A 4.00 m length of light nylon cord is wound around a uniformcylindrical spool o

Question

A 4.00 m length of light nylon cord is wound around a uniformcylindrical spool of radius 0.500 m and mass 1.00 kg. The spool ismounted on a frictionless axle and is initially at rest. The cordis pulled from the spool with a constant acceleration of magnitude2.60 m/s2. (a) How much work has been done on the spoolwhen it reaches an angular speed of 7.20rad/s?
1 J

(b) Assuming that there is enough cord on the spool, how long doesit take the spool to reach this angular speed?
2 s

(c) What length of cord is pulled from the spool when the angularspeed is reached?
3 m (a) How much work has been done on the spoolwhen it reaches an angular speed of 7.20rad/s?
1 J

(b) Assuming that there is enough cord on the spool, how long doesit take the spool to reach this angular speed?
2 s

(c) What length of cord is pulled from the spool when the angularspeed is reached?
3 m

Explanation / Answer

Let's examine energy. The only thing changed is kinetic energy sothe work done, W is: W = K = Kf - Ki =(1/2)If2 -(1/2)Ii2 =(1/2)I(f2-i2) The spool can be modeled as a cylinder, whose moment of inertia is(1/2)MR2 So a) W = (1/2)(1/2)MR2(f2-i2) =(1/4)MR2(f2-i2) = (1/4)(1.00 kg)(0.5m)2( (8.10 rad/s)^2 - 0) = 4.1 J b) Since a = v / t, t = v / a. At the pointwhere the spool unwinds, a= 2.95 m/s2. The speed at thisspeed is given by v = r, where r is the radius of thespool. So v = r = (8.10 rad/s)(0.5 m) = 4.05 m/s and t = v / a = (4.05 m/s) / (2.95 m/s2) = 1.373 s c) First we find the total angle that it rotated, given by: = it + (1/2)t2 But i = 0, i = 0, and = a/ r So f = (1/2)t2 =(1/2)(a/r)t2 = (1/2)( (2.95 m/s2) / (0.5m))(1.373 s)2 = 5.56 rads And the total distance is d = r = (5.56 rads)(0.5 m) = 2.78m

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