moment of inertia of spherical shell?
922680
The moment of inertia of a sphere is:\n\n(2/5)* M * R^2\n\nFor a hollow sphere (spherical shell) you can subtract the hollow part from the sphere part, but since the M (mass) term will change based on diameter it is easier to substitute:\n\nM = p * 4/3 * PI * R^3     (p = density)\n\nSo.. in the end you get  I = 0.5333 p ( R2^5 - R1^5 ) \n\nWhere R2 = Outer Radius, R1 = Inner Radius
432088
*copied from previous post\n\nYour question is unclear about what set of assumptions you want to use. You seem to imply a perfectly constructed stool, with a perfectly centered force, using incompressible material. You want the stool to have no mass, and you imply that you want to use statics to solve this problem. Strictly applying statics, this system is indeterminate, NOT the ridiculous answer of 0 [N] in two legs and W/2 [N] in the others. There are only 3 equations of equilibrium here, 1 for vertical forces and 2 for moments. Since you have four legs, this system cannot be solved by statics alone. Treating the stool as a deformable structure, you can use geometric compatibility to get a fourth equation. I challenge you to find a competently built stool that has 0 [N] in any leg (a physical stool built of real materials), or an incompetently built stool with 0 [N] in any two legs (unstable system that is obviously not static). The answer indicated as "correct" requires an incorrect application of statics. A large part of studying physics is learning when to use a particular tool, and the physical limits of that tool.
877852
Inertia with outstretched hands is represented as I_o\nInertia with wrapped hands is represented as I_w\nInertia of the body only is represented as I_b = 0.40\nm is the mass of the arms\nL is the length of the outstretched arms\nr is the radius\n\nI of a rod = (mL^2)/12\nI of a thin ring = mr^2\n\nI_o = I_b + (mL^2)/12\nI_o = 0.40 + (8.0*1.8^2)/12\nI_o = 2.56\n\nI_w = I_b + mr^2\nI_w = 0.40 + 8.0*0.25^2\nI_w = 0.90\n\nUsing conservation of angular momentum:\n\nI_w * speed_w = I_o * speed_o\n0.90 * speed_w = 2.56 * 0.40\nspeed_w = 1.14 rev/s
665570
I assume delta(x, y, z) = 1 is the mass density ... isn't rho usually used for that?  Anyway, you need to integrate mass times square of perpendicular distance from axis of rotation.\n\nLooks like it will be messy.\n\nMaybe outer x from 0 to 1, middle y from 0 to 1 - x^2, inner z from 0 to (12 - 4x - 3y)/2, of delta(x, y, z) (x^2 + y^2) dz dy dx.
895868
Oh dear.  You should be able to do this one.  You simply need to calculate the mass of the core and divide by the total mass.  You have a density and a radius.\n\nMultiply the density of the core by the volume of the core (be sure you use the same units!)  I will leave it to you to look up/remember the equation for the volume of a sphere.\n\nTake the result and divide by the total mass. Are you sure that is the density, btw?  It seems awfully light for iron at the center of the Earth.  Even water has a mass of 1000 kg/m^3.
904251
Let's say F1 and F2 are the tensions (forces) in the inner and outer sections of the rod respectively, w is the angular velocity, and r the distance to the center (half the length of the rod). Equating the radial force to (mass x centripetal acceleration) for each mass gives:\n\nr*w^2*m1 = F1 - F2 = 2F2\n(2r)*w^2*m2 = F2\n\nDividing the second equation by the first all variables cancel except for the ratio of the masses:\n\nm2/m1 = 1/4
652894
the amount of pressure exerted on an object
919628
The volume of a spherical shell that has inner radius r and outer radius R is given by:\n\nV = 4/3 * pi * (R^3 - r^3)\n\nThe heights of the boundaries between the various layers of the atmosphere are not constant -- for instance, the top of troposphere is at about 7 km above the Earth's surface at the poles, and ~17 km at the equator.  For the sake of this calculation, I'll assume the following average values for the heights of the various layers (see source), as measured from the surface of the Earth:\n\nTroposphere: 0 - 12 km\nStratosphere: 12km - 50 km\nMesosphere: 50 km - 85 km\nThermosphere: 85 km - 640 km\nExosphere: 640 km - 10,000 km\n\nTo get the radii corresponding to these heights, we need to add the average radius of the Earth (6378 km) to each of these values.  Then we can plug those radii into the equation above to calculate the volume of each layer.  If you do this, you should get:\n\nV(troposphere) = 6.15*10^9 km^3\nV(stratosphere) = 1.96*10^10 km^3\nV(mesosphere) = 1.83*10^10 km^3\nV(thermosphere) = 3.17*10^11 km^3\nV(exosphere) = 1.70*10^13 km^3
