falling ball viscosity measuring principle?
341043
Previous answerer had it correct.  Falling-ball viscometers are based on Stokes' Law.  See sources below for a complete description.
147491
The previous idea of doing an experiment or two is a good idea. Here are some more ideas:\n\nThe ratio of the height you drop the ball from to its rebound height is called its coefficient of restitution. Different types of squash balls have different coefficients of restitution. \n\nThe square root of the coefficient is also the ratio of the maximum speed of descent to maximum speed of rebound (converting potential energy into kinetic energy, and vice versa).
905434
You can think of viscosity as referring to how much drag is exerted on a surface as the liquid flows by.  Generally, a thicker fluid has higher viscosity so I agree with your choice of honey.\n\nAs particles (molecules) become more loosely connected, the liquid flows more easily, so should be obvious how this relates to viscosity.
851431
MASS / VOLUME\n that's way you measure it in g/mL
250107
Viscosity is a fluid property characterized by its resistance to flow. Non-Newtonian fluids are a special property of viscosity. Fluids such as peanut butter and tooth paste are non-Newtonian fluids.  They can form pointed surfaces. Fluids such as water and air which flow “normally”  (Newtonian fluids) do not form pointed surfaces. That point is a singularity that when differentiated may appear to give an infinite viscosity.
1010100
ok, so if we're assuming that it was a head on elastic collision, and all the energy was transferred then there are two possible situations: if the 25lb ball was sliding then the total energy transferred will be equal to the amount it had originally (since we assumed elastic collision) that is equal to 1/2mv^2...where mass would usually be measured using the metric system. \nif the ball was rolling then it has not only translational energy (1/2mv^2) but also rotational energy which is equal to 1/2Iw^2 where I = inertia of a ball...if it is a solid sphere the moment of inertia is equal to 2/5mr^2 (m = mass, r = radius) and w (omega) = rotational speed in radians per second...which is equal to velocity/radius. so the amount of energy transferred would be equal to 1/2Iw^2 + 1/2mv^2.
762251
If 2 billiard balls of mass m (ma and mb) are perfectly elastic and ball A approaches from the left with velocity 2 m/s and hits ball B which has velocity 1 m/s, and is also going to the right, what's the status after the collision. \n\nA is going to the right with velocity 1 m/s and B is going to the right with velocity 2 m/s.\n\nma*2 m/s + mb*1 m/s = ma*1m/s + mb*2m/s\nmomentum is conserved
980974
Yeah....but then how to distinguish a draw from a dive?  \n\nI see your point but this would be hard to keep track of and assess.
859137
first thing u should do is do draw the path of the projectile,  using only imagination, the ball will definetly fall down as it moves away from the table edge until it finally hits the floor.\n\nprojectile path is always parabolic, and the direction of the velocity vector is tangent to the path of the projectile at any point.\n\nat the initial state, the velocity vector was pointing left or right according to your assumption, but as the parabolic path falls down, the velocity vector points down with it, when the ball hits the floor the velocity vector will be perpendicular to the horizontal in the downward direction.\n\nso at 45 degrees downward, the velocity vector has two components, on in the horizontal V*cos(theta) and another downward V*sin(theta), to get how high the ball at this instant apply the same laws but use V*sin(theta) instead of V ,  and to find how far from the table use V*cos(theta) instead of V.\n\nhope this helped.
