Long string wrapped around cylinder. Find cylinders angular speed?

[kerbyjonsonjr]

Homework Statement

A long string is wrapped around a 6.3-cm-diameter cylinder, initially at rest, that is free to rotate on an axle. The string is then pulled with a constant acceleration of 2.0 m/s² until 1.5 m of string has been unwound. If the string unwinds without slipping, what is the cylinder's angular speed, in rpm, at this time

Homework Equations

The Attempt at a Solution

I am not even sure how to really start this problem. I used C=2pi*r and got .1979m so I know that is the distance of each revolution. So then I did 1.5/.1979 and got 7.5788 total revolutions. I am not sure if I even need to do that but that is all I have come up with. I am not sure where to go from here.

 

[verty]

So then I did 1.5/.1979 and got 7.5788 total revolutions.

This is almost right. You want to relate the angular velocity of the cylinder to the velocity of the string. Once you have the velocity of the string in m/s, you can convert it to rad/s (and then rpm or whatever).

 

[kerbyjonsonjr]

This is almost right. You want to relate the angular velocity of the cylinder to the velocity of the string. Once you have the velocity of the string in m/s, you can convert it to rad/s (and then rpm or whatever).

Could I use x=1/2at² and then solve for t and get t= 1.22. Then do 7.5788 rev/1.22s and then convert to rpm?

 

[RTW69]

tangential acceleration=radius * angular acceleration therefore you can find angular acceleration

Torque=I*angular acceleration; I is for cylinder therefore you can find torque

But torque also equals Force*radus; setting the torques equal allows you to find force

Work=force*distance spring is pulled

but work is also= .5*I*w^2 therefore you can find w (angular velocity)

Change w from rad/sec to RPM's

 

[verty]

Could I use x=1/2at² and then solve for t and get t= 1.22. Then do 7.5788 rev/1.22s and then convert to rpm?

No, you want instantaneous velocity, not average velocity.