Angular Velocity of Disk After 0.85 Revolutions

[dalitwil]

Q: A disk of radius 2.40 cm and mass 1 kg is pulled by a string wrapped around its circumference with a constant force of 0.36 Newtons. What is the angular velocity of the disk to three decimal places after it has been turned through 0.85 of a revolution?

So I started by obtaining my degrees from the 0.85 revolutions (S=rθ). This equals 2.225 degrees.

Next I used ∆x=R∆θ
(2.40)*(2.225)
=5.340 radians

W=F∆x
(0.36)*(5.340)
=1.9224

Using conservation of work (W=∆K):
K=.5Iω^2 (where I of a disk=.5MR^2)

W=1.9924=.5(.5*1*2.4^2)ω^2
ω=1.15542

WRONG. I can't figure out where I am going wrong, please help!

 

[Crosson]

You have some pretty heavy duty mistakes there.

First, do not use degrees at all whatsoever.

∆x=R∆θ At this point, you use degrees for ∆θ (wrong, should be radians) and solve for ∆x in radians (wrong, should be meters).

The answer with Work and Kinetic energy looks correct, but the reason they don't match is that you didn't convert ∆x correctly.

 

[charng]

I would first check the units of the given values to ensure that they are all in the correct units (e.g. mass in kilograms, force in Newtons, radius in meters). Then, I would double check my calculations to make sure they are accurate. It is possible that a small error was made during the calculations, leading to the incorrect answer. I would also consider the possibility of any external factors that may have affected the experiment, such as friction or air resistance. If necessary, I would repeat the experiment or consult with a colleague to confirm the results.