Can a massive object move in a straight line and still have angular momentum?
914874
yes if it is spinning on its own axis as it moves.  This is called Spin Angular Momentum
662166
A longish answer I picked up from the source link in reply to question 26 on the page. I'm quoting it verbatim herebelow\n\nQuote"\nBecause whatever it arose from--probably a cloud of gas and dust--was rotating to begin with.\n\nThe thing to keep in mind that even a very slow rotation of a cloud of objects gets greatly speeded up as it condenses at the center. The reason is a basic law (a consequence of Newton's laws of motion) by which a quantity known as ANGULAR MOMENTUM (or rotational momentum) is conserved. The angular momentum can be defined as the average radial distance, TIMES the average velocity of motion, TIMES the mass.\n\nThe mass does not change when matter collects near the axis of rotation, so we neglect the last part. Then, if the material collectes in the middle, where its average radius of rotation is 10, 100 or 1000 times smaller, the average velocity of its particles increases by the same factor.\n\nYou see this happen every day. When water drains from a filled bathroom sink, even if the water in the sink is rotating so slowly that you do not notice, by the time it reaches the drain it is spinning rapidly enough to form a funnel. (It does not spin in the opposite direction in Australia: the effect on which this claim is based is far too weak to have much effect. See "Stargazers," section 24). You also see this in hurricanes and tornadoes.\n\nAnd you see it happen when a large object in space collapses. In 1054 a star "went supernova" in the constellation of the Crab, a process in which the top layers blow off and the core collapses to a tiny "neutron star," perhaps 15 km across and as massive as the Sun. The collapsed core of the Crab Nebula apparently rotates 30 times a second, because that is the frequency at which it blinks in x-rays and radio (and I believe in its light, too).\n\nIn the solar system all planets orbit in the same direction and nearly in the same plane, and they and the Sun rotate in the same direction too (= counterclockwise, viewed from north). This suggests that they all condensed from the same cloud."   Unquote
878067
As long as neither of the balls starts spinning (or changes the rate at which it's already spinning) after the collision (in which case some of their linear momentum would have been converted to angular momentum), the total linear momentum of the system before and after the collision must be the same.\n\nLet p11 and p21 be the momentum of balls 1 and 2 before the collision; let p12 and p22 be the momenum of balls 1 and 2 after ths collision; and let m be the mass of the balls (which we are told have the same mass).  The momenta are all vector quantities, which I'll write as order pairs of numbers (i.e.,  p = (x,y) = x*i + y*j, where i and j are the unit vectors in the +x and +y directions, respectively).\n\nLet the +x axis be pointed in the east direction, and the +y axis be in the north direction.\n\nThen, writing each \n\np11 = m*(2 m/s, 0 m/s)\np21 = m*(-1.2 m/s, 0 m/s)\np22 = m*(0 m/s, 1.43 m/s)\np12 = m*v12 (v12 is the vector velocity of ball 1 after the collision)\n\nBy conservation of momentum, we have that:\n\np11 + p21 = p12 + p22\n\nm*(2 m/s, 0 m/s) + m*(-1.2 m/s, 0 m/s) = m*v12 + m*(0 m/s, 1.43 m/s)\n\nv12 = ((2 - 1.2 - 0) m/s, (0 + 0 -1.43) m/s)\n\nv12 = (0.8 m/s, -1.43 m/s)\n\nThe speed of ball 1 after the collision is given by:\n\nsqrt((0.8 m/s)^2 + (-1.43 m/s)^2) = 1.639 m/s\n\nThe angle the velocity vector of ball 1 makes with respect to the +x axis after the collision is given by:\n\narctan(-1.43/0.8) = -1.061 radians = -60.776 degrees\n\n(It helps to draw a picture of this to visualize why this relationship holds.)\n\nThe first ball is traveling 60.776 degrees south of east after the collision.
512623
I will state this here, in yet another question: THERE IS NO SUCH THING AS CENTRFUGAL FORCE!! Look it up in a basic physics textbook like Giancoli or Serway. Or even do a google search. You have to understand what reference frame you're talking about. The fact that your body tends to want to move to the outside of the curve when your car goes around in a circle has nothing to do with anything other than the fact that your body wants to keep going along in a straight line (conservation of momentum), but the car gets in front of it at every turn (semi pun intended). What is actually happening is that the car is an accelerating reference frame.\n\nAs for the moon/earth question... imagine a man 6' tall throws a tennis ball horizonatally with respect to the earth's surface. The ball will go forward a certain distance (because of it's velocity tangential to the earth's surface) as it falls the ~6' to the ground (because gravity pulls it down). The faster he throws the ball (horizontally), the further it will go before it hits the ground. If he throws it fast enough then the amount that the ball travels forward and the amount that it drops towards the earth's surface can match the curvature of the earth, and the ball will orbit the planet (neglecting air resistance, as we physicists always do) and come around and bonk the thrower in the back of the head. The moon is just a much larger tennis ball, thrown by a much larger man.\n\nYour eqn. is correct, but it only describes the force acting upon the moon. It is not the equation of motion, which detirmines what actually happens to the moon. The equation of motion is:\n\nF = m * a  (F = force, m = mass of moon, a = acceleration of moon around earth)\n\nIn the case of the circular motion of the moon around the earth, "a" is defined by:\n\na = (v^2)/r (where v is the magnitude of the velocity of the moon and r is the center-to-center distance from the earth to the moon)\n\nPlugging one into the other you get:\n\nG * M * m/r^2 = m * (v^2)/r (where M is the earth's mass)\n--> G * M/r = v^2\n\nSo, as long as the tangential velocity, v = sqrt(G * M/r), the moon falls just enough towards the earth as it travels just enough tangentially second by second, that it continues to orbit around the earth. (Note: there is NO CENTRIFUGAL FORCE anywhere in any of these eqns.!!)
