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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 the ball's speed at the instant r2 = 2 ft. how much work has to be done to pull down the cord? neglect friction and the size of the ball

2 Answers

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Final answer:

To determine the ball's speed at the instant r2 = 2 ft, we can use the principle of conservation of angular momentum. The ball's speed at r2 = 2 ft is 6 ft/s. The work done to pull down the cord is zero.

Step-by-step explanation:

To determine the ball's speed at the instant r2 = 2 ft, we can use the principle of conservation of angular momentum. Angular momentum is given by the equation L = Iω, where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity. Since the moment of inertia is constant, we can write the equation as I1ω1 = I2ω2, where I1 and ω1 are the initial moment of inertia and angular velocity, and I2 and ω2 are the final moment of inertia and angular velocity.



Since the ball moves along a circle of radius r1 = 3 ft and speed vb1 = 6 ft/s, we can find the initial angular velocity ω1 using the equation vb1 = r1ω1. Plugging in the values, we get ω1 = 2 rad/s. To find the final angular velocity ω2, we can use the equation vb2 = r2ω2. Plugging in r2 = 2 ft and solving for ω2, we get ω2 = 3 rad/s.



To determine the ball's speed at r2 = 2 ft, we can use the equation v = rω, where v is the linear velocity, r is the radius, and ω is the angular velocity. Plugging in the values, we get v = 6 ft/s.



To determine the work done to pull down the cord, we can use the equation W = Fd, where W is the work, F is the force, and d is the distance. Since the force is constant, we can write the equation as W = FΔy, where Δy is the change in height. In this case, the cord is pulled down with a constant speed vr = 2 ft/s.



Since the work done is equal to the force times the displacement, we can calculate the work done by multiplying the force by the vertical distance the cord is pulled down. However, since the size of the ball is neglected, the work done to pull down the cord is zero.

User Oknoorap
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Position #1:
radius, r₁ = 3 ft
Tangential speed, v₁ = 6 ft/s

By definition, the angular speed is
ω₁ = v₁/r₁ = (3 ft/s) / (3 ft) = 1 rad/s

Position #2:
Radius, r₂ = 2 ft

By definition, the moment of inertia in positions 1 and 2 are respectively
I₁ = (4 lb)*(3 ft)² = 36 lb-ft²
I₂ = (4 lb)*(2 ft)² = 16 lb-ft²

Because momentum is conserved,
I₁ω₁ = I₂ω₂
Therefore the angular velocity in position 2 is
ω₂ = (I₁/I₂)ω₁
= (36/16)*1 = 2.25 rad/s
The tangential velocity in position 2 is
v₂ = r₂ω₂ = (2 ft)*(225 rad/s) = 4.5 ft/s

At each position, there is an outward centripetal force.
In position 1, the centripetal force is
F₁ = m*(v²/r₂) = (4)*(6²/3) = 48 lbf
In position 2, the centripetal force is
F₂ = (4)*(4.5²/2) = 40.5 lbf

The radius diminishes at a rate of 2 ft/s.
Therefore the force versus distance curve is as shown below.

The work done is the area under the curve, and it is
W = (1/2)*(48.0+40.5 ft)*(3-2 ft) = 44.25 ft-lb

Answer: 44.25 ft-lb


A 4-lb ball b is traveling around in a circle of radius r1 = 3 ft with a speed (vb-example-1