Spockie wrote:
ariel font is used here
Wooden Normal Linear Frictional
Diameter: Block Radius Force Velocity Force
m Rpm: grams m N m/s N
0.4 15 198 n/a 1.9404 0.314159265 n/a
33 198 0.179 1.9404 0.691150 0.528393256
45 198 0.097 1.9404 0.942477796 1.813158252
78 198 0.024 1.9404 1.63362818 22.0171135
Frictional Percent
Force Difference
Coefficient For Cofficient
Percent
n/a Difference
0.272311511
0.934424991
11.34668805 2.646539156
first of all the coefficient of frictional force should be the same
right
I'm assuming the lab involved putting wooden blocks on a disk spinning
at various speeds, and measuring how far you can put them from the
center without them flying off. If so, then yes, the coefficient
should be about the same in each case.
the formulas i use for all my calculations are acurate , i used
m*v^2/radius to get frictional force then from that force i
divided
by the normal force which is mass * gravity
to get the v for the frictional force formula i used
(rpm * pi * 198 * .001)/60 60 is for a minute and .001 * 198 is
grams in kilograms
That's clearly wrong; the linear velocity doesn't depend on the mass.
Plugging your data into this formula, I can also see it isn't what you
used.
It looks like you calculated the linear velocity by dividing the
circumference of the disk by the time it took to make one revolution,
or some equivalent calculation. This will give you the linear velocity
of the edge of the disk, but parts of the disk closer to the center
aren't moving as fast. The center of the disk has no linear velocity
at all.
In order to calculate the linear velocity of the wooden block, you need
to divide the circumference of the circular path it takes by the time
it takes the disk to make one revolution.
if my formulas are correct and everything else, then i do not
understand
why the percent difference is so big
to get percent difference i took absolutevalue(range (highest -
lowest))/
average of all three numbers
frictional force coefficient should be the same for all right?
Yes. It looks like you did everything else correctly. If you fix your
formula for the linear velocity, you will get practically the same
coefficient of friction (I'm getting 0.22) for the second and third
trials. The coefficient from the fourth trial is off, but you'd expect
some error, given the conditions for that trial.
.
