Question #89cc5

Question

Question #89cc5

1 Answer

Impact of this question

1807 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-04T01:28:25

We know that the moment of inertia $I$ of a disk is related with its mass $M$ and radius $R$ as follows

$color(blue)(I=1/2MR^2.......................[1])$

Now $M=4/3piR^3drho$

where $m->"mass" , d->"thickness" and rho ->"density"$

$=>R=((3M)/(4pidrho))^(1/3)$

$=>R^2=((3M)/(4pidrho))^(2/3)$

Hence equation [1] becomes

$color(green)(I=1/2Mxx((3M)/(4pidrho))^(2/3).......................[2])$

Now if $M and d$ are remaining constant as per given condition then

$color(red)(Iprop1/rho^(2/3).......................[3])$

If the moment of inertia of first disk of density $rho_1=7.2"g/"cm^3$ be $I_1$ and the moment of inertia of 2nd disk of density $rho_2=8.9"g/"cm^3$ be $I_2$ then the ratio of moment of inertia of two disks is given by

$color(blue)(I_1/I_2=(rho_2/rho_1)^(2/3)=(8.9/7.2)^(2/3)~~1.15)$

Science

Math

Humanities

... and beyond

Question #89cc5

1 Answer

Related questions

Impact of this question