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
Science Advisor
Homework Helper
Insights Author
Gold Member
38,017
7,714
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
535
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
 
  • #4
95
18
Yes, but you did not need to separate MA from m. Since friction is not mentioned, I would have assumed it zero.

I have a small doubt wrt this question itself. The solution assumes that external force equals zero yet we are to find instantaneous acceleration of A. Could you explain why this is so?
 
  • #5
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
38,017
7,714
I have a small doubt wrt this question itself. The solution assumes that external force equals zero yet we are to find instantaneous acceleration of A. Could you explain why this is so?
The system being considered is the cart of current mass M plus the elemental increase in mass, dm, in time dt. These start at different velocities but end at the same velocity; each undergoes an acceleration. The total momentum is conserved.
 

Related Threads on Mass flow and rolling carts (Kleppner 3.11)

  • Last Post
Replies
2
Views
6K
  • Last Post
Replies
2
Views
895
Replies
8
Views
4K
Replies
2
Views
517
  • Last Post
Replies
13
Views
3K
  • Last Post
Replies
4
Views
13K
  • Last Post
Replies
3
Views
3K
Replies
4
Views
2K
Replies
5
Views
12K
Replies
2
Views
5K
Top