Precise Ingredients

Classical Mechanics Level 3

You want to pour exactly $300 \text{ g}$ of sugar into a bowl. You place the bowl on a scale and zero it. Then you pour sugar, from a constant height at a constant rate. When should you stop pouring to ensure that you have exactly $300\text{ g}$ of sugar?

Details and Assumptions :

Ignore air resistance during the motion.
Pouring means that the sugar initially has zero velocity.
Once the sugar reaches the scale, it is brought to a full stop.
The time needed to pour the 300 g of sugar is longer than the falling time of the sugar.

Bonus : Will the answer change if the sugar bounces up and down a bit before finally settling?

Just before the scale indicates

300\text{ g}

When the scale indicates

300\text{ g}

Just after the scale indicates

300\text{ g}

More information is needed

1 solution

Arjen Vreugdenhil
Mar 27, 2016

Let $t$ be the time it takes for the sugar to fall to the scale (in s), and $\mu$ the rate at which it is poured (in kg/s).

The reading of the scale is higher due to the impulse of the stopping sugar. During time interval $dt$ , a total mass of $\mu\:dt$ at speed $v = g\:t$ is stopped by the scale. This makes for an impulse of $dp = \mu\:g\:t\:dt,$ so that the impact force on the scale is $F_i = \mu\:g\:t$ . Thus the actual mass on the scale is less than that indicated on the scale, in the amount $\Delta m_i = -\mu\:t.$

The reading of the scale does not include the mass of the sugar in the air. After we stop pouring, the scale will collect an additional amount of sugar in the amount $\Delta m_a = \mu\:t.$

Interestingly, these two effects cancel each other exactly: $\Delta m_i + \Delta m_a = 0$ . Therefore we can stop pouring at the moment that the scale indicates the precise value we desire!

Oh wow, that was unexpected!

Calvin Lin Staff - 5 years, 2 months ago

Nice solution. Though I choose more info needed. How do we know the pour is not at a rate and height that it does not zero out? Say 500g per second at 30 ft?

bruce merritt - 5 years, 2 months ago

I will exclude that possibility by adding another assumption :)

Arjen Vreugdenhil - 5 years, 2 months ago

For Bonus question:- No, I don't think that the answer will change because the sugar might bounce up and down a bit due to air resistance.. If I am wrong, pls correct me..

Parag Zode - 5 years, 2 months ago

We already ruled out air resistance (see Assumptions above).

The sugar may bounce a bit, but I don't believe that will affect the answer (or at least, not significantly). Here is why:

A more elastic bounce will increase the impact force.
It will also cause there to be more sugar in the air instead of on the scale.

It would be interesting to work out the details!

Arjen Vreugdenhil - 5 years, 2 months ago

I agree with you Sir! Thanks.

Parag Zode - 5 years, 2 months ago

Ya but, don't you need to account for the mass of the bowl? The question didn't say whether or not you zeroed the scale beforehand.

Sagar Kothari - 5 years, 2 months ago

"You place the bowl on the scale and zero it." In that order. :)

Arjen Vreugdenhil - 5 years, 2 months ago

Precise Ingredients

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