1086779
1. You need to know the initial height of the ball from the ground.  I will assume from ground level.\n\n2. Calculate the vertical and horizonal components of the speed.\na^2 + b^2 = c^2 where c = 80 and since it is a 45 angle a=b\n2a^2 = 80^2\na = 56.6 mph (both vertical and horizontal speeds are the same)\n\n2. Now take the vertical speed (56.6mph) and find time it takes to hit the ground.\nSince acceleration due to gravity is -32.2 ft/(s^2)\n56.6 mi/h * 1/60 h/min * 1/60 min/s * 5280 ft/mi = 83 ft/s\nNow using\na=(v2-v1)/t where a is acceleration, v1 = inital velocity, and v2 = final velocity, and t = time it took\n-32.2 = (0 - 83)/t\nt = -83 / -32.2 = 2.58sec\nsince that is the time for the ball to finally reach the top of the arc, multiply by two to get the complete arc\ntotal time in the air = 5.16sec\n\n3. Calculate distance travelled in that time.\nI am leaving air resistance out due to incomplete data (you need mass of the ball, resistance will also change due to the shape, the barometric pressure, etc.)\n5.16s * 83 ft/s = 428.3 ft travelled\n\nEdit:\nSince the distance from the ground is small compared to the velocity of it falling, we can estimate just an additiona 0.04 secs so total time is now 5.20 secs\n\n4. Drag equation\nD = Cd * A * 0.5 * r * V^2\nCd = Drag coefficient\nA = reference area (half surface area of baseball 2pi * r^2 = 3.53 ft^2)\nr = density of air (20 C and sea level is 1.2 kg/m3)\nr = 1.2 kg/m^3 * 2.2 lb/kg * 0.0283 m^3/ft^3 = 0.0747 lb/ft^3\n\nCd can't be calculated since it is different for every object and situation.  Only way for Scientist to find it is using a wind tunnel.\nI cheated and looked it up for a standard baseball it is 0.007 lb/ft\nD = 0.007 * 3.53 * 0.5 * 0.0747 * V^2\nD = 0.0009229 * V^2\nDi = 0.0009229 * 83^2 initial force from drag \nDi = 6.36\nuse F = ma to get the negative initial acceleration\n6.36 = 0.3125 * ai\nai = 20.4\n\n-20.4 = (d/5.20 - 83)/5.20\nd = 120 ft\nThis is only true if the drag stays constant, but it decreases as the velocity decrease (exponentially).\n\nAn estimation can be done by calcing the acceleration after each second, then decreasing velocity due to that acceleration.\nV(i) = 83.0, Drag(i) = 6.36, a(i) = -20.35\nV(1) = 62.7, Drag(i) = 3.62, a(i) = -11.59\nV(2) = 51.1, Drag(i) = 3.41, a(i) = -7.70\nV(3) = 43.4, Drag(i) = 1.74, a(i) = -5.55\nV(4) = 37.8, Drag(i) = 1.32, a(i) = -4.22\nV(f) = 33.6\nAfter adding up the distances travelled after each second (including the 0.2 sec at the final velocity), I come up with an estimate of 284.6 ft travelled.  \n\nThis is still a desent estimate, but there is still additional factors to consider.  The ball may reach terminal velocity on its way down which will increase the time returning to the ground.  This problem at first glance looks simple, but as you can see, it gets complex very quickly.
195717
dba_gregg has missed the point that both are same weight\n\nThe moment of Inertia of a hollow sphere  is 1.6 times greater than the solid sphere and so it rolls down slowly.\n\nThe speed of solid sphere will be 1.09 times the speed of hollow sphere.
927274
Yes - roll a ball down a hill. It has linear inertia (the balls center of mass moving down the hill) and rotational inertia (the ball is spinning while it moves down the hill).
221114
I understand what you're saying, but you forgot one small thing:\n\nThe bowling ball has more mass, and therefore needs more pull to reach the same speed because of momentum. You know, it's harder to push a car than push a cup. The momentum pull and the pull of gravity completely cancel each other out, so all things fall at 9.81 m/s/s.
426540
A ball is made up of a large number of particles, say 10^24.\nWhen one measures the speed and position of the ball, one measures the *average* speed and *average* position of all these particles. This is not in contradiction with Heisenberg's uncertainty principle which states only that it is impossible to measure velocity and position of the same, given particle.
206966
Viscosity: \nmax. 100 Cp at 40ºC measured on the feed to be atomized. Measurements on concentrate not older than 15 min. and kept under vacuum between evaporator and sampling for viscosity measurements. Method Brookfield viscometer model LVT with spindle 2, rotation 60 rpm, measured at 40ºC.
528388
You can go about this problem a few ways. A force exherted on an object will give it a certain velocity. If F=m*a, and you have the force along with the mass, you can determine the acceleration. The acceleration, of course, is the change in velocity/change in time. You'll get a velocity change per second. If you know the ball is in contact for 0.03 seconds, you can determine the final velocity, v. \n\nKinetic energy is just 1/2 * m * v^2 so that will be a simple calculation once you figure out v.
1053965
Simple parabolic path formula \n\na hit ball is given a velocity v at and angle theta.\n\nAssuming no air resistance the balla travels through the air along its horizontal vector un abstructed, while acted on by gravity in the vertical direction.\n\nthe equation\n\nd=vt + 1/2gt^2 can be solved for d which will be the total vertical distance at time t, where t is the time at which the ball is the highest, lets say about 1/2 of the total time, g is the gravitational constant and v in the velocity in the vertical direction