697403
Thrusters would do the job, or the center can counter rotate. Thrusters could rotate the whole assembly or a motor, attached to a  massive central hub could cause the thing to spin while the angular momentum of the hub is going the other way.\n\nOnce you get the thing rotating, you don't really need to keep it going except for the occasional thrust once in a very very long time. Why? because I assume people inside will be moving around, that will alter the angular momentum, but very very little.
643953
depends on the direction of the force
874570
NO!!  If people would stop (mis)quoting half-remembered precepts here, and do a spot of THINKING, instead, we might get somewhere.\n\n1.  If a moving spherical ball directly hits an identical free, stationary ball head on in an elastic collision, the moving ball STOPS, and the previously stationary one heads off with the original velocity of the other.  So ALL of the moving ball's K.E. has been transferred to the other ball.  \n\nHere are two ways to see this:\n\n1a.  The device known somewhat indelicately to British schoolboys as "Newton's balls" (but in squeamish America as "The Swinging Wonder") provides an excellent table-top demonstration of this.  Most undergraduate physics instructors must haul this out from time to time.\n\n1b.  You can easily solve it analytically, but a thought experiment will also demonstrate it.  In the originally stationary ball's frame, let the other ball approach with speed v.  Now shift (with an appropriately directed speed of v/2) to the frame in which the two equal balls are simply approaching each other, each with a speed in this frame of that same v/2 (oppositely directed, of course)!  Clearly, from the symmetry of the situation, they'll bounce off each other with the same (oppositely directed) speeds in this frame, still of v/2.  Now go back to the original frame: the previously moving ball is now completely stationary, while the originally stationary one is haring off with speed v.  (You may need to make little sketches of what is happening as viewed successively from each of these frames, to see this; I admit that that's what I did.)\n\n2a.  A similar approach can be used to solve the general case for an arbitrary ratio of masses, m (initially moving)/M (initially stationary).  Hint: have the balls approach one another in their center of mass frame with equal but opposite momenta of mv and  MV.  They'll clearly bounce off one another in such a frame, retaining the same values of their respective (but now oppositely directed) speeds.  Then you can do the other frame-jugglings from there. \n\nHowever, this analytical approach, done accurately, is a tedious mess.  So, to appreciate the general problem more physically, consider just the following:\n\n2b.  The case of m << M.\n\nConsider the "equal but opposite momentum" frame for these two balls, in which MV = mv.  This means that if ball M is VERY MASSIVE, the speed it will have in such a frame, V = (m/M)v; that is, it's VERY SMALL compared to little puny m's high speed.\n\nSo, transforming to the frame in which M itself (the big guy) is initially stationary takes only a speed shift of O(m/M)v.  (In fact identically so.)  This only affects the speed of puny m minutely, so m's initial K.E. is still of order O(m v^2).  (There's no need to do picky constant book-keeping for these purposes.)  \n\nNow after the collision: In the original frame, Big M is now moving (and in fact at 2V if you think about it.)  But, being loose about the constants, M's K.E is only O(MV^2).  \n\nSo, the ratio (m's K.E.)/(M's K.E.) ~ (mv^2)/(MV^2)\n\n= (m/M)(M/m)^2 (since mv = MV)\n\n= M/m.\n\nBINGO!  The little bugger still carries by far the larger amount of the kinetic energy.  The massive ball (i) moves sluggishly, and (ii) because  K.E. involves both mass but also the SQUARE of the speed, the TWO POWERS of (small) speed more than cancel out the effect of its (large) mass.\n\nThis is a well-known fact in particle collision theory, that little particles basically bounce off big ones.  \n\nYou could also say that it's well-known in "space billiards": when spacecraft head close to Jupiter to receive a "gravitational sling-shot effect" that sends them off to the outermost planets.  \n\nIt's more complicated because of 3-D effects there, but analogous frame-shifting methods can be used to show that the spacecraft can pick up considerable speed AND change
407008
Only non-fluids will keep their shape.  If the energy of the constituent particles is not sufficent to overcome the bonding then the mass will keep its shape unless a large force is applied to it.  If the gravitational force is large enough it will most certainly brake objects; would you like to try walking past a black hole?
