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Find kinetic energy from potential energy and moment of inertia?

  • Thread starter cc2hende
  • Start date
  • #1
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Homework Statement



A solid cylinder of mass 0.600kg is released from rest and allowed to roll down a board that has an incline of 30degrees to the horizontal floor. If the solid cylinder is released 0.60m above the floor, what is the cylinder's linear speed when it reaches the lowest end of the board? Take into account translational and rotational kinetic energy. Assume no sliding of the cylinder/no friction)

Homework Equations



Rotational KE= (1/2)*I*ω^2
Translational KE= (1/2)*m*v^2
PE1 + KE1 = PE2 + KE2

The Attempt at a Solution



PE1 + KE1 = PE2 + KE2
mgh + 0 = 0 +1/2 mv^2
mgh=1/2mv^2
√2gh=v
√2(9.8m/s2)(0.60m)
=3.43 m/s

I don't know if this is right though because the question throws me off when it says to take translational and rotational KE into account.
 

Answers and Replies

  • #2
1,197
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You have forgotten the rotational KE.
 
  • #3
1,197
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Because the object is rolling, that means it has translation as well as rotation. Therefore the potential energy is shared between two types of kinetic energy. What you have done only applies to a block sliding down a slope with no friction. You must account for the rotational KE. You have the formula for it. Omega can be related to linear velocity.....
 

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