what is spherical pendulum ?
491270
In contrast to a plane pendulum, which can just swing in one plane, a spherical pendulum can swing in all directions except that the distance to the suspension keeps constant, so the swinging mass is always on the surface a shpere around the suspension.
1139262
A "sundial" casts a single shadow from its gnomon (the point that comes out of the top). \n\nAn "armillary sphere" however is a different device altogether. It helped in the understanding of the earth and stars. Many armillary sheres were created with the sun in the middle, therefore casting numerous shadows depending on which point in space you were.\n\nWhile on earth at 1:38pm a sundial will cast one shadow.\n\nCheck out these links for more information:
490529
For circular membranes, the radial part of the vibrations is a bessell function while the angular part is an ordinary trigonometric function. This comes from writing the wave equation in polar coordinates and doing a separation of variables. The radial function turns out to satisfy Bessell's equation. The frequencies then turn out to correspond to the zeros of the bessell functions.
865877
Check this out\n1. http://en.wikipedia.org/wiki/Spherometer\n2. http://online.physics.uiuc.edu/courses/phys402/exp/Sph/SPH.pdf\n\nRef 2 provides you the information you are looking for.
279317
gravity.\n\nany shape of mass that is not spherical is gravitationally unstable
631468
Remote controlled movable ball amusement device\n\nA remote controlled movable ball amusement device includes a hollow sphere having a propulsion mechanism within the sphere. The propulsion mechanism includes a receiver and a drive unit. The drive unit causes the sphere to move when actuated by the receiver. A remote transmitter sends signals to the receiver for causing the actuation of the drive unit.
703506
A simple gravity pendulum or bob pendulum (plural pendulums or pendula), is a weight (or bob) on the end of a rigid rod (or a string/rope), which, when given an initial push, will swing back and forth under the influence of gravity over its central (lowest) point.\n\nApplications;\n\n1.Timekeeping:\nThe most widespread application is for timekeeping. A pendulum whose time period is two seconds is called the seconds pendulum since most clock escapements move the seconds hands on each swing. Clocks that keep time with the use of pendulums lose accuracy due to friction.\n2.Gravimetry\n3.Seismology\n4.Schuler tuning\n5.Coupled pendulums\n6.Two coupled pendulums form a double pendulum. Many physical systems can be mathematically described as coupled pendulums. Under certain conditions these systems can also demonstrate chaotic motion.\n7.Pendulums for divination and dowsing:\nPendulums (these may be a crystal suspended on a chain, or a metal weight) can also be used in divination and dowsing. See pendulums for divination and dowsing for a more detailed discussion.\nFor more information pl. click link bellow:\nhttp://en.wikipedia.org/wiki/Pendulum
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The simple pendulum requires a massless connection between the pivot and the oscillating mass.  Also, the mass must be concentrated at one discrete radius (can have no radial thickness).  Add air and pivot drag to these and you essentially have the difference between theory & reality........
801554
Assume spherical approximation for balloon:\n\nVolume of weather balloon at time, t is:\n\nV(t) = 4/3 pi r(t)^3\n\n(1).....dV(t)/dt = 4 pi r(t)^2 x dr(t)/dt  (chain rule)\n\nlet t be in minutes\n\nr(t)= .04t m,  that is (500/1000)*0.08*t m\n\nd r(t)/dt = .04m/min\n\nSo when r(t) = 90cm = .9m, we have from (1)\n\nd V(t)/dt = 4 pi (0.9)^2 x .04 m^3/min\n\n=0.4072 m^3/min
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As the mass increases so does the gravity. A spherical object is shaped this way because of its gravity. The larger the sphere gets the smaller the curvature of its surface becomes. It is related but curvature is more of a side effect of a large rotating object. Gravity is dependant upon the mass of the object not the curvature of it.
641595
The term is oblate spheroid I believe. Get that extra credit!
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Coriolis Fairy is a myth about the Coriolis Force wherein it is believed that in the northern hemisphere, water goes down the toilet clockwise. And it twirls counterclockwise in the southern hemisphere.\n\nTechnically speaking, the Coriolis is not even a real force, but a fictitious one resulting from a rotating frame of reference. It's an apparent deviation caused by the inertia of an object traveling above the earth's surface while the planet spins eastward below; it's fictitious in the sense that if the globe were not spinning, there would be no deviation. This effect was first described by nineteenth-century French mathematician Gustave-Gaspard Coriolis, who evidently became interested in the motion of rotating bodies after writing "The Mathematical Theory of the Game of Billiards."\n\nTo understand why it works, imagine that you are looking down at the globe from above. The earth would look like an old-fashioned vinyl record player, with the North Pole for the axis. At the center, the record moves so slowly that it appears to stand still, while at the far edge the record is traveling quickly, where it has more distance to cover in each revolution. Similarly, at the equator, the earth is moving from west to east at 1,040 miles per hour, while at the latitude of Long Island it is moving closer to 700 miles per hour.