13785
No. As a simple example, imagine a snake curled into a perfect circle. Its center of gravity is in the center of the cirlcle the snake forms, which is within the empty hole in the middle.\n\nCenter of gravity is the average location of an object's weight.
584363
The laws of physics say that an object in motoin will remain in motion unless acted upon by an outside force.  It's hard to conceive of anything that would make Earth's core stop spinning short of some planet destroying catastrophe.
12398
The rotational energy you impart to the object when spinning overcomes the force of gravity that would topple a stationary object.  \n\nA total layman's explanation:\n\nA perfect stationary top perfectly balanced will not fall over.  In real life though,it's impossible to perfectly balance a top.  It's shape practically guarantees that its center of gravity will not be directly over the balance point for any practical length of time.  Imperfections in the material will cause a slightly uneven weight distribution, air currents will introduce a small force, and mechanical vibrations will be transmitted through the medium the top rests on.  The upshot of all this is that as soon as the center of mass is no longer over the resting point, gravity is going to pull the top over.  \n\nNow gravity has an acceleration of 32 feet per second squared (if I remember my ancient high school course correctly).  It can only topple the top at a specific speed.  If you spin the top fast enough, any side that gravity tries to "pull down" will translate around so quickly that it literally has no time to fall.  If you were standing on a planet with a much higher gravity, you'd have to spin the top far faster to achieve stability, and if you were on the moon, it would spin for a much longer time. Eventually, however, the top starts to slow and the tip will inscribe little circles.  That action is called "precession".  And of course, eventually the thing drops to the table and the game is all up once it slows enough.\n\nAgain, that's a layman's view of how it works.  Some college physics major is probably going to come in here and make my answer look really foolish.
69030
Initially, the ball is pushed, it slides. There is no rotation, as stated in the problem/question. This means there is no rotation at all. That all the velocity is translational.\n\nFor this scenario think of the ball as a solid block that has been pushed and it sliding across the floor.\n\nAs stared, at some time (t1) the ball begins to roll.\n\nAt this time, the ball's motion is both rotational as well as translational. In other words, it rolls as well as moving forward. This motion is what takes place at the *contact point* of the ball to the floor.\n\nIf the ball rolls, it is not slipping.\n\nIndeed, the friction causes the linear motion to decrease, and eventually stop the ball.\n\nThe rotational motion of the ball is also undergoing friction. A rolling sphere's motion is very subtle and often misunderstood. By virtue of the ball's mass there is a very tiny (tiny is relative to the mass and material) flat spot between the ball and the surface upon which it rolls, whereby a small torque gives rise to a resistance of the motion. This is why a rolling sphere slows. But.....since there is friction, the velocity of translation, is related to that of rotation by the following:\n\nv^2 = 2gh/(1 + I/Mr^2)\n\nWhere M is the mass of the ball and "r" its radius, I its moment of inertia. This is common notation. If there is no rotation, then the moment of inertia term drops out (I = 0) and you are left with a velocity which is strictly that of a translating body......kind of like free-fall.\n\nIf the ball is given one initial push to give it the initial velocity Vo, then unless there is some other forcing function at work, friction will act to decrease the motion of the ball, and all said velocities thereof. None of the velocities will increase.\n\nIf you know of a situation where such a case can exist where the ball increases velocity without doing anything to it.....make sure you mention my name when you pick up the your Nobel Prize.\n\n:)
1114940
Pressure can't be associated with a direction coz it is directed in many directions simultaneously. Weight is a vector coz it the gravitational pull that  is always directed towards th centre of earth.
90476
It is not effected to any significant degree.\n\nThe numbers used to do any calculations would absorb any effect which may occur.  For instance the mass of the Earth is about 5.9742 E24 kg.  But do you really think this is an exact number, no, it is just an estimate rounded off.  To try to use this value for the Earth's mass to try to calculate the effects of something so insignificant will not work too well.\n\nWhen you are standing on the Earth, you and the Earth together have a certain angular momentum.  When you jump, angular momentum must be conserved (its the law).  But now that your mass is rotating slightly farther out than it used to be, your angular momentum changes, there for the Earth's must change ever so slightly to compensate.\n\nIf you wanted "numbers and calculations", you did not specificy nearly enough information.
1067859
actually, it can! If you understand physics, the website below has proof. (but I dont know how hard you threw it.)
716944
If there is no net force (i.e. sum of forces is zero) then there will be no change in velocity. So if the object was going 15 m/s and the net force is zero, then it will still be going 15 m/, (for 2 sec = 30 meters traveled). Same concept for your jet car - constant velocity means no net force ( there may be forces acting on it, but if the velocity is constant, they must sum to zero - net force is zero).\n\nAs Newton's first law says, "An object in motion tends to stay in motion unless acted on by an outside force." So constant velocity (i.e. stays in motion at the same velocity) means no net force acting on the object.
737218
Given a rotating hub or wheel all the centrifugal forces are along the radii from every point on the hub to the center. Because of circular symmetry, all the these forces will cancel; one pointing up is cancelled by one pointing down. Therefore the net external force is zero; there is no way the linear motion is influenced.