Question

Imagine a metal rod 0.4 m long with a mass of 2 kg. You attach the rod at one end by a lightweight 3.0-m-long cord and twirl the rod around your head. Which expression below gives the best estimate of the moment of inertia of the rod-cord system?
a) I=mr^2, where m = 2kg and r = 3.2m
b) I=1/3mr^2, where m = 2kg and r = 3.4
c) I=1/12mr^2, where m = 2kg and r = 1.7m
d) I=mr^2, where m= 2kg and r = 3.4m

Answer 1

Answer:

Explanation:

Given that

a metal rod 0.4 m long

mass of 2 kg

lightweight 3.0-m-long cord

a) I=mr^2, where m = 2kg and r = 3.2m

b) I=1/3mr^2, where m = 2kg and r = 3.4

c) I=1/12mr^2, where m = 2kg and r = 1.7m

d) I=mr^2, where m= 2kg and r = 3.4m

$I=\int\limits^{3.4}_3 \lambda dx \,x^2\\\\=\lambda\frac{x^3}{3} |^{3.4}_{3}$

$I=\frac{M}{L}[\frac{3.4^3}{3}-\frac{3^3}{3} ] \\\\I=\frac{2}{(0.4\times3)}[\frac{3.4^3}{3}-\frac{3^3}{3} ]\\\\I=\frac{2}{1.2}(39.304-9)\\\\I=1.67\times12.304\\\\I=20.51$

$a)I_1=mr^2\\\\I_1=2\times3.2^2\\\\I_1=20.48kgm^2\\\\b)I_2=\frac{1}{3} mr^2\\\\=\frac{1}{3} \times2\times3.4^2\\\\I_2=7.71kgm^2\\\\c)I_3=\frac{1}{12} mr^2\\\\I_3=\frac{1}{12} \times2\times1.7^2\\\\I_3=0.461kg/m^2\\\\d)I_4=mr^2\\\\I_4=2\times3.462\\\\I_4=23.12kg/m^2$