Date: Fri, 16 Feb 2001 16:29:23 -0800

Author: Eric Ayars

Subject: Re: physics question

Post:

>PS I bet that the cart would not move. I might be out 10 bucks.

You're out ten bucks.

Simplest solution:

There is no external force, so momentum is conserved. Initial
velocity of center of mass is zero, so initial momentum is zero, so
the center of mass never moves. Since the water moves from one
location (center of cart) to another location (below off-center
drainpipe) and the CM doesn't move, the cart must move. Once the
water stops draining, the cart stops: final momentum equals initial
momentum. The final position of the cart (measured from x=0) is
x(cart) = - x(water) * m(water) / m(cart)

Complex solution:

Calculate the change in momentum of an infinitesimal volume of water
as it starts moving from some arbitrary location in the cart to the
drain. You may ignore the vertical component, since the cart is on a
track and we're only interested in horizontal motion. F=dP/dt.
Integrate over the entire volume of the cart to find the total force
supplied by the cart on the water during the drain time.

Now calculate the change in momentum of an infinitesimal volume of
water as it stops moving horizontally and moves vertically down the
drain. F=dP/dt again. Integrate over the entire volume of water
drained to find the total force supplied by the water on the cart.

Compare the two total forces. They will have the same magnitude, and
the same duration, so the final momentum of the cart equals the
initial momentum of the cart (which is zero) so you might think the
cart never moves, BUT...

The first of the two forces starts BEFORE the second force, and ends
before the second. The time shift between the two is (on average) the
distance between the center of the tank and the drain, divided by the
flow velocity.
delta t = x(water)/v(water)
The net force is not zero for a brief time at the beginning of flow,
during which the cart starts moving. The net force is zero during the
bulk of the flow, so the cart continues moving at constant velocity.
The net force is not zero (and equal and opposite the initial force)
for a brief time as the flow stops, during which time the cart stops.

I haven't worked it out, but I'd bet considerably more than \$10 that
the final position of the cart, as calculated with this method, works
out to
x(cart) = - x(water) * m(water) / m(cart)

Have a good weekend - I'm off to do something useful. Like figure out
how to start water flow from an off-center drain in a tank on an
air-track glider without giving the glider any unwanted momentum.
I've got a couple of students looking for a challenging project lab
anyway...

Eric
--

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| Dr. Eric Ayars ayarer@wwc.edu |
| Assistant Professor of Physics, Walla Walla College (509) 527-2476 |
| http://homepages.wwc.edu/staff/ayarer |
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