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

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SUMMARY

The correct translational velocity of a disk rolling down an incline is derived using the principles of conservation of energy and the moment of inertia. Given a vertical distance of h=15m and gravitational acceleration g=9.8m/s², the total kinetic energy at the bottom combines both translational and rotational components. The moment of inertia for a disk is I=1/2MR², leading to the equation Kf = 1/2Mv² + 1/2(1/2MR²)(v/R)². Solving this yields a translational velocity of approximately 17.15 m/s at the bottom of the incline.

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

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