Answered

A 1.5 kg bicycle tire of radius 0.33 m starts from rest and rolls down from the top of a hill that is 14.8 m high. What is the translational speed (forward speed) of the tire when it reaches the bottom of the hill? Assume that the tire is a hoop with I=mr2 .



Answer :

MathPhys

Answer:

12.0 m/s

Explanation:

Energy is conserved. The initial potential energy is equal to the sum of the final translational and rotational kinetic energies. Translational kinetic energy is half the mass times the square of the speed, and rotational kinetic energy is half the moment of inertia times the square of the angular speed.

PE = KE + RE

mgh = ½ mv² + ½ Iω²

For a hoop, I = mr². For rolling without slipping, ω = v/r.

mgh = ½ mv² + ½ (mr²) (v/r)²

mgh = ½ mv² + ½ mv²

mgh = mv²

v² = gh

Plugging in values:

v² = (9.8 m/s²) (14.8 m)

v = 12.0 m/s

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