974496
Equatorial surface gravity \n1.35 m/s²\n0.14g
826539
Moment of inertia:  2 opposite panes behave as 1 of 150 kg mass and 2.4 m width:\nInertia of above about center of mass = (1/12)mL² = 150(2.4²)/12 = 72 kg.m².  For 2 such panes, Itot = 144kg.m²\n\nTorque = Fr = 52N*1.2m = 62.4 Nm\n\nα = T/I = 62.4/144 = .433 rad/sec²
818010
Angular momentum is conserved,\n\nAngular momentum is (inertia * angular velocity)\n\nInertia for a disk = .5*mass*radius*radius = 0.5 * 5.6 * .2 * .2\n\nThe initial angular momentum is then: 0.5 * 5.6 *.2 * .2 * 1.8 (note that is in rev/s)\n\nThe inertia for the clay is: 0.5 * 3.1 * .04 * .04\n\nThe final angular momentum of disk and clay is then: 0.5 * ((3.1 * .04 * .04) + (5.6 * .2 * .2)) * f   where f is the answer you are looking for.\n\nInitial angular momentum = final angular momentum:\n0.5 * 5.6 *.2 * .2 * 1.8 = 0.5 * ((3.1 * .04 * .04) + (5.6 * .2 * .2)) * f or\n\nf = (5.6 *.2 * .2 * 1.8)/((3.1 * .04 * .04) + (5.6 * .2 * .2)) in rev/sec
966128
use the universal law of gravitation to find the force of gravity at the inner radius.  Use 1kg as your second mass (mass is irrelevent in this next part anyways), and use this force to find the accel of gravity.  Using the force and accel of gravity, you can find the kinetic energy needed in order to overcome the potential gravity of the planet (m cancels).  with kinetic energy you can find velocity, and with velocity and radius, period.  Hope that helps.\n\nfor the second part, repeat, except with the outer radius as D.
324486
Add v2/R.\nv=speed of rotation at the suface of the earth.\n(it depends on the geographic altitude fi)\nR = radius of the earth.
943735
a) T = dL / dt = 16\nL = 16 t \n\nI don't think you can do any better than that since they don't ask for the angular momentum at a particular moment in time, and it certainly is not constant.\n\nb) L = Iw\nw = L / I = 16 t / I = 16(33 * 10^-3) / (1.2 * 10^-3) = 440 rad/s
827798
You haven't told us what point it's pivoting about.\nThis is a Physical Pendulum problem:\n\nT = 2π√(I/mgL)......for small angular oscillations....\n\nI = mR²  where R is the mean radius of the hoop\nL = the distance from the pivot to the center of mass of the hoop\n\nGOTTA have L............
871488
Torque=F*D\nT=18*.35\nT=6.3\nWe know that T=a*i     [a=angular acc.\n                                       i=moment of inertia]\n         T=a*mr^2       [m=mass\n                                  r=radias]\n         6.3=a*2.65*(.35)^2\n         a=19.4 radian/sec^2\n\nProblem solned. Have a nice day.
535952
Assuming a big ring that rotates, we need to fix the radius and the velocity out on the edge.  The acceleration toward the middle (which causes the feeling of gravity) is given by\n\na = v^2/r\n\nTo match gravity on earth, assume 9.80 m/s^2\n\nYou can make the radius what ever you want.  Assume 100 meters.\n\nThen 9.80 = v^2/100.  So, v^2 = 980.  v = 31.30495.\n\nSo, how fast is it spinning?\n\nThe circumference is 2*pi*r = 2*100*3.14159 = 200*pi.\n\nThe time it takes for it to do one rotation is:\n\n200*pi (meters)\n--------------------\n31.30495 (meters/sec)\n\n= or 20.07088 seconds.\n\nThe angular velocity is 360 deg/20.07088 seconds = 17.9364 deg/sec\n\nor, 2pi radians/20.07088 = .3130495 rads/sec.\n\nNotice, this is exactly what you would have gotten by taking v/r and getting the angular velocity.  I don't know if you are studying that part of rotational dynamics yet.\n\nSo, the bottom line is, as long as v^2/r = 9.80, you can then get the rotational velocity in radians from v/r.