A child is twirling a 0.0108-kg ball on a string in a horizontal circle whose radius is 0.116 m.?
820207
Srry, too lazy to do my own homework let alone someone else's.
823010
As I write this, the question has been posted for 3 hours without an answer, so I'll give it a try.  When you suspend a slack (not taut) string between two points, the string, having weight, will naturally form a parabolic shape between the points.\n\nEarlier I did the math, figuring out how long the string would be if strung across the 8-inch diameter and falling parabolically to a depth of 10 inches in the center.  This is a "length of curve" problem that involves some very messy calculus.\n\nThat's not the only problem.  Having done the math for the longest string -- the one that crosses the diameter -- I realized that the same problem would have to be repeated for each of the chords that will have shorter distances between the endpoints and shorter depths.\n\nAs a result, after reading your question again, I decided to take a new tack, focusing on the "how do you make" part of your question.  There's still some math, but not nearly as much as there would've been otherwise.\n\nTo begin, think of a circular tennis racket with loose strings sinking in the center.  That's the kind of net we're going to design.  For orientation, think of the 8-inch circular rim as the face of a clock. with 12:00, 1:00, ... 11:00 positions.\n\nFor the general shape, long strings will connect the 12:00 and 6:00 positions, falling parabolically to a central depth of 10 inches, and also connect the 3:00 and 9:00 positions, falling to the same depth.\n\nThere can be shorter strings connecting 1:00 and 5:00, 11:00 and 7:00, 10:00 and 2:00, and 8:00 and 4:00, and falling to a shallower depth (less than 10 inches).  These four strings will all be the same length.  Another set of four still-shorter strings will connect 2 and 4, 10 and 8, 11 and 1, and 7 and 5, again all of equal length and falling to a still-shorter depth.\n\nAll strings will be parabolic.  The longer and deeper strings will be in the center, and the shorter and shallower strings (the chords) will be toward the sides.  There will be an odd number of both vertical (e.g., 12-6) and horizontal (e.g., 9-3) strings because there's one central string in each direction and an even number of strings on the sides due to symmetry.\n\nTo form the net, the strings will all be woven as they interlace with each other.  I imagine that your net may be constructed with fishing line.\n\nHaving described the general structure of the net, we now have to do some math to figure out the spacing of the strings.  Going back to the tennis racquet idea, let's suppose we want the strings to be equally spaced <when viewed from above.>\n\nFor an example, suppose the strings are a half-inch apart.  Let the 12-6 line be the y-axis and the 9-3 line be the x-axis.  We want vertical strings at x-values of 0, 0.5, 1.0, 1.5, ... 3.5 (plus and minus) for a total of 15 strings.  There'll also be 15 horizontal strings for a total of 30.\n\nWe've already said that the central strings will attach across diameters (plus and minus 4) and descend to a depth of 10.  In three dimensions, suppose the bottom of the net is at (0,0,0), and the top circle is in the plane z = 10.  The equation of the parabola for the center strings is z = ax^2, passing through (4,10), so 10 = 16a, or a = 5/8, and the final equation in the x-z plane is\n\nz = (5/8) x^2  (Equation 1)\n\nLet's look at the placement of the string at x=3.  In the x-y plane, the circle has radius 4, and by the Pythagorean Theorem, the y-value is sqrt(4^2 - 3^2) = sqrt 7.  This string will be attached at (3, sqrt 7, 10) in 3 dimensions, with the other end at negative sqrt 7 in the y-dimension.  The distance between the attachment points is 2 sqrt 7.\n\nHow far down will this string go?  To get that, we evaluate Equation 1 at x=3:\n\nz = (5/8)(3^2) = 45/8 = 5 5/8 = 5.625\n\nThis means the <height of the netting> at x=3 is z= 5.625 inches.  But the top of the netting is at z=10, so the <drop> is 10 -
574466
A heavy pendulum with a ballpoint
433102
Since the piece of string is not necessarily straight, it is insufficient to simply measure the distance from one end to the other.  Furthermore, the length of the string depends on the observer - if the string were moving at relativistic speeds (that is v/c is not << 1) compared to the observer, the string would not be nearly as long as an observer that moved with the string.  To measure the proper length of the piece of string, consider an observer that moves with the piece of string and that the entire string itself moves at a constant speed (since if part of the string were spinning very quickly and another were motionless, the observer on the motionless piece would observe the spinning piece to have length contraction as above).\n\nSide note:  My girlfriend is telling me that she wants to kick me in the face for responding like this.. hope you enjoy! (:\n\nAnyhow, the appropriate way to find the length of the string is to move as an observer sitting on the string and integrating over the path at an instant in time.  This will give you the proper length of the string which is the longest possible measurable length for the string.  Any other measurement would find a value that is smaller than this and would vary depending on the speed of the observer compared to the string as well as any relativistic gravitational effects.
