KE. ===================== 4. But the situation of the top mass is different. The energy is calculated to be 1/2 m (2u^2) = 2mu^2. This seems that the mass at the top do not have any rotaional kinetic energy and it is purely translational kinetic energy. That is the velocity is not contributing to the rotation. Or If we consider that the mass has rotational kinetic energy alone then it will result that the mass is not moving translationally. However, it is actually moving horizotally. Ans: See the answer to 2 above. It will be clear. 5. Mathematically telling u^2 + u^2 = 2u^2 But {u +u} ^2 = 4u^2 The top point has a linear velocity u as well and angular velocity ω. After a small interval of time, it will have a velocity u in the forward direction and another u tangential to its path. Ans: Which frame of reference? In the frame of reference of the instantaneous point of contact, the entire system has only angular velocity of ω. 6. My opinion is The energies should be calculated separately and these two u should not be squared afrer summing up. They should be first squred and then should be added. Ans: Note that the formula total KE = translational KE + rotational KE works only if the translational KE considers the speed of the center of mass and the rotational KE is calculated with respect to the axis through the center of mass. In this problem, if you want to use this formula, then you need to find the coordinates of the center of mass of the entire system, find the speed of the c.m., find the moment of inertial of the system with respect to the c.m. This will work but it will be unncecerrarily very complicated. It is important to note that the center of the ring is not the same as the c.m. of the entire system. ============ @Moon, You are making the same mistake as Pearlsawme. You are using the formula Total KE = translational KE + rotational KE You are taking rotational KE around the axis of rotation through the center of the ring. This is where you are making the mistake. While using this formula, the rotational KE must be the rotational KE around the axis through the center of mass of the system, which, in this problem, is different from the center of the ring. What if you need to find the kinetic energy of only the mass m on the top? At the given instant, this mass is moving towards right with speed u. If any particle of mass m is moving with speed 2 u, then its KE = 1/2 * m * (2u)^2 = 1/2 * m * 4 u^2 = 2 m u^2 But your method witll give KE = 1/2 * m u^2 + 1/2 * m u^2 = m u^2, which is obviously wrong. _______________. Note that even if some object is rotating if you add the translational kinetic energy of all its particles, you will get its total kinetic energy. @Rohan and anybody else who may be interested, try the following:- For simplicity, ignore the masses attached to the ring in this problem. Consider only the ring, which is rolling without slipping with its center's velocity = u towards right. Add the translational KE of all of the parts of the ring. You will get m u^2, which is what we got for the KE of the ring. Let me give some hint:- Fix origin at the center of the ring and x axis towards right. From the x axis, measured counterclockwise, consider an element of the ring between angles theta and theta + dtheta Find the mass of this element. It has a velocity component u towards right and another velocity component (again u) along the tangent. The angle between the two velocity components = 90 deg - theta (as you can verify by drawing a diagram). Find the speed of the element by finding the magnitude of the resultant of these velocity vectors (each of magnitude u and of angle 90 deg - theta between them). Find the mass of the element (note that the element has length R dtheta). Having found the mass and the speed of the element, you can get its KE by 1/2 * mass * speed^2 Integrate this from theta = 0 to 2 pi You will get m u^2 ============== Thanks Madhukar Sir.Considering your knowledge, your support indeed adds value to an answer. ============== @Pearlsawme, In your latest edit, you have written that the correct answer is 4 m u^2. But that is wrong for two reasons:- You have found the energy of the three masses as 1/2 m (2u) ^2 + 1/2mu^2 + 1/2 mu^2 = 3 mu^2. Note that one of the masses is 2m, which you did not consider. Secondly, all masses do not have speed u. The masses on the left and on the right have speed u sqrt2 each. We cannot assume that the masses are not attached to the ring because the question says that they are attached. The farther we go from the point of contact with the ground, the greater will be the speed. For example, if a mass were attached to the ring at its point of contact with the ground, its speed would be zero. @Pearlsawme: Finding the moment of inertia of the entire system about the point of contact and allowing the system to rotate with an angular velocity of ω, will give only the total energy of the system if the ring rotates about the point of contact and with out rolling about its center. We ignore the rolling of the ring and masses about its center in such calculation. But it is given that the ring rolls about its center with out slipping. Ans: We are not ignoring anything. The system is indeed in pure rotation about the instantaneous point of contact with the ground. ___________________.

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