What is the frequency of the wheel after the clay sticks to it?
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
445486
Because the up/down motion is independent of the side-side motion. The stone is already going at whatever speed the car is and that doesn't change. The only thing that changes is the up/down part of the motion. For someone outside the car, the stone looks like it goes along a parabolic arc, not just up and down.\n\nThis is related to the relativity of motion. If you are moving uniformly, it is impossible to tell whether you are moving or not unless you look outside the car.
959353
All of the power from the motor goes to friction because they just let the wheel coast and burn up all the energy.  But, I'm sure that wasn't the intended answer for the problem\n\nYou need to make a few preliminary calculations:\nFirst you need the Inertia of the flywheel which you can find given the mass and radius of the wheel.\n\nThen you need to calculate the angular acceleration required to bring an object at rest up to 5.88 rad/s in 0.531 sec.  Using this, you can get the torque of the motor - friction.\n\nFinally, calculate the angular acceleration required to bring an object spinning at 5.88 rad/s down to rest in 26.8 revolutions.  Using this, you can get the torque of the friction by itself.\n\nAdd the second result from the first to get the torque of the motor.  Take the friction torque and divide by the motor torque times 100 to get percentage of friction.
804155
a) T = 0.5 sec\n\nb) freq = 1/T = 2 Hz\n\nc) w = 2pi*freq = 4 pi\n\nd) w^2 = k/m\nk = m*w^2 = 0.5*16 pi^2 = 8*pi^2\n\ne)  V_max = A*2*pi*f = 0.35*2*pi*2 = 1.4 pi m/sec\n\nf) A_max = A*(2*pi*f)^2 = 0.35*(2*pi*2)^2 = 5.6*pi^2 m/sec^2
676957
E = 1/2 M V^2 where V is the velocity.  Either we need to know the distance of travel of the load or the speed of the vibrating load.\n\nA load  hitting the surface 3 times a second going back and forth 1 ft each way will exert greater energy than a load traveling only 1 inch 3 times a second.\n\nTherefore, if the distance of travel is "d"\nE= 1/2 [M (df)^2]
26747
ok.  call the downward translational velocity of the whole thing v.  this means that the tangential velocity at the edge of the poles (at the radius r) is equal to v.  the reason for this is that the strings are wound AROUND THE POLES, so this winding and the condition that the string has to unwrap for the thing to move, sets the angular velocity w = v/r.\n\nnow, if you look at the main wheel part, the angular velocity has to be equal to the angular velocity of the poles, since they are one solid piece.  however, as you note the radius of the wheel R > r.  so, the fact that v / r = w = V / R implies that V > v.  in other words, the motion of the outer edges of the wheel is faster than the motion of the outer edges of the rods.\n\ni think you are getting confused by saying that V is the translational velocity of the wheel.  it's not.  v is the translational velocity of the wheel, and of the rods.  the fact that the string is wound *around the rods* and not around the wheel sets the angular velocity of the whole apparatus equal to v/r.  V, the tangential velocity of the outer edges of the wheel, is much faster than the translational velocity.\n\nthat clear?
499147
The period of a pendulum is given by\nT = 2*pi*sqrt(L/a)\nWhere T is period in seconds, L is the length, and a is the acceleration (in this case, due to gravity). Frequency, if I remember correctly, is 2*pi/T. Use this information to solve.\nInitial speed doesn't matter.
393990
It's not a characteristic of the human eye, which does not have a "frame rate". The effect occurs with devices like TV and movie cameras which have frame rates of 16 to 60 Hz. Any data sampling system is prone to such aliasing. If spokes pass a given point at exactly the frame rate, or at exactly 2, 3, etc. times tha frame rate, the wheel appears to be stopped. A slight change in wheel speed plus or minus makes the apparent wheel speed that plus or minus amount.
1044821
Frequency 2Hz\n\nTime period = 1/2 s\n\nAmplitude = 10 cm.
374463
Assuming no air resistance, the first equation is:\n\n400=1/2 a t^2  (a, as you said, is 32.2 ft/sec^2)\n\nt, the time calculates to be abot 4.98 seconds\n\nthen, v=at, gives v=(32.2)*(4.98)=160 ft/sec\n\nIn the real world, the speed after 400 feet of fall will be less than that due to air resistance, which would slow a brick considerably at 160 ft/sec (over 100 miles/hr)
134621
This can be solved by conservation of angular momentum.\nSince net external torque on the merry go round is zero (as it moves with constant angular velocity), conservation of angular momentum can be applied.\n\nInitial Momentum = Final Momentum\n2350 * .6 = [(2350 + 4*65*5.2*5.2)*(New angular velocity)]\n\n1410 = [(2350 + 7030.4)*(New angular velocity)]\n1410 = 9380.4 * (New angular velocity)\nThis gives the new angular velocity to be 0.15 radian per second.
