A marble is placed at the bottom of a semi-spherical bowl, as shown in the figure. The marble is then displaced from the bottom
of the bowl to a position about halfway from the top of the bowl. The marble is then released from rest such that the marble always remains in contact with the bowl. Students observe that the marble rolls back and forth as it oscillates about the bottom of the bowl. Which of the following statements best explains why the marble undergoes oscillatory motion?
a)The sides of the bowl become steeper at positions farther from the bottom of the bowl.b) The net force exerted on the marble always has a component directed toward the bottom of the bowl. c) The normal force exerted on the marble decreases with increasing distance from the bottom of the bowl. D ) The gravitational force exerted on the marble is constant in magnitude and direction.
a)The sides of the bowl become steeper at positions farther from the bottom of the bowl.b) The net force exerted on the marble always has a component directed toward the bottom of the bowl. c) The normal force exerted on the marble decreases with increasing distance from the bottom of the bowl. D ) The gravitational force exerted on the marble is constant in magnitude and direction.
1 answer:
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Answer:
v₁ = 37.5 cm / s
Explanation:
For this exercise we can use that angular and linear velocity are related
v = w r
in the case of the spool the angular velocity for the whole system is constant,
They indicate the linear velocity v₀ = 25.0 cm / s for a radius of r₀ = 1.00 cm,
w = v₀ /r₀
for the outside of the spool r₁ = 1.5 cm
w = v₁ / r₁1
since the angular velocity is the same we set the two expressions equal
v1 =
let's calculate
v₁ =
v₁ = 37.5 cm / s