Two masses, one changing, hung over a pulley

[Samuelb88]

Homework Statement

Two objects are connected by a massless rope that passes over a massless, frictionless pulley. One of the objects has a fixed mass (m), while the other is a bucket filled with water (M) with initial mass of 45. kg. The objects are releases from the rest (shown below). There is a hole in the bucket and the water leaks out at the rate given (dM/dt).

(a) Determine when the fixed mass (m) reaches its maximum height, this maximum height, and the mass of the bucket (M) at this time.

(b) Determine when the fixed mass (m) reaches the ground, its final speed, and the final mass of the bucket (M).

Homework Equations

http://img3.imageshack.us/img3/4852/93084392.jpg

The Attempt at a Solution

I set up my coordinate system with x directed upward and let M(t) function the changing mass of the bucket.

$$M(t) = M_0 + \frac{dM}{dt}\right)$$

Since the objects are bound by a massless rope, the acceleration for both objects is the same and the sum of the forces of the system is

$$SUM(F_x) = M_t_o_t_a_la_x = M(t)g - mg$$

Solving for acceleration

$$a_x = ( \frac{M(t) - m}{M(t) + m} \right) )g$$
$$a_x = ( \frac{(45-t) - 40}{(45-t) + 40} \right) )g$$
$$a_x = ( \frac{(5-t}{(85-t} \right) )g$$

Now I've explicity defined an equation for the acceleration of the system.

The fixed object m will be at its max height when the velocity v = 0 m/s. Noticing velocity is at a max. at t = 5s, I integrate from 5 to t, and set v = 0 m/s.

$$0 = g \int_5^t \frac{5-t}{85-t}\right) dt = 5\int_5^t \frac{1}{85-t}\right) dt - \int_5^t\frac{t}{85-t}\right) dt$$

(i divided by g, above)

$$0 = (-5(ln|85-t|) - (-t - 85ln|85-t|))$$ (from 5 to t)
$$0 = 80ln|\frac{85-t}{80}\right)|+ t - 5$$

Taking t = 0 s to be the moment when the objects are released from rest, the time when the projection of the velocity wrt time re-obtains this value (is at rest again) is at t = 9.8 s. Therefore, the fixed mass (m) is at its maximum height at t = 9.8 s.

$$\int_0^xdv = \int_0^{9.8} (80ln|\frac{85-t}{80}\right)| + t - 5) dt$$
$$x(t) = \frac{1}{2}\right) (t - 170)t + 80(t- 85)ln|\frac{85-t}{80}\right) |$$ (from 0 to 9.8)
$$x(9.8) = ... -.482572$$

Since the bucket M started 10. m above the ground, I added 10 to x(9.8) which equals 9.51743m which I took to be the maximum height. Could anyone check my work? The whole thing seems wrong imo, when I graphed the velocity wrt to t, v(0) = -.15. I determined x_max occurred at t=9.8s because v(9.8) = -.15. But shouldn't at v(0) = 0m/s? (neither object was initially moving...) therefore, v(9.8)=0m/s aswell. I'm quite confused about the whole thing.

Any help would in answering part (a) or (b) be greatly appreciated. Thanks.

 

[kuruman]

Wait a minute! It is true that at t = 5s the acceleration is zero and the velocity is maximum, but that maximum is not zero. The masses are still moving in the same direction after t = 5s except now the speed is decreasing. The expression that you integrated,
$$0 = 80ln|\frac{85-t}{80}\right)|+ t - 5$$
is v(t), not zero. You need to integrate once more to get x(t), then maximize x with respect to time.

I am perturbed by the fact that the mass of the empty leaking bucket is not given. If there is still water left in it when the fixed mass reaches the ground, you re OK. If not, then you need to know that mass. In other words, the answer to part (b) must be less than 9 s.

 

[Samuelb88]

Alright, I understand when a=0m/s^2 (@ t = 5s), the objects will still have a velocity in the direction they were originally moving, therefore, x_max will occur at some time t > 5s.

I integrated both sides of the acceleration equation

$$\int_0^v dv = \int_5^t a_xdt$$
$$v(t) = g( 80ln|\frac{85-t}{80}\right) | + t - 5)$$

Now, whether or not if I integrate again, I can still find the critical point in which v(t) = 0, therefore, at that time t+5 (since I integrated from 0 to 5, correct?), x will be at its max.

From the expression above, I set v = 0m/s, and i find that v=0/ms when t=5, therefore, x_max occurs at t=5+5=10s.

integrating both sides of the velocity equation

$$\int_0^x dx = \int_5^t v(t)dt$$
$$x(t) = g(\frac{t-170}{2}\right) 2 + 80(t-85)ln|\frac{85-t}{80}\right) |)$$ (from 5 to t)
$$x(t) = g(\frac{t-170}{2}\right) 2 + 80(t-85)ln|\frac{85-t}{80}\right) | + 412.5)$$
$$x(10.) = ...-2.63889 m$$

Therefore the max height = x(10) + h = -2.63... + 10. = 7.3611 m ?

So, do you see what I am saying? Sorry if I misunderstood your post, but to my understanding, you are saying I need to integrating 80ln(...)+t-5, which is my expression for velocity to get x, which I did. Then to maximize it, you would differentiate x(t) and set x'(t) equal to zero, which is what I did, only before finding an expression for x.

I still think I'm doing something wrong... My professor hands out answer keys with answers to parts of the problem. On the key, it is given that it takes 9.8 for the fixed mass m to reach its maximum height. But from my calculations, when I set v(t) = 0m/s, the equality 0=0 is true when t=5, therefore, it takes 5 seconds to velocity to speed up to its max speed, and 5 more seconds for velocity to slow down to 0 m/s.

I am perturbed by the fact that the mass of the empty leaking bucket is not given. If there is still water left in it when the fixed mass reaches the ground, you re OK. If not, then you need to know that mass. In other words, the answer to part (b) must be less than 9 s.

I don't think this requires me to optimize based on the constraint of the original mass, and the final mass of the leaking bucket M but rather (for part (a)), determine the critical points for the position function x(t) and plug in the critical values of t into x and... wah-lah! the answer. unfortunately that's not happening for me.

Any (more) help/ideas? :)

 

[ehild]

You are right, the maximum height is reached at 9.8 s from the beginning, when the bucket is released, and the maximum height is about 9.6 m. You do not need to change the limits of integration, write both the velocity and the height as functions of the total elapsed time.

$$v=g(t+80\ln(1-t/85))$$

$$x=g(t^2/2-6800((1-t/85)(\ln(1-t/85)-1)+1))$$

ehild

 

[Ingenue1]

A problem just like this is solved step by step at this college solutions website.

http://zebu.uoregon.edu/~probs/mech/circ/Rot5/Rot5.html