Arolling (without slipping) hoop with a radius of 0.27 m and a mass of 1.80 kg climbs an incline. at the bottom of the incline, the speed of the hoop's center-of-mass is vi = 8.90 m/s. the incline angle is not needed in this problem. what is the angular speed of the hoop's rotation? enter a number rad/s (5 attempts remaining)input 1 status what is the hoop's translational kinetic energy at the bottom of the incline? enter a number j (5 attempts remaining)input 2 status what is the hoop's rotational kinetic energy at the bottom of the incline ? the hoop's moment of inertia is mr². (since its mass is concentrated on the edge). enter a number j (5 attempts remaining)input 3 status find the maximum (vertical) height h the hoop reaches on the incline.

Explanation:

Angular speed of hoop ω = v / r

= 8.90 / .27

= 32.96 rad / s

Translational kinetic energy = 1/2 mv²

= .5 x 1.8 x 8.9²

= 71.29 J

Rotational kinetic energy = 1/2 Iω²

= 1/2 mR²x ω²

= 1/2 mv²

= 71.29 J

Total kinetic energy

= 2 x 71.29

= 142.58 J

This energy will be used to attain height

If h be the height attained

mgh = 142.58

h = 142.58 / mg

= 142.58 / 1.8 x 9.8

= 8.08 m .

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