A bright physics student purchases a weather vane for her father\'s garage. The
ID: 2014676 • Letter: A
Question
A bright physics student purchases a weather vane for her father's garage. The vane consists of a rooster sitting on top of an arrow. The vane is fixed to a vertical shaft of radius 0.0361 in and mass 1.22 kg that is free to turn in its roof mount as shown in the figure. The student sets up an experiment to measure the rotational inertia of the rooster and arrow. String wound about the shaft passes over a pulley and is connected to a 2.58 kg mass hanging over the edge of the roof. When the 2.58 kg mass is released, the student determines the time t that the mass takes to fall through a distance 0.545 m. The acceleration of gravity is 9.8 m/s . Find the rotational inertia I of the rooster and arrow, if t = 14.7 s . Answer in units of kg m2.Explanation / Answer
Consider the 2.58 kg mass:
It travels 0.545 m from rest in 14.7 s, x = vot + 0.5at2
0.545 = 0 + 0.5a(14.72)
a = 0.005044 m/s2
ΣF = mg - T = ma
(2.58 kg)(9.8 m/s2) - T = (2.58 kg)(0.005044 m/s2)
T = 25.27 N
The 2.58 kg moves 0.545 m, the shaft will rotate sweeping 0.545 m of circumference
the shaft will rotate θ=x/r = 0.545 m/0.0361 m = 15.1 rad
With 25.27 N force acting on the peripheral of the shaft, the shaft rotates by 15.1 rad in 14.7 s
θ = ω0t +0.5αt2
15.1 = 0 + 0.5α(14.72)
α = 0.140 rad/s2
Since τ = Iα = Fr
I = Fr/α = (25.27 N)(0.0361 m)/(0.140 rad/s2) = 6.53 kgm2.
The moment of inertia of the shaft and the rooster is 6.53 kgm2
The moment of inertia of the shaft (cylinder) = 0.5MR2 = 0.5(1.22 kg)(0.0361 m)2 = 0.000795 kgm2.
∴ The moment of inertia of the rooster is 6.53 - 0.000795 = 6.53 kgm2.
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