287697
The ball would execute simple harmonic motion in addition to being hugely affected by Coriolis "force" due to the spinning Earth. \nIf the tunnel were not spinning and the ball was dropped straight down the center, the equation governing the ball's motion would be ma = GM(r)m/r^2 where M(r) refers only to the mass of the solid sphere under the ball, since uniform spherical shells do not exert any force on objects inside them. Assuming uniform density p, we have M(r) = (4/3)*Pi*r^3*p, so a(r) = G*(4/3)*Pi*r*p. If we let k = G*(4/3)*Pi*p, which is a constant, we get the ubiquitous equation a(r) = k*r, or d^2r/dt^2 = k*r, which we all know the solution to as the simple harmonic oscillator r(t)=A*cos(Sqrt[k]*t + B) where in our case A is the radius of the Earth and B=0.\nTo complicate matters, due to the presence of some pretty dense air, it will be necessary to add dampening conditions to the differential equation, giving a damped oscillator.\nAlso, the Earth is spinning and since the ball is not rigidly connected to it, it experiences a Coriolis "force" towards the tunnel walls in addition to its oscillation, which will surely cause it to at least stop prematurely.\nIt is more likely the ball will be destroyed by hitting the sides of the tunnel at high velocity. ;)
840035
Ok, we're going to assume the 4.6268 cm is in addition to the 58.8cm, so the ball doesn't stop until it has fallen 63.4268 cm.  Therefore it has lost .0833*.634268*9.8 = .51778 Nm of energy.\n\nSince all this energy goes into compressing the spring, and its energy is .5kx^2, k = 2*.51778/.046268^2 = 483.74 N/m
757387
(a) Assuming the base is a half-circle and that there are no other forces acting on the mass (friction, drag, etc), then the kinetic energy at the base is equal to the potential energy at the top. Note that the change in potential energy is not affected by the semi-circular path the mass follows, but rather the net change in height. This net change is simply the radius of the half-circle container:\n\nKE = mgR = (.71 g * 1 kg/1000 g)(9.8 m/s²)(25 cm * 1 m/100 cm)\n\nKE = .00174 J\n\n(b) The kinetic energy associated with rotation is given as:\n\nKER = (1/2) I w², where w is the rotational velocity. Note that since the sphere rolls without slipping, the translational velocity of its cener of mass is simply:\n\nv = w * r\n\nSubstituting this into the expression for the rotational kinetic energy:\n\nKER = (1/2) (I / r²) v²\n\nNote that the expression for the translational kinetic energy is:\n\nKET = (1/2) m v²\n\nThus, the total kinetic energy of the ball at the base of the hemisphere is;\n\nKE = KER + KET = (v²/2)( I/r² + m)\n\nThe fraction of the rotational kinetic energy over the total is:\n\nf = KER / KE = (v²/2)( I/r²) / (v²/2)( I/r² + m)\n\nf = ( I/r²)/( I/r² + m)\n\nFinally, the moment of inertia, I, of a sphere is:\n\nI = (2/5)mr²\n\nSubstituting this into the expression for f:\n\nf = (2/5)m/( (2/5)m + m)\n\nf = (2/5)/( (2/5) + 1)\n\nf = 2/7\n\n(c) At the base of the hemisphere, there are two forces acting on the ball: the force of gravity and the centripetal force. At the base, these forces are coincident (act in the same direction). Thus the normal force, which opposes these two, is simply:\n\nN = Fgrav + Fcent\n\nN = mg + mv²/R\n\nTo find v, recall that KET / KE is 2/7. Using this, we can manipulate the terms to get v:\n\n(1/2) (I / r) v² = (2/7)(mgR)\n\n(1/2) ((2/5)m) v² = (2/7)(mgR)\n\n(1/5) v² = (2/7)(gR)\n\nv = sqrt((10/7)(gR))\n\nv = sqrt((10/7)(9.8 * 25 cm * 1m/100 cm))\n\nv = sqrt((10/7)(9.8 * 25 cm * 1m/100 cm))\n\nv = 1.871 m/s\n\nFinally, the normal force is computed as:\n\nN = mg + mv²/R\n\nN = (.71 g)(9.81 m/s²) + (.71 g)(1.871 m/s)²/(25 cm)\n\nN = (.71 g)(9.81 m/s²) + (.71 g)(1.871 m/s)²/(25 cm)\n\nN = .0169 Newtons\n\n\nGood Luck! :)
648691
Divide all 9 balls into 3 groups of 3 balls.\n\n1) weigh any 2 groups:\n         - if they are equal, the crooked ball is in the 3rd group\n         - if not, you have found the group where the ball is\n\n2) repeat the same experiment with that group until you find the ball
755334
Analyze the forces at the top of the loop:\n\nThe force of gravity pulling the ball downward,\nthe upward centrifigal force keeping the ball on the track,\nand the force of the track pushing the ball downward.\n\nThe answer to the question is when the centrifigal force is exactly equal to the weight, so the force of the track equals zero.\nmg=mw^2R\nw is the angular velocity\nw=sqrt(g/R)\n\nOnce you know the angular velocity, compute the equivalent linear velocity by mutlipying by R\n\nthen use mgh=1/2mv^2\nto compute h\n\nWhere is point Q?\n\nj
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.
870731
A Rube Goldberg machine
841543
Are you taking into account the fact that the moment of inertia changes once the child jumps on?  In that case you would get that the intial moment of inertia is 1/2MR^2, then once the child jumps on (treating the child as a particle) the moment of inertia becomes 1/2MR^2 + McR^2 where Mc is the mass of the child.  \n\nFrom the information you've provided, that seems to be where you went wrong.  You could also probably convert the linear velocity that you are given to angular velocity and then just say \n\nI(initial) w(initial) = I(final) w(final)\n\nHope that helped.
146548
Kid this isn't a magic b-ball.
196815
tie a piece of string round the egg, and then measure the string
217916
the diameter cannot be 9cm when the radius is 12cm because Raduis (r) = Diameter (d)/ 2 = 9/2 = 4.5 .\nRadius is half of diameter
15239
The correct answer is 190. It would involve any spare (all pins down on two balls) in the first frame, then single-pin spare conversions in frames 2-9 and a 9-count, spare and another 9-count in the tenth and final frame.
834149
The period T is 0.9s\nT = 1/f = 2pi/w so w = 2pi/T\n\nAs the movement is circular, the total of the forces is in this plan and to the centre and the value is mw²R.\n\nSo P + S = mw²R where P is the weight and S the tension of the string and R the radius and a the angle between the string and the vertical\nIn projection, we have :\nmw²R = S sin a\nmg = S cos a\n\nDividing the 2 equations :\ntg a = w²R/g\nand R = L sin a\n\nso w² = g/(Lcos a)\nso cos a = g/Lw² = g T²/(L 4 pi²) \n\nso cos a= 9.8 0.9²/ (0.3 4 3.14²) = 0.6702\n so a = 47.91°  or a =0.83 rad\n\nBut this angle is the angle between the string and the vertical so angle between the string and the horizontal is 90-a = 42.09°