A thin cylindrical shell and a solid cylinder have the same mass and radius. The two are released side by side and roll down, without slipping, from the top of an inclined plane that is 1.8 m above the ground. Find the final linear velocity of the thin cylindrical shell. The acceleration of gravity is 9.8 m/s 2 . Answer in units of m/s.

Explanation:

Let the velocity be v

Total energy at the bottom

= rotational + linear kinetic energy

= 1/2 Iω² + 1/2 mv² ( I moment of inertia of shell  = mr² )

= 1/2 mr²ω² + 1/2 mv² ( v = ω r )

= 1/2 mv² +1/2 mv²

= mv²

mv² = mgh ( conservation of energy )

v² = gh

v = √gh

= √9.8 x 1.8

= 4.2 m /s

15.3 cos 65 degrees trig

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