Question

A long string is wrapped around a 6.1-cm-diameter cylinder, initially at rest, that is free to rotate on an axle. The string is then pulled with a constant acceleration of 1.6 m/s2 until 1.3 m of string has been unwound. If the string unwinds without slipping, what is the cylinder's angular speed, in rpm, at this time?

Answer 1

v^2 / r is tangential/centripetal acceleration. NOT angular acceleration.
Anyhow, First you can find out how much time is spent pulling the string using this kinematics equation:
x = (1/2)* a * t^2 => t = sqrt[(2*x)/a]
then you can use the time to solve for the linear velocity of the string using this:
V = t * a
Then you can convert the linear velocity into angular velocity using this:
w = V / C where w = angular velocity and C = circumference = pi*Diameter
solving these equations you have:
t = sqrt[(2*x)/a] = sqrt[(2*1.3)/1.6] = 1.275 sec
V = t * a = 1.275 sec * 1.6 m/s^2 = 2.04 m/sec
w = V / C = V / pi*D = (2.04 m/sec) / (3.1416 * 0.061 m) = 10.65 revolution / sec
to convert to RPM multiply by (60 sec / min)
(10.65 revolution / sec)*(60 sec / min)=639 RPM