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

Step-by-step 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(x^3)/(3) |^(3.4)_(3)`

`I=(M)/(L)[(3.4^3)/(3)-(3^3)/(3) ] \\\\I=(2)/((0.4*3))[(3.4^3)/(3)-(3^3)/(3) ]\\\\I=(2)/(1.2)(39.304-9)\\\\I=1.67*12.304\\\\I=20.51`

`a)I_1=mr^2\\\\I_1=2*3.2^2\\\\I_1=20.48kgm^2\\\\b)I_2=(1)/(3) mr^2\\\\=(1)/(3) *2*3.4^2\\\\I_2=7.71kgm^2\\\\c)I_3=(1)/(12) mr^2\\\\I_3=(1)/(12) *2*1.7^2\\\\I_3=0.461kg/m^2\\\\d)I_4=mr^2\\\\I_4=2*3.462\\\\I_4=23.12kg/m^2`