A thin cylindrical shell and a solid cylinderhave the same mass and radius. The two arereleased side by side and roll down, withoutslipping, from the top of an inclined planethat is 1.3 m above the ground.Find the final linear velocity of the thincylindrical shell.The acceleration of gravity is9.8 m/s2

Respuesta :

Answer:

v = 5.05m/s

Explanation:

H = 1.3m

initial velocity = 0

final velocity = v = ?

g =9.8 m/s^2

we apply the conservation of energy; all potential energy is comletely converting to kinetic energy

[tex]mgh = \frac{mv^{2}}{2}[/tex]

mass is same; it cancels out

[tex]v =\sqrt{2gh} = \sqrt{2(9.81)(1.3)}[/tex]

v = 5.05m/s

Answer:

the final linear velocity is 3.56931 m/s

Explanation:

the solution is in the attached Word file

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