Recent content by phyllis

Thank you, I get it now!

sorry I meant .375

0.375

total KE = 0.8mv²

3/10

Homework Statement A spherical object with moment of inertia 0.6mr2 rolls without slipping down an incline. At the bottom of the incline. Homework Equations Rotational KE = ½Iω² The translational kinetic energy = ½mv² The Attempt at a Solution Rotational KE = ½Iω² = ½ * 0.6mr² (v/r)² =...

This is where I get lost, I understand that total KE is 1/2mv^2+.3mr^2

Rotational KE = ½Iω² In my problem moment of inertia 0.6mr^2 = 1/2 * 0.6mr^2 (v/r)^2 = 0.3mv^2

liner KE in terms of m and v would be 1/2mv^2, rotational KE in terms of 1/2Iω^2, and I know that v= ωr

HI I'm trying to work out the same problem but I don't get where the total kinetic energy = 0.285mv² + ½mv² = 0.785mv²