MadSci Network: Physics

Re: Calculate power required to deliver a curling stone

Date: Tue Sep 25 09:43:18 2007
Posted By: John Link, Senior Staff Physicist
Area of science: Physics
ID: 1190714238.Ph

The question: "Calculate power required to deliver a curling stone. I work with wheelchair curlers, who deliver the stone using only a 'shot stick' with which they essentially push the stone, from a seated position, to send it on its way down the ice. There is no sweeping in wheelchair curling so the delivery is the only factor that influences where the stone ends up. I am trying to work out the forces (ultimately power) associated with delivering the stone. I think this is very similar to questions posed here on 'how to calculate horsepower from acceleration and weight', but as it was a long time since I did Physics at school I'm struggling a bit with the calculations.

The things I know already or that can easily be measured:

The weight of the stone (19.96kg)

The speed of the stone for the first few seconds (2 or 3 seconds) following delivery (ranges between 2 and 3 meters per second)

The friction coefficient of curling ice has been reported to be ยตk = 0.0168 although I'm not sure how valid this is as I can only find a web reference and not the research paper (Babcock, David D. The Coefficient of Kinetic Friction for Curling Ice. 8 April 1996.)

The regulation size of the stones is 91.4 cm (36 in) in circumference, and 11.4 cm (4 1/2 in) in height. Although I suspect drag is not going to be too relevant?

I'm sure that with this information I should be able to work out force and therefore ultimately power."

There is one piece of information missing, which is the amount of time during which the stone is being accelerated. But if we make some reasonable assumptions about that we can come up with a reasonable answer.

From your past Physics training you probably will recall that the work provided by the "deliverer" can be calculated from the kinetic energy of the stone as it is "released", and that the kinetic energy is
0.5 m v2
where "m" is the mass and "v" is the velocity, or, better stated, the speed. Power is work per unit time, which is why we need the time during which the stone is pushed. See energy transfer at Wikipedia and power at Wikipedia. We will calculate the average power.

The work is thus 0.5 * 19.96kg * (2.5m/s)2 = 62 J.

If the work is applied during one second the average power is thus 62 J/s (62 Watts), but if the work is applied during 2 seconds the power is only 31 J/s (31 Watts).

But what if one wants to know the force required to propel the stone? Ah, that is a slightly different question!! If one assumes a constant force then the force is best calculated from the change in momentum of the stone rather than the kinetic energy. See impulse at Wikipedia. The change in momentum is simply mass times velocity (because the stone starts at rest), and the force is
F = m v / t
And again we must make some assumptions about the amount of time during which the force is applied! If the stone is propelled during a 1-second interval the force is
F = 19.96kg * 2.5m/s / 1 s = 50 Newtons
but if the force is applied during 2 seconds the force is only 25 Newtons.

In actuality in a situation like this the force is not applied uniformly but rather starts at zero, ramps up to a maximum, perhaps is held more or less constant for a while, and then ramps down to zero. So the peak force will be bigger than the average force we just calculated. If the force profile is more or less triangular (the peak force is held for only a very brief moment) then the peak force will be roughly twice what we calculated. The same kind of argument also applies to power.

John Link, MadSci Physicist

Current Queue | Current Queue for Physics | Physics archives

Try the links in the MadSci Library for more information on Physics.

MadSci Home | Information | Search | Random Knowledge Generator | MadSci Archives | Mad Library | MAD Labs | MAD FAQs | Ask a ? | Join Us! | Help Support MadSci

MadSci Network,
© 1995-2006. All rights reserved.