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Conserve both momentum and total energy (i.e., potential energy plus kinetic energy).\n\nMy coordinate system will have the pendulum lifted to the left and the block on the right moving left, so that the initial momentum is - M2 Vo2.\n\nIt will also have the block at height 0, so the initial potential is M1 g h where h is the initial height of the pendulum, which looks like L(1 - sin(10 degrees)).\n\nThe initial kinetic energy is of course (1/2) M2 (Vo2)^2.\n\nSo ...\n\ninitial momentum: - M2 Vo2\ninitial energy: M1 g h + (1/2) M2 (Vo2)^2\n\nAfter the collision the potential energy has decreased to zero, and if w is the combined velocity of the two masses then ...\n\nfinal momentum: (M1 + M2)w\nfinal energy: (1/2)(M1 + M2)w^2 + [the energy "loss" from the slight heating and deformation of the two masses sticking].
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1.\nPhases:\nf1 = (t/T1)*2*pi, where T1 = 1 sec\nf2 = (t/T2)*2*pi, T2 = 1 sec\nIn phase: f1 = f2 + 2*pi*n, where n is integer (0, +-1, +-2, ...)\n(t/T1)*2*pi = (t/T2)*2*pi + 2*pi*n\nt(1/T1 + 1/T2) = n\nT2 = 0.98 = 98/100 = 49/50, 1/T2 = 50/49\nt(49/49 - 50/49) = n\n-t/49 = n\nt = 49 seconds, n = -1\nNumber of oscillations if second, shorter pendulum (shorter pendulum has shorter period): N = t/T2 = 49/(49/50) = 50\n\n2.\nCoordinate: y(t) = 3*sin(20*pi*t)\nSpeed: v(t) = y'(t) = 20*pi*3*cos(20*pi*t)\nMean position: y(t) = 0. Take t0 = 0.\nMaximum displacement: y(t1) = amplitude = 3, speed = v(t1) = 0\nv(t1) = 0\ncos(20*pi*t1) = 0\n20*pi*t1 = pi/2\nt1 = 1/(2*20) = 1/40
759467
The pendulum mechanism  consists of 4 parts. The weight, the pendulum arm, the suspension spring which is at the top of the arm and an adjusting counterweight. The pendulum is connected via a pivoting device to the escapement mechanism at the top of and behind the pendulum suspension spring. The pendulum weight is suspended from the pendulum arm. The pendulum swings through an arc with a central axis which is adjustable in most pendulum clocks.The reason your clock is stopping is because the escapement with its ascociated timing parts, is not working properly. This is because the escapement operates as a balanced mechanism via the balanced operation of the pendulum. This adjustment is generally via a counterweight on the pendulum arm and usually looks like a small circular weight about half way up the pendulum arm which can be moved either left or right to correct for offsets in the balance of the clocks frame and corrects for level of the clock, the clock should be level if it is a mantle clock and evenly level across the face if it is a wall type clock. The adjustment of the pendulum is critical for the timing of the clock as it adjusts the operation of the escapement mechanism which is its heart. When timing the clock ensure that the sound of the "ticking" is even and not fast on one side and slow on the return of the pendulum swing. Before adjusting the pendulum make sure the clock is actually level both front to back and across the face, you can measure front to back by looking at how vertical the pendulum is and measure left to right by setting a level across the face from the 3 to the 9. After levelling then do the pendulum adjustment. Sometimes the pendulum weight itself can be adjusted left to right in its fastener in relation to the pendulum arm.The axis is the pendulum arm's swing arc and the pendulum's total swing arc is known as its moment.The axis can be divided into 2 sides from the centre at rest position of the pendulum arm, left and right, and these 2 sides need to be the same amount of swing if the clock is to work correctly. If there are no adjustments on the pendulum arm or the pendulum weight, as can be the case on the cheaper models of clock then the timing is only adjustable by lifting one end of the clock off from level either on the left or the right depending on the longest "tick". If the uneven tick is longest on the left pendulum swing then that left side needs to be lifted up to extend the pendulum swing to the other side. If there is an adjustable counter weight on the pendulum arm then the same principle applies. If the "tick" is longest on the left swing then the counterweight needs to move slightly to the right to counteract the out of balance swing moment this will make the pendulum swing further to the right on the next "tick".I hope that this information is useful. I used to service clocks as a hobby and have some at home that are still running nicely after some years in service.
914856
OK as the ball is swung around in a horizontal circle the scale inside the tube is measuring the tension on the string. By knowing the time to complete one revolution  linear velocity can be found from v=r*omega(angular velocity). in the second part the v is given and needs to be converted into radians/second by using 3.7=.5*omega where omega=3.7/5=.74rad/s then using the mass radius and angular velocity you can find the tension in the string. With a= acceleration (into center) omega^2*r=.74^2*.5=.2738m/s^2and using f=ma the tension in the string will equal f which equals .2*.2738=.0548newtons Unless i have a silly math mistake somewhere.
574466
A heavy pendulum with a ballpoint