# Angular motion problem

#### BirdyNamNam115

Problem
A long string is wrapped around a 5.00-cm-diameter cylinder, initially at rest, that is free to rotate on an axle.
The string is then pulled with a constant acceleration of 1.50 m/s2 until 1.25 m of string has been unwound.
Assume the string unwinds without slipping.
(a) What is the angular displacement of the cylinder after 1.25 m of string has been unwound?
(b) What is the angular acceleration of the cylinder as the string is unwound?
(c) What is the final angular velocity of the cylinder, expressed in rpm?

Knowns
tangential acceleration(at) =1.25m/s
arc length(s)=1.25m
radius of the cylinder(r) = .025m

Attempt at solving the problem
a.)
Θ=s/r
Θ=1.25/.025

b.)
at=αr
1.25=α(.025)
1.25/.025=α

c.)
Using the equation wf=wi+2αΘ
wf=2αΘ
wf=2(60)(50)

This is where I am stuck. I'm not really quite sure if my reasoning that the tangential acceleration is 1.25m/s^2, or if my final angular velocity is reasonable at all. I'm also having trouble wrapping my head around the conversion to RPM. Thanks for any help in advance.

#### skeeter

(a) ok

(b) $a = 1.50 \, m/s^2$
you typed in 1.25, but arrived at the correct solution

(c) $\omega_f^2 = \omega_0^2 + 2\alpha \Delta \theta$

conversion of radians/sec to revolutions/min ...

$\dfrac{rad}{sec} \cdot \dfrac{1\, rev}{2\pi \, rad} \cdot \dfrac{60 \, sec}{1 \,min} = \dfrac{rev}{min}$

#### BirdyNamNam115

Wow I can't believe that I forgot that angular velocity was squared, thanks!