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Alecsey [184]
3 years ago
13

A thin cylindrical shell and a solid cylinder have the same mass and radius. The two are released side by side and roll down, wi

thout 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.
Physics
1 answer:
Vitek1552 [10]3 years ago
8 0

Answer:

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

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Then note that,

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Then, the x component of the force is

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