Moment of inertia about a perpendicular axis through its center?

[sp3sp2sp]

Homework Statement

A 15-cm-diameter CD has a mass of 21 g .
What is the CD's moment of inertia for rotation about a perpendicular axis through its center?

Homework Equations

I = (1/2)MR^2

3. The Attempt at a Solution
I = (1/2)(0.21g).((15/2)^2)
= 5.9*10^-4 kg*m^2

cant see what I am doing wrong but mastering physics says this is the wrong answer. thanks for any help

 

[stockzahn]

Apart from the mixed up units, you seem to use 0.21 g instead of 21 g. What's the book's answer?

 

[sp3sp2sp]

thanks i fixed the units but i don't have answer because its mastering physics. I've used 1 attempt so far and my answer is definitely wrong in some way

I = (1/2)(0.21kg).((0.15m/2)^2)

= 5.9v*v10^-4 kg*m^2

 

[stockzahn]

thanks i fixed the units but i don't have answer because its mastering physics. I've used 1 attempt so far and my answer is definitely wrong in some way

I = (1/2)(0.21kg).((0.15m/2)^2)

= 5.9v*v10^-4 kg*m^2

Again there seems to be a problem with the mass: 1000 g = 1 kg $\rightarrow$ 21 g = ...

 

[sp3sp2sp]

wow that was a dumb error. Ok so it should be

I = (1/2)(0.021kg).((0.15m/2)^2)

= 5.9v*v10^-5 kg*m^2

But mastering-p didnt say i have a "rounding error" or something..it just said its wrong. So is this answer correct though befiore I submit again? thanks

 

[stockzahn]

wow that was a dumb error. Ok so it should be

I = (1/2)(0.021kg).((0.15m/2)^2)

= 5.9v*v10^-5 kg*m^2

But mastering-p didnt say i have a "rounding error" or something..it just said its wrong. So is this answer correct though befiore I submit again? thanks

The formula seems to be correct to me, my calculator says: 5.90625 ⋅ 10^-5, but I don't know how you are expected to round or if there is something else to consider, but the formula yields the moment of inertia for a solid cylinder rotating around its vertical axis ...