# Buckets on a pulley - Dynamics Problem

• Shreya
In summary, the conversation discusses the calculation of initial velocity and momentum conservation in a system consisting of two buckets attached to a string and a putty. There is confusion about the given answer and the correct calculation of pre-impact velocity, as well as the consideration of the total mass of the system. Ultimately, the correct calculation is determined to be V=10/7 m/s, considering the signs of the momentum.
Steve4Physics said:
A rigorous analysis (i.e. not using any ‘tricks’) would need to include the resulting external impulse, which sounds messy.
I would not say it is messy. Since the pulley's velocity does not change, there is a simple equation relating that impulse to ##\int T.dt##.

But I would like to expand on my comment (post #10) about the flaw in the question. Inextensible does not imply inelastic; in principle it could be perfectly elastic.
My approach to such idealisations is to take them as the limits of a more realistic set up. Here we could suppose some elasticity and a small extensibility, analyse that in conjunction with a completely inelastic collision between the putty and the mass it hits, then take the limit as the spring constant tends to infinity and the elasticity tends to perfect.

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• Lnewqban and vcsharp2003
vcsharp2003 said:
But I was looking only at the mass of putty since you metioned ##mV - mv## as change in momentum and surely the impact force on this mass will decrease its speed, which will decrease it's momentum. The change in momentum equation is only depending on initial and final momentum of putty, so I focused only on the mass of putty.
The putty’s change in momentum (mV – mv) is the same as the system’s change in momentum (because the net momentum of the buckets is always zero).

The system's momentum-change is ultimately produced by an external impulse on the pulley.

If that’s not clear, imagine you are holding the pulley when the putty hits the bucket. You will feel a sudden downwards impulsive force. You would need to supply an upwards impulsive force to keep the pulley stationary.

The momentum that the system loses is transferred (through the pulley and whatever supports it) to the entire earth.

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haruspex said:
But I would like to expand on my comment (post #10) about the flaw in the question. Inextensible does not imply inelastic; in principle it could be perfectly elastic.
I don't get it. Don't 'inextensible' and 'inelastic' (when applied to string) mean the same thing - that the length doesn't change under tension?

What am I missing?

Steve4Physics said:
The system's momentum-change is ultimately produced by an external impulse on the pulley.
Isn't this the same that I explained in post#35?

Steve4Physics said:
I don't get it. Don't 'inextensible' and 'inelastic' (when applied to string) mean the same thing - that the length doesn't change under tension?

What am I missing?
Idealisations need careful handling. The way to make them valid is, as I wrote, to treat them as the limiting cases of continua of realistic scenarios. If the result tends to a limit then the idealisation is valid.
A real string is extensible and elastic with some elastic constants ##k_{stretch}>k_{relax}>0##. (This is already an idealisation because I am taking each of those to be constant.)
Inextensible is the limit as constant ##k_{stretch}## tends to infinity. Independently, perfectly elastic is the limit as ##k_{relax}\rightarrow k_{stretch}## and perfectly inelastic is the limit as ##k_{relax}\rightarrow 0##.
Inextensible and perfectly elastic might need extra care; the relative rates of two limit processes may matter.

It is illuminating to compare this with colliding bodies. We take billiard balls to be incompressible yet highly elastic, and putty to be quite the opposite.

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• • Lnewqban, jbriggs444 and Steve4Physics
vcsharp2003 said:
Isn't this the same that I explained in post#35?
Yes. Apologies. For completeness, can I add that the change in momentum of the putty on impact can be traced through a series of internal (impulsive) forces: putty on bucket, bucket on rope, rope on pulley, pulley on external support. That’s the mechanism.

