Calculating Rotational Kinetic Energy of a Baton

[sp3sp2sp]

Homework Statement

A drum major twirls a 94-cm-long, 500 g baton about its center of mass at 150 rpm. What is the baton's rotational kinetic energy?

Homework Equations

KE_rot = 0.5 I w^2

The Attempt at a Solution

w = (150rpm*2pi)/60 = 15.708rad/s

I = (1/12)*0.5kg * 0.94m = 0.039167kg*m^2

KE_rot = 0.5 I w^2

KE_rot = 0.5(0.039167)(15.708rad/s)^2
= 4.8J

says wrong answer.
Cant figure out what I am doing wrong
thanks for any help

 

[Mr rabbit]

Take a look at the moment of inertia: you do kg times m and you obtain kg m^2.

 

[sp3sp2sp]

thanks..ok forgot to square so I = (1/12)MR^2
I = (1/12) 0.5kg ((0.94m)^2) = 0.03682kg*m^2

so KE_rot = 0.5(0.03682)(15.708rad/s)^2 = 4.5J

(i just want to confirm correct before I submit answer again)
thanks for the help

 

[Mr rabbit]

I think it is correct.