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A solid sphere of radius 27 m is positioned at the top of an incline that makes a 27 degree angle with the horizontal. This initial position of the sphere is a vertical distance of 2.3 m above the incline. The acceleration of gravity is 9.8 m/s squared. The moment of inertia of a sphere is 2/5 mr squared. Calculate the speed of the sphere when it reaches the bottom of the incline in the case where it rolls whithout slipping. Also calculate the speed of the sphere when it reaches the bottom of the incline in the case where it slips frictionlessly without rolling ?

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msakhile

Answer:

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Step-by-step explanation:

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The speed of the sphere when it reaches the bottom of the incline is; 5.675 m/s

What is the speed of the sphere object?

From law of conservation of Energy, we know that;

P.E at the top = Translational kinetic energy at bottom - Rotational kinetic energy at bottom

The formula is;

mgh = ¹/₂mv² + ¹/₂Iω²

where I is moment of inertia = ²/₅mr² and ω is angular velocity = v/r

Thus;

mgh = ¹/₂mv² + ¹/₂(²/₅mr²)(v²/r²)

m will cancel out and r² will also cancel out to give;

gh = ¹/₂v² + (¹/₅v²)

gh = 0.7v²

v = √(gh/0.7)

v = √(9.8 * 2.3/0.7)

v = 5.675 m/s

Read more about Speed of Object at; https://brainly.com/question/10962624

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