Justin asked:
"Is this because the only mass that is actually undergoing an acceleration
is the timy bit of mass dm? In other words, is the first term zero
because dv/dt = 0 for the m in the cart."
Yes, The scale must exert an upward force (dp/dt) on each dm to stop its
downward motion. This is a totally inelastic collision.
Bob Sciamanda (W3NLV)
Physics, Edinboro Univ of PA (em)
trebor@velocity.net http://www.velocity.net/~trebor
----- Original Message -----
From: "Justin Parke" <FIZIX29@AOL.COM>
To: <PHYS-L@lists.nau.edu>
Sent: Monday, November 25, 2002 3:27 PM
Subject: Impulse and force
| I would appreciate some help clarifying some ideas:
|
| Consider a scale with sand falling in at a constant rate, say C kg/s.
The reading on the scale as a function of time consists of two parts: the
weight of the accumulated sand as a function of time + the rate of change
of the sand's momentum, that is dp/dt. It is the dp/dt term in which I am
interested.
|
| dp/dt= d(mv)/dt = m(dv/dt) + v (dm/dt)
|
| In all the treatments I have seen the first term is never included, so
that the force from the sand is just v (usually sqrt(2gh) or some such
thing) times dm/dt, or v*C. Is this because the only mass that is
actually undergoing an acceleration is the timy bit of mass dm? In other
words, is the first term zero because dv/dt = 0 for the m in the cart.
|
| This seems obvious to me as I write, but strange that I have not seen it
mentioned in any of the well known texts.
|
| Thanks
|
| Justin Parke
|
| This posting is the position of the writer, not that of SUNY-BSC, NAU or
the AAPT.
This posting is the position of the writer, not that of SUNY-BSC, NAU or the AAPT.