Question

A rolling (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.

Answer 1

Answer:

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 .