A 4.00-m length of light nylon cord is wound around a uniform cylindrical spool
ID: 1444594 • Letter: A
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 2.66 m/s2.
(a) How much work has been done on the spool when it reaches an angular speed of 7.30 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
a) Work Done on the spool = I ^2
where I is moment of inertia = = MR^2/2
I = 0.5 * 1 * 0.5 * 0.5
I = 0.125 kgm^2
Work done W = I * angular velocity
Work = 0.5 * 0.125* 7.3* 7.3
W = 3.33 Joules
--------------------
b) angular accleration = a/R
Alpha = 2.66/0.5
Alpha = 5.32 rad/s^2
time t= / = 7.3/5.32
t = 1.372 sec
----------------------------------------
c) angular displacemnt = 0.5 *A * t^2
theta = 0.5 * 5.32 * 1.372 * 1.372
theta = 5 radians
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.