If the 250-lb block is released from rest when the spring is unstretched, determine the velocity of the block after it has descended 5 ft. the drum has a weight of 50 lb and a radius of gyration of k0 = .5 ft about its center of mass 0

1080 rpm

Explanation:

The drum will follow a constant acceleration for a given time and then maintain its speed.

The constant acceleration will follow:

w(t) = w0 + γ * t

Where

w0 = initial angular speed (zero in this case)

γ = angular acceleration

The angular acceleration will depend on the torque applied and the inertia moment of the drum

T = γ * J

The block will apply a torque of:

T = P * r

T = 250 * 0.5 = 125 lb*ft

The drum weights 50lb, so it has a mass of

m = P/g = 50/32.2 = 1.55 lb*s^2/ft

The drum has a moment of inertia of

J = 1/2 * m * r^2

J = 1/2 * 1.55 * 0.5^2 = 0.194 lb*s^2*ft

(remember that lb is a unit of force not mass, but lb*s^2/ft is an indirect unit of mass)

So, the angular acceleration is

γ = T/J

γ = 125/0.194 = 644 rad/s^2

Now we need to know how long will it take the block to fall 5 ft.

Since wee know the drum has radius r = 0.5 f, we need to find how much it needs to spin to release 5 ft of rope.

The circumference is 2π*r = 3.14 feet, it will take it 5/3.14 = 1.59 turns to release 5 feet of rope. 1.59 turns is 2π*1.59 = 10 radians.

The equation for the rotation is:

a(t) = a0 + w0 * t + 1/2 * γ * t^2

a0 and w0 are zero

10 = 1/2 * 644 * t^2

t^2 = 0.031

t = 0.176 s

Then, the angular speed is:

w(0.176) = 644 * 0.176 = 113.3 rad/s

Which is 113.3/2π = 18 turns per second or

18 * 60 = 1080 rpm

do you have any options?

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