6- In the experimental procedure- why is a path length longer than 1 m.docx

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A longer path length is suggested for measuring the motion of a falling mass on a string to improve the accuracy of measurements for calculating acceleration, tension, torque, and angular acceleration. Data was collected for masses between 0.05-0.3 kg with path length 1.434 m and hub radius 0.04 m. Calculations were performed to determine the values for each variable and a linear fit was done to find the moment of inertia I as the slope and frictional torque Tf as the intercept.

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6. In the experimental procedure, why is a path length longer than 1 m suggested for the motion
of the mass on the string? The following data were taken with a systern like the one described in
this laboratory. The path length of the falling mass was x-1.434 m, and the radius of the hub
around which the string was wrapped was r- 0.040 m. For the values of mass m on the string
listed, the times to accelerate the distance x are given in the table. Use these data to calculate the
values for the acceleration a, the tension T, the torque t, and the angular acceleration ?. Perform a
linear least squares fit with ? as the vertical axis and x as the horizontal axis. Record the slope as
I the moment of inertia, the intercept as ?, the frictional torque, and the correlation coefficient r.
7Â· m (kg) 0.050 0.100- 0.150 0.200 0.250 0.300 a (m/s2) t (s) 10.60 7.40 6.10 5.20 4.60 4.20 T
(N) ? (rad/s?) kg-m
Solution
Q7.
path length, l = 1.434 m
hence
time taken = t
l = 0.5*a*t^2
hence
a = 2l/t^2
mg - T = ma
hence
T = m(g - a)
T' = T*r = 0.04T
also

l/2*pi*r = 0.5*alpha*t^2
hence
alpha = l/pi*r*t^2
hence the complete table is as under
0.05 10.6 0.025525098 0.001021004 0.101561138
0.1 7.4 0.052373996 0.00209496 0.208389507
0.15 6.1 0.077076055 0.003083042 0.306675878
0.2 5.2 0.106065089 0.004242604 0.422019579
0.25 4.6 0.135538752 0.00542155 0.53929156
0.3 4.2 0.162585034 0.006503401 0.646905296
from the plot of T' vs alpha we get
T' = 0.1755*alpha + 0.0028
I = slope = 0.1755 kg m^2
Tf = 0.0028 Nm

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