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Translational velocity help

  • Thread starter alexito01
  • Start date
  • #1
5
0

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
 

Answers and Replies

  • #2
hage567
Homework Helper
1,509
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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.
 
  • #3
54
0
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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