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arlik [135]
2 years ago
10

A solid, uniform sphere with a mass of 2.5 kg rolls without slipping down an incline plane starting from rest at a vertical heig

ht of 19 m. If the sphere has a radius of 0.60 m, what is the angular speed of the sphere at the bottom of the incline plane
Physics
1 answer:
kolezko [41]2 years ago
8 0

Answer:

1/2 m v^2 + 1/2 I ω^2 = m g h       conservation of energy

I = 2/5 m R^2     inertia of solid sphere

1/2 m v^2 + 1/5 m ω^2 R^2 = m g h

1/2 v^2 + 1/5 v^2 = g h

v^2 = 10 g h / 7 = 1.43 * 9.80 * 19 m^2/s^2 = 266 m^2/s^2

v = 16.3 m/s

v = R ω

ω = 16.3 / .6 = 27.2 / sec

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A ball is thrown horizontally from the top of a 60 m building and lands 100 m from the base of the building. How long is the bal
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A ball is thrown horizontally

That means the vertical component of the initial velocity u_{y}=0

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That means the horizontal displacement x = 100 m

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Divide both sides by 3.5

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Substitute these values in the rule

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→ v_{y} = -34.3 m/s

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The final velocity v is the resultant vector of  v_{x} and v_{y}

→ Its magnetude is v=\sqrt{(v_{x})^{2}+(v_{y})^{2}}

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→ Its direction tan^{-1}\frac{-34.3}{28.6}=-50.18

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