858881
12.25cycles/20sec = 0.6125 cycles/sec, = 0.6125Hz, which is the frequency.  The period is the reciprocal of the frequency, which equals approx. 1.6327 seconds.\n\nI didn't count that first half cycle because it looks longer than the rest.  I started counting at the first positive zero-crossing at time 2.6, and counted 12.25 cycles from there to time 22.6\n\nThis is coming from someone who has made his living for the last few decades looking at and interpreting thousands of such displays.
1120959
You are looking for a plasticine clay.  It does not get hard and is very maliable.
338364
Wagon wheel effect read about it at http://en.wikipedia.org/wiki/Reverse_rotation_effect
958811
You physics types ALWAYS have to delete something, eh? Like mass, or air resistance or friction! lol\n\nAnyway, I think the ball would be a) either moving forward or stationary, depending on the wagon's rate of acceleration, relative to the ground, and b) rolling backward (due to inertia), relative to the wagon.
256104
Read Newton's 3rd law of motion.\nAlso, use F=ma to figure out your acceleration.
791539
Wow. Your classmate just asked this question. But here's your answer anyways...\n\nThere are a few things to keep in mind here: all of the energy in the initial system is rotational, and the initial energy is equal to the energy lost to friction. Also note that the energy is lost to friction at a constant rate.\n\n(a) The initial energy of the system is:\n\nE = (1/2) * I * w²\n\nWhere I is the moment of inertia and w is the angular velocity. The power transferred from the system due to friction is equal to the time derivative of the energy loss:\n\nP = dE/dt = I * w * a\n\nwhere a is the angular decceleration.\n\nSince the power is transferred at a constant rate and removes all of the energy from the system, we can take this as:\n\nP = -E/t (negative sign denotes power loss)\n\nCombining the power equations yields:\n\nI * w * a = -E/t\n\nSolving for a:\n\na = -E / (I * w * t)\n\nFinally, substitute the expression for E to obtain:\n\na = -I * w² / (I * w * t)\n\na = -w/(t)\n\nSubstituting the values yields:\n\na = -(150 rev/min * 2*pi rad/rev * 1/60 min/sec)/(2.2 hr * 3600 sec/hr)\n\na = -(15.71 rad/s)/(7920 s)\n\na = -.001983 rad/s²\n\n(b) The rotational behavior of the fly wheel can be modeled as simply:\n\nR(t) = -1/2*a*t² + w*t + Ri\n\nwhere R(t) is measured in radians and Ri is the initial angular displacement. Assuming that Ri = 0 (we'll measure the number of rotations from its initial position), then the flywheel will revolve:\n\nR(t) = -1/2*a*t² + w*t (radians)\n\nR(t) = (-1/2*a*t² + w*t)/(2 * pi) (revolutions)\n\nSubtituting the values:\n\nR(2.2 hr) = R(7920 s) = (-.5*.001983*(7920 s)² + (15.71 rad/s)(7920 s))/(2*pi)\n\nR = 9900 revolutions\n\n(c) The tangential component of the linear acceleration of a point on a rotating body is given as:\n\nA-tangent = a * r\n\nwhere A = linear acceleration. The angular acceleration is constant, so for all w (including w=75 rev/min), the linear acceleration will also be constant. Thus:\n\nA-tangent = (-.001983 rad/s²)(.5 m)\n\nA-tangent = -9.915 x 10^(-4) m/s²\n\nA-tangent = -0.9915 mm/s²\n\n(d) The magnitude of the net linear acceleration on the particle is composed of a tangential and a normal component. The equation is of the form:\n\nAnet = [ (A-tangent)² + (A-normal)² ]^(1/2)\n\nOr:\n\nAnet = [ (a*r)² + (v-tangent²/r)² ]^(1/2)\n\nNote that the tangential velocity is given by:\n\nv-tangent = w * r\n\nThus, the magnitude of net linear acceleration is:\n\nAnet = [ (a * r)² + (w² * r)² ]^(1/2)\n\nSubstituting the known values yields (note that for this case, the angular velocity of 75 rev/min is equal to 7.854 rad/s):\n\nAnet = [ (-.001983 * .5)² + (7.85² * .5)² ]^(1/2)\n\nAnet = 30.81 m/s²\n\nFINAL ANSWERS:\n\n(a) a = -.001983 rad/s²\n(b) R = 9900 revolutions\n(c) A-tangent = -0.9915 mm/s²\n(d) Anet = 30.81 m/s²