What Is the Correct Translational Velocity of a Disk Rolling Down an Incline?

[alexito01]

Homework Statement

A disk is released from rest from the top of an incline. The bottom of the incline is a vertical distance h=15m below the top. The wheel rolls without slipping. The moment of inertia of the disk is given by I=1/2MR². What is the translational velocity of the disk at the bottom of the incline? (use g=9.8m/s2)

Homework Equations

PE= mgy and KE= 1/2mv^2 and ω=v/r

The Attempt at a Solution

I ended up with 17.1 m/s but it keeps telling me its wrong

 

[hage567]

Are you using rotational kinetic energy anywhere in your calculations? Notice that you were given the moment of inertia of a disk.

 

[visharad]

Ki = 0
Kf = translation kinetic energy + rotational kinetic energy = 1/2 Mv^2 + 1/2 I w^2
In the above, substitute I = 1/2 MR^2 and w = v/R to find Kf in terms of M and v.

Ui = Mgy
Uf = 0
Kf + Uf = Ki + Ui
Substitute values and solve for v.