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Hi, thanks in advance for any help and info. Have two conditions.

The first condition is a mass of water spinning as a fixed mass. Say we have a rectangular mass .25 inches tall, 2 inches long and .50 thick spinning at 3000 RPM with its center of mass at R = 2.00 inches. (Please see first attached jpg) The centripetal force of this mass is then F = m*V^2/R

To calculate mass the density is 1.94 slugs/ft^3. The volume is .25 X 2 X .50 = .25 in^3. The mass is then 1.94 slugs/ft^3 * ft^3/(12 in)^3 *.25 in^3 = .00028 slugs

The tangential velocity at R = 2.0 is 3000 REV/min * pi * 4in / (12 in/ft * 60s/min) = 52.4 ft/s

Therefore the centripetal force = .00028 slugs * (52.4 ft/s)^2/(2.0 in / 12in/ft) = 4.6 lbs.

For the second condition we open the ends of the rectangular tubes and place 90 degree elbows at the periphery as shown in the second attached jpg (similar to a lawn sprinkler)

Question is:

Is the centripetal force the same for condition two or does this change dependent on the mass flow out of the tubes?

The first condition is a mass of water spinning as a fixed mass. Say we have a rectangular mass .25 inches tall, 2 inches long and .50 thick spinning at 3000 RPM with its center of mass at R = 2.00 inches. (Please see first attached jpg) The centripetal force of this mass is then F = m*V^2/R

To calculate mass the density is 1.94 slugs/ft^3. The volume is .25 X 2 X .50 = .25 in^3. The mass is then 1.94 slugs/ft^3 * ft^3/(12 in)^3 *.25 in^3 = .00028 slugs

The tangential velocity at R = 2.0 is 3000 REV/min * pi * 4in / (12 in/ft * 60s/min) = 52.4 ft/s

Therefore the centripetal force = .00028 slugs * (52.4 ft/s)^2/(2.0 in / 12in/ft) = 4.6 lbs.

For the second condition we open the ends of the rectangular tubes and place 90 degree elbows at the periphery as shown in the second attached jpg (similar to a lawn sprinkler)

Question is:

Is the centripetal force the same for condition two or does this change dependent on the mass flow out of the tubes?