Mass flow and rolling carts (kleppner 3.11)

  • #1
535
72

Homework Statement


Material is blown into cart A from cart B at a rate b kilograms per
second. The material leaves the chute vertically down-
ward, so that it has the same horizontal velocity u as cart B. At the
moment of interest, cart A has mass M and velocity v. Find dv/dt,
the instantaneous acceleration of A.

Homework Equations


Momentum

The Attempt at a Solution


[/B]
System studied: cart A and the chute.

Let's put that ##M_A## is the mass of cart A (unloaded), ##m(t)## is the material's mass in cart A at time ##t##, and ## \triangle m = m(t+\triangle t) - m(t)## is a small amount of material falling from the chute into cart A in a ##\triangle t## seconds.

At a given time ##t##, the horizontal momentum will be:
## P(t) = (M_A + m(t)) v(t) + \triangle m \ u(t) ##
## P(t+\triangle t ) = (M_A + m(t) + \triangle m) v(t+\triangle t) ##

So that in time ##t##:
## \frac{dP}{dt} = (M_A + m) \frac{dv}{dt} + \frac{dm}{dt} (v - u)##

In this system, only friction from the wheels of cart A contribute to horizontal external force :
##f_{ext} (t) = -\mu g (M_A +m(t)) ##

Since at time of interest ##t_i##, we are given:
##M_a + m(t_i) = M##,
## v(t_i) = v##,
##u(t_i) = u##,
## \frac{dm}{dt} = b##,

the acceleration should be:
##\frac{dv}{dt}(t_i) = \frac{1}{M_A + m(t_i)}(f_{ext}(t_i) + \frac{dm}{dt} (u - v) = -\mu g +\frac{b}{M} (u-v) ##


Do you agree with this solution ?
 

Answers and Replies

  • #2
haruspex
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Do you agree with this solution ?
Yes, but you did not need to separate MA from m. Since friction is not mentioned, I would have assumed it zero.
 
  • #3
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72
Thank you haruspex!
I did separate because it wasn't so clear at first, but now I see it would work just the same ! Thanks
 

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