MadSci Network: Physics |

Posted By:

Area of science:

ID:

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 µk
= 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 v^{2}

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

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

MadSci Network, webadmin@madsci.org

© 1995-2006. All rights reserved.