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 2.58 m/s2.

(a) How much work has been done on the spool when it reaches an angular speed of 7.85 rad/s?

(b) How long does it take the spool to reach this angular speed?

(c) How much cord is left on the spool when it reaches this angular speed?

Explanation / Answer

1) Force acting on spool = ma = 1(2.9) =2.9N let v be the final velocity of the spool, u be initial velocity of the spool, t be the time taken, ? be the angular speed of the spool v = r ? v = (0.5)(6.3) = 3.15 m/s v = u + at 3.15 = 0 + 2.9t t = 1.086s Time taken to accelerate from rest to an angular velocity of 6.3 rad/s = 1.086s Therefore, total work done = Ft = 2.9 * 1.086 = 3.15J 2) Time taken = 1.086s (as shown) 3) Let the length of cord be s (v^2) = (u^2) + 2as (3.15^2) = 0 + 2(2.9)s s = 1.71m

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.