Atwood's Machine with one side holding a variable mass

In summary: Then differentiate it with respect to time. The result will be the expression for the rate of change of acceleration as a function of time. That's what they're looking for.In summary, the problem involves an Atwood's machine where at time t = 0, container 1 has a mass of 1.30 kg and container 2 has a mass of 2.80 kg, with container 1 losing mass at a constant rate of 0.200 kg/s. The acceleration magnitude of the system at t = 0 is 3.59 m/s^2. The problem asks to find the rate of change of acceleration at t = 0 and t = 3.00 s, as well as the
  • #1
coffeebird
15
0

Homework Statement



Figure 5-47 shows Atwood's machine, in which two containers are connected by a cord (of negligible mass) passing over a frictionless pulley (also of negligible mass). At time t = 0, container 1 has mass 1.30 kg and container 2 has mass 2.80 kg, but container 1 is losing mass (through a leak) at the constant rate of 0.200 kg/s. At what rate is the acceleration magnitude of the containers changing at (a) t = 0 and (b) t = 3.00 s? (c) When does the acceleration reach its maximum value?



Homework Equations



Fnet=ma



The Attempt at a Solution



m1= 1.30-.200t m2=2.80

Fnet(at t=0) = Tension= m1g- m2g=27.44-12.74=14.7N

acceleration of system (at t=0)= Fnet/mass= 14.7/ 4.1= 3.59 ms^2
 

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  • #2
You've found the acceleration for time t=0, but they're looking for the rate of change of acceleration (with respect to time). What does 'rate of change of a function' suggest to you?

Why don't you begin by finding an expression for the acceleration assuming that m1 and m2 are constants (don't plug in any numbers yet, use symbols only). Then write an expression for m1 as function of time and plug it into the acceleration expression. What does that give you?
 
  • #3
so, m1= 1.30-.200t, but i don't know how to figure out an acceleration expression really
 
  • #4
gneill told you you don't want an "acceleration expression", you want the rate of change of acceleration. Have you taken Calculus?
 
  • #5
HallsofIvy said:
gneill told you you don't want an "acceleration expression", you want the rate of change of acceleration. Have you taken Calculus?

Well to be fair, the OP will need to begin with an expression for acceleration with respect to time, and then proceed to find its rate of change with respect to time.
 
  • #6
coffeebird said:
so, m1= 1.30-.200t, but i don't know how to figure out an acceleration expression really

Start by taking the work you did to find the acceleration at time t=0 and write it out symbolically (don't substitute any numbers). What the expression for a(t = 0) in terms of m1, m2, and g?
 
  • #7
HallsofIvy said:
gneill told you you don't want an "acceleration expression", you want the rate of change of acceleration. Have you taken Calculus?
in response- yes i know that i am looking for the rate of change- and yes i have taken calc classes, but i don't really see how that clarifies anything in this case, because- i guess I'm going to be super duper specific here - i don't know how to find that rate of change without first finding an expression for the acceleration--and there's probably something I'm forgetting here.

so...would that be the Fnet/mass= a? where do i go from there though...find the derivative of this, but how if i have the two variables, because they're both going to be changing...i'm sorry if my questions seem too stupid. is there like a rule against against lowly questions on this site, because i'll leave no prob
 
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  • #8
coffeebird said:
in response- yes i know that i am looking for the rate of change- and yes i have taken calc classes, but i don't really see how that clarifies anything in this case, because- i guess I'm going to be super duper specific here - i don't know how to find that rate of change without first finding an expression for the acceleration--and there's probably something I'm forgetting here.

so...would that be the Fnet/mass= a? where do i go from there though...find dF/dm?

i'm sorry if my questions seem too stupid to some people, is there like a rule against against lowly questions on this site, because i'll leave no prob
Here's a link for the Wikipedia entry for Atwood's machine. Link

It gives the acceleration in terms of m1 & m2 .

Just plug-in your values for m1 & m2, including m1's time dependence, then take the derivative w.r.t. time.
 
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  • #9
The only stupid question is the one that isn't asked. If you need help clarifying concepts then this is the place to be!

An Atwood machine is driven by gravity. So in general the idea behind finding the acceleration is to proceed to find the tension in the string and thus the net forces acting on the masses. But a shortcut of determining the net "unbalanced" force acting on the system is fine, too.

Going the traditional route, draw a Free Body Diagram for each mass and solve for the tension. Use the tension to find the overall acceleration.

-or-

Using the shortcut, determine what the unbalanced force is (i.e., m2*g - m1*g) and use that as the net force acting on the total mass.

Either way, write an expression for the acceleration in terms of g, m1, and m2.
 

Related to Atwood's Machine with one side holding a variable mass

1. How does the mass of one side affect the acceleration of Atwood's Machine?

The mass on one side of Atwood's Machine affects the acceleration of the system in a linear relationship. This means that as the mass on one side increases, the acceleration of the system decreases, and vice versa.

2. What is the purpose of using a variable mass on one side of Atwood's Machine?

The variable mass on one side of Atwood's Machine allows for the manipulation of the system to study the relationship between acceleration and mass. It also allows for the application of different masses to test the validity of the equations used to calculate acceleration.

3. How does the angle of the string affect the acceleration of Atwood's Machine?

The angle of the string does not affect the acceleration of Atwood's Machine. The acceleration is solely determined by the mass on one side and the difference in mass between the two sides.

4. What is the relationship between the mass on one side and the tension in the string of Atwood's Machine?

The mass on one side and the tension in the string of Atwood's Machine have a direct relationship. As the mass on one side increases, the tension in the string also increases. This is due to the fact that the tension in the string is what allows for the acceleration of the system.

5. How does changing the mass on one side of Atwood's Machine affect the acceleration of the system?

Changing the mass on one side of Atwood's Machine affects the acceleration of the system in a linear manner. This means that as the mass on one side increases, the acceleration decreases, and vice versa. This relationship is described by the equation a = g * (m2 - m1)/(m1 + m2), where a is acceleration, g is the acceleration due to gravity, m1 is the mass on one side, and m2 is the mass on the other side.

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