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A 4-lb ball B is traveling around in a circle of radius r1 = 3 ft with a speed (vB)1 = 6 ft>s. If the attached cord is pulled down through the hole with a constant speed vr = 2 ft>s, determine how much time is required for the ball to reach a speed of 12 ft>s. How far r2 is the ball from the hole when this occurs? Neglect friction and the size of the ball.

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jolis1796

Answer:

r₂ = 1.5 ft

t = 0.75 s

Explanation:

W = 4 lb    ⇒  m = W/g = 4 lb / 32.2 ft/s² = 0.12 slug

v₁ = 6 ft/s

r₁ = 3 ft

v₂ = 12 ft/s

r₂ = ?

t = ?

vr = 2 ft/s

Conserving angular momentum we have:

m*v₁*r₁ = m*v₂*r₂    ⇒     r₂ = v₁*r₁ / v₂

⇒     r₂ = (6 ft/s*3 ft) / (12 ft/s) = 1.5 ft

Now, we get the time as follows:

vr = d / t    ⇒   t = d / vr

where d is

d = Δr = r₁ - r₂ = 3 ft - 1.5 ft = 1.5 ft

finally

t = 1.5 ft / 2 ft/s

⇒ t = 0.75 s

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