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Kinetic Energy and Moment of Inertia

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


A hoop with mass m= 10 kg, radius r=0.6 m, and moment of inertia I=mr^2 is rolling without slipping with a linear velocity of 8.5 m/s. What is the total kinetic energy (in J).


Homework Equations


K=1/2 m v^2
Kr=1/2 I ω^2


The Attempt at a Solution


The kinetic energy of the object is
K = 1/2 m v^2
K = 1/2 (10) (8.5) = 42.5 J

The rotational kinetic energy
Kr = 1/2 (10 * 0.6^2) (8.5/0.6)^2 = 361.25 J

Total kinetic energy
42.5 + 361.25 = 403.75 J

I am not sure about my solution since the answer I end up with is not included in the choices of the question.
 

Answers and Replies

  • #2
818
67
Have a look at this part again:

K = 1/2 m v^2
K = 1/2 (10) (8.5) = 42.5 J
Also, Kt = K + Kr = 1/2*m*v2 + 1/2*m*r2*(v/r)2 = ?
 

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