Let a thin rod of length a have density distribution function p(x)=10e^(-0.1x),

Question

Let a thin rod of length a have density distribution function p(x)=10e^(-0.1x), where x is measured in cm and p in grams per centimeter. (a) If the mass of the rod is 30g, what is the value of a? (b) For the 30g rod, will the center of mass lie at its midpoint, to the left of the midpoint, or to he right of the midpoint? Why? (c) For the 30g rod, find the center of mass, and compare your prediction in (b). (d) At what value of x should the 30g rod be cut in order to form two pieces of equal mass?

Explanation / Answer

a)   Integral from 0 to a (p(x) dx) = 30 g
integral from 0 to a (10*e^(-0.1x) dx) = 30 g
-100*e^(-0.1*x) from 0 to a = 30 g
-100*e^(-0.1*a) + 100*e^0 = 30 g
-100*e^(-0.1*a) + 100 = 30 g
-100*e^(-0.1*a) = -70
e^(-.1*a) = 0.7
-.1*a = ln(0.7)
-.1*a = -0.357
a = 3.57 cm

b) So, the density decreases with increasing distance, meaning more of the mass is closer to x=0. So, the center of mass will be to the left of the midpoint.

c) The center of mass is is thus
[0,a] 10xe^(-0.1x) dx = [(100x+1000)*e0.1x] (limit 0 to a) = (-949.6 + 1000) = 50.4 cm

d)  split the integral and find k where

[0,k] 10e^(-0.1x) dx = [k,a] 10e^(-0.1x) dx

Now solve this!!!!

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