strahler19
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If a steel, thin-shelled wheel of radius r and mass M is moving along the road at 2 m/s, what is the total kinetic energy?

1) 2M
2) 3M
3) 1M
4) 5M
5) 4M

Answer :

The moment of inertia for rolling motion is
I = Mr²

The linear velocity is v = 2 m/s. The rotational velocity is
ω = 2/r 

The total KE (kinetic energy) is
KE = (1/2)Mv² + (1/2) I ω²
     = (1/2)(M)(4) + (1/2)(Mr²)(4/r²)
    = 2M + 2M
    = 4M

Answer:  4M

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