What is the angular acceleration of the grindstone?
964637
750 rev/sec = 760*2pi radian/sec\n\n1/11th of that is "taken away" by the ax lady every second\n\n(760/11)*2pi radians/s^2
723260
Question 1:\n\nTorque is a force applied at a distance,\nT = F * r\nWhere T is the torque, F is the force, and r is the radius the force is applied away from the center or rotation.\n\nT = 40 Newtons * .15 meters = 6 Newton meters\n\nA torque causes an angular acceration,\nT = alpha * I\nwhere T is the torque,  alpha is the angular acceleration, and I is the rotational inertia ("moment of inertia").\n\nAn angular acceleration relates to a linear acceleration,\nalpha = a / R\nwhere alpha is the angular acceleration, a is the linear acceleration, and R is the radius of rotation (the wheel's radius in this case).\n\nThe linear acceleration can be found by finding the change in linear velocity with respect to time,\na = delta v / delta t\na = 2.25 m/s / 3 s = .75 m/s^2\n\nalpha = .75 m/s^2 / .5 meters = 1.5 radians per second^2\n\nT = alpha * I \n--> I = T / alpha\nI = 6 Nm / 1.5 rad/s^2 = 4 kg m^2\n \nDo your own homework next time (like on question 2).\nhttp://sln.fi.edu/qa97/spotlight3/spotlight3.html
970347
e, because of the Newton's 1st law: "An object at rest will remain at rest unless acted upon by an external and unbalanced force . An object in motion will remain in motion unless acted upon by an external and unbalanced force". Assuming inertial frame of reference.\nThe object can still have angular acceleration if there are two forces cancelling each other but are applied to an object at different points. \n\nIn real life there is always something that will slow down the object even in space.
77860
hey, didnt i answer something very similar to this for you? you need to do F=ma so 43000000N/2100000kg then you get an acceleration, but then you have to subtract gravity from this because thats the acceleration produced by the rocket but there is still gravity pulling with 9.8 m/s^2 in the opposite direction. so just do (43000000/2100000) - 9.8, and that's your answer :)
140474
Use the formula Ac = v^2/r = r w^2, where w = angular speed.\n\nSince you already have the angular speed, you can use the second form of the equation above.\n\nAc= r w^2 =(1.0 m) (2.0 rad/s)^2
728902
This can be done easily using a scientific calculator or there are many angular / decimal tables in Trig text books or many machinists handbooks or machinists guides.\n\nwww.industrialpress.com/en/tabid/63/Default.aspx
1126388
Surface tension and elasticity of the water are just about the same thing, for all practical purposes. The water can not get out of the way fast enough to let the stone sink before the motion and deflection cause the stone to bounce. The gyroscope motion of the stone is what gives it enough balance to not lose much of it's momentum when it hits something, like water. That's why flat, smooth stone work best.
1066934
After 1s, velocity of baloon is -2m/s (- as it is going against earth gravity). As stone is dropped from the baloon, it will have same velocity.\nDistance travelled by baloon in 1s is: 1m\nTherefore: s = ut + 1/2 a*t^2\n1 = -2*t + 1/2*10*t^2 (taking acc due to gravity, g = 10)\n1 = -2t + 5t^2\n5t^2 - 2t - 1 = 0\nt = (2 +/- sqrt(4 + 20))/10\nt = (2 +/- 4.9)/10\nt = (2 + 4.9)/10 = 0.69 sec\n(neglecting - coz negative t is not possible)
37325
first thing you need to do is calculate the moment of inertia of the wheel.  this can be done by treating the poles and the main wheel separately.  the moment of inertia of a circular disc (the wheel) is (1/2)MR^2, and of the rods (also a circular disc!) is (1/2)mr^2.  add them together to get the total moment of inertia.  \n\nyou'll need to use conservation of energy.  specifically, when the wheel drops a height h, the potential energy lost is -mgh.  this energy gets split into two parts, the rotational energy of the wheel and the translational energy (since it's moving downwards).  you can then write an equation setting the potential energy loss equal to the sum of the rotational and kinetic energies:\n\n-E_p = E_rot + E_trans\n-mgh = (1/2)Iw^2 + (1/2)(m + M)v^2\n\nhere I used (1/2)Iw^2 for the rotational kinetic energy and (1/2)mv^2 for the translational kinetic energy, where I substituted (m + M) for the total mass of your object.\n\nthe key point is now to realize that because the wheel unwinds along a string that is wrapped around the pole, the angular velocity w = v/r where r is the pole's radius and v is the translational velocity of the pole.  so you can substitute that into your expression.  it's not clear from your question whether you want the angular acceleration or the translational acceleration.  so you can either substitute the rw in for v, or v/r in for w.  \n\nin either case you then have a formula for the velocity as a function of the height of the wheel, which you can then solve for the velocity, and then take the derivative to get the acceleration.\n\ngood luck!
100113
no there isnt. in skateboarding/snowboarding the rotations are:\n 90*\n180*\n 270*\n360*\n 450*\n540*\n 630*\n720*\n 810*\n900*\n\nand it keeps going. the indented degrees are generally what you will spin into a jib/grind
965191
This isn't very hard - you just need to do each of the little pieces. I'll give some hints.\n1) When the block is at the extreme from the rest position all the system energy is stored in the spring. The stored energy is spring force (k*x) times spring deflection distance (x). So it's k*x^2 where x is 8.5. \n\nThe block has maximum speed when the block is in the center of the oscillation range (the spring is fully relaxed at that point), so all the energy is in kinetic energy. Energy is conserved, so\nMax spring energy = Max KE\nk*x^2 = (1/2)*m*v^2\n\n2) Best hint here: think, no math req'd.\n\n3) Spring force F = k*x. Acceleration = F/m\n\n4) Acceleration is max (see #3) at x = 8.5 and zero at x = 0. In the formula\nx = (1/2)*a*t^2,\nyou can use the average of the accelerations those 2 points for a.
852750
T=I*w’, T=16 torque, w’= angular acceleration, I=1.2/1000; In time t (SECONDS) its angular speed w=w’*t;\nOr w’=w/t or w’=T/I; thus w/t=T/I or I*w=T*t is angular momentum, & w=T*t/I
229137
There should be a pop out snap button on the right side about 18 inches or so down. pop that out and use a quarter inch rod about 10 inches long and slowly go in horizontally until you touch the spindle shaft. Now hold slight in-ward pressure while slowly turning the spindle/grinding wheel by "HAND ONLY" You should feel the rod fall into one of the spindle locking holes. Use Caution and make sure you turn off your main power before you proceed !! P.S. Try to visually center the spindle with this hole for easier access to the spindle lock holes.
1036388
Torque is given by the product of moment of Inertia and angular acceleration. \n  \nR - Radius = 13/2 inch =0.1651m\nMoment of inertia of the disc is ½ M R^2. = ½ x 15 x 0.1651 x 0.1651 \n=0.204435 kg m^2.\n\nAssuming that the disc is stopped in t seconds by the applied torque, the angular acceleration is 1800 x 2 x pi/ 60t radian per second^2 =188.49556/ t (radian per second^2)\n\nTorque = 0.204435 kg m^2 x 188.49556/ radian per second^2\n..............= 38.535/t Nm.
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
325473
Measure the weight of 1 cup of sand and figure it from there.\n\n :-)
1151145
Calculate acceleration a=(24-10)/16 = 14/16\nF=ma=1200*(14/16)=1050N