A spider with mass 0.70 g is hanging from rest on a 20 cm long thread of 0.0020
ID: 3893382 • Letter: A
Question
A spider with mass 0.70 g is hanging from rest on a 20 cm long thread of 0.0020 mm diameter silk. Spider silk density is 1290 kg/m^3.
a) If the small child holding the other end of the silk moves their end of the silk, how long will it take for the spider to feel it?
b) If the string is whirled in a circle at a rate of 2 revolutions per second, what is the minimum and maximum time for vibrational infromation from one side of the string to reach the other?
Show all steps, conversions and work please.
Explanation / Answer
cross sectional area = pi*(1*10^-6)^2 = 3.141*10^-12 m^2
mass per unit length mu = density*area
= 1290*3.141*10^-12
= 4.0526*10^-9 Kg/m
Wave velocity V = sqrt(T/mu) = sqrt(m*g/ mu)
= sqrt(0.0007*9.81 / 4.0526*10^-9)
= 1301.7 m/sec
Time taken = distance/speed = 0.2/1301.7 = 1.536*10^-4 sec = 0.153 ms
centipetal force = m*r*omega^2
= 0.0007*0.2*(2*pi*2)^2
= 0.0221 N
maximum tension = 0.0221 + mg = 0.0221 + (0.0007*9.81)
= 0.0289 N
V = sqrt(T/mu) =sqrt(0.0289/4.0526*10^-9) = 2673.89 m/sec
minimum time = 0.2/2673.89 = 0.0747 ms
minimum tension = 0.0221 - mg = 0.01523 N
V = sqrt(T/mu) =sqrt(0.01523/4.0526*10^-9) = 1938.76 m/sec
maximum time = 0.2/1938.76 = 0.103 ms
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