• vcsharp2003 and Lnewqban
For whatever it's worth, here is my take on the solution to this problem. Let ##v_0## be the downward velocity of the putty at the beginning of the collision, ##T_0=Mg## equal the tension in the string prior to the collision, v(t) be the downward velocity of the putty at time t during the collision, V(t) be the downward speed of the downward-moving bucket at time t during the collision (and the upward speed of the upward-moving bucket at time t during the collision), F(t) be the contact force between the putty and the downward-moving bucket at time t during the collision, and ##\tau(t)## be the perturbation to the string tension at time t during the collision. Then, during the collision, the downward force balance on the putty, the downward force balance on the downward-moving bucket, and the upward force balance on the upward-moving bucket are, respectively, $$mg-F=m\frac{dv}{dt}$$$$-(T_0+\tau)+F+Mg=F-\tau=M\frac{dV}{dt}$$$$(T_0+\tau)-Mg=\tau=M\frac{dV}{dt}$$
If we add there three equations together, we obtain:$$mg=m\frac{dv}{dt}+2M\frac{dV}{dt}$$Integrating this with respect to time yields: $$mgt=m(v-v_0)+2MV$$At the end of the collision, ##t=t_c## and ##v = V## so that $$v=\frac{mv_0}{m+2M}+\frac{mgt_c}{m+2M}$$Since the time for the collision is assumed negligible,, this reduces to $$v=\frac{mv_0}{m+2M}$$

• vcsharp2003
vcsharp2003 said:
I tried to apply your straightening trick to first pulley masses system and came up with following straightened string.
View attachment 300435

View attachment 300430
Is your graphic solution referring to that image?
If so, you need to consider the effect of the moving pulley on the mechanical advantage or ratio of output force (FG1) to input force (FG2) in this system (mass 1 having half of the speed of mass 2).

vcsharp2003 said:
That's a nice trick that helps to solve problems on Atwood machine. But I was wondering if it has a wider applicability when considering masses connected via pulleys like on an incline or more complex cases of Atwood machine as shown in images below. If it could, then this would be an awesome trick in solving many difficult problems.

I tried to apply your straightening trick to first pulley masses system and came up with following straightened string.
You cannot indiscriminately stretch any string with masses and expect the straightening trick to work. Its value depends on the fact that the acceleration of the masses has the same magnitude, in which case you can apply Newton's second law to the entire system and ignore the internal tension. In the multiple pulley examples you have shown (first two figures in #32), the accelerations of the two masses do not have the same magnitude.

You can, however, straighten the string in the last three figures which are the same problem with different incline angles. The "generalized" Atwood machine has two masses each on its own inclined plane. Before straightening, on has to be careful to use the angle of the component of the weight parallel to the original direction of the string as the driving force. In that case, for the generalized Atwood machine, Newton's second law gives $$m_2 g\sin\!\theta_2-m_1g\sin\!\theta_1=(m_1+m_2)a\implies a=\frac{(m_2 \sin\!\theta_2-m_1\sin\!\theta_1)g}{m_1+m_2}.$$To find the magnitude of the acceleration using the expression above,
for the original vertical Atwood machine, set ##\theta_1=\theta_2=90^o##;
for the half-Atwood machine shown in your third figure, set ##\theta_1=0,~\theta_2=90^o##;
for the fourth figure, set ##\theta_1=0,~\theta_2=\theta##;
for the fifth figure, set ##\theta_1=\theta,~\theta_2=90^o##.

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• Steve4Physics, Lnewqban and vcsharp2003
kuruman said:
You can, however, straighten the string in the last three figures which are the same problem with different incline angles.
Thankyou, that's a very useful trick for quicker problem solving when connected masses are having the same acceleration magnitudes. I wasn't aware of this neat trick.

vcsharp2003 said:
Thankyou, that's a very useful trick for quicker problem solving when connected masses are having the same acceleration magnitudes. I wasn't aware of this neat trick.
It's not really a trick. It's the realization that if you write the one-dimensional equations for the two free body diagrams, you get
##m_2g\sin\!\theta_2-T=m_2a## and ##T-m_1g\sin\!\theta_1=m_1a##. Add the equations to get
##m_2g\sin\!\theta_2-m_1g\sin\!\theta_1=(m_2+m_1)a##. By straightening the string you eliminate the tension graphically instead of algebraically.

• Steve4Physics, Lnewqban and vcsharp2003
Lnewqban said:
Post #1 shows that the answer given is 10/7 m/s.