A 4.00 m length of light nylon cord is wound around a uniform cylindrical spool

Question

A 4.00 m length of light nylon cord is wound around a uniform cylindrical spool of radius 0.500 m and mass 1.00 kg. The spool is mounted on a frictionless axle and is initially at rest. The cord is pulled from the spool with a constant acceleration of magnitude 2.80 m/s2.

A 4.00 m length of light nylon cord is wound around a unifo cylindrical spool of r 00 kg he spool a constant acceleration of magnitude 2.80 m/s2. (a) How much work has been done on the spool when it reaches an angular speed of 6.70 rads? (b) Assuming that there is enough cord on the spool, how long does it take the spool to reach this angular speed? (c) What length of cord is pulled from the spool when the angular speed is reached? e cord is pulle d from the spool with

Explanation / Answer

(a) Ok, first we need to find the tension of cord. Using rotational dynamics,

= a / R = 2.8 / 0.500 = 5.6 rad/s2, where a = linear acceleration

I = moment of inertia of a hollow cylinder (uniform cylindrical spool) = 0.5 M*R2

Tension in the nilon cord T, T*R = I* => T*0.500 = 0.5 * 1.00*(0.500)2*5.6 => T = 1.4 N

To find the work done by the cord on the spool, we need to find the angle , made by the cylinder.

Wi = 0 rad/s, Wf = 6.7 rad/s, alpha = 5.6 rad/s2

=> 6.72 -0= 2*5.6* => =4.01 (rad)

Work done W = T*(*R) = 1.4*(4.01*0.500) => W = 2.81 J

b) wf = t = at/R => t = Rw/a = 0.500*6.7/2.8 = 1.20 s

c) = (1/2) t2 = (a/2R)t2 => R* = (a/2)t2 => L (2.8/2) * (1.2)^2 = 2.02 m

L = R* = the arclength subtended by angle , so it will be the total length of cord unwound after spool rotates radians.

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