Rotational Inertial and Torque question

[skinnyabbey]

Can anyone help me out?

The top has a moment of inertia
of 0:0002 kg.m2 and is initially at rest. It is
free to rotate about a stationary axis AA'. A
string, wrapped around a peg along the axis
of the top, is pulled in such a manner as to
maintain a constant tension of 5.22 N in the string.
If the string does not slip while wound
around the peg, what is the angular speed
of the top after 82.1 cm of string has been
pulled off the peg? Answer in units of rad/s


I have a problem with this question. I don't know how to calculate the linear distance given to an angular distance. I found the acceleration first using Torque=(I*acc). But then I don't know what to do next after this. My acceleration turned out to be 26100 rad/s^2

 

[Chi Meson]

in order to check your torque, we knweed to know the radius of the peg.
edit:

aha! by reverse engineering I have discovered that you have used the force as the torque (unless the radius of the peg is 1 meter, which I doubt).

If the radius of the peg (that the string is wrapped around) is not given, then there is not enough information to solve this problem.
$$\tau = I \alpha$$
$$\tau = Fr$$
$$Fr = I \alpha$$

 

[hoseA]

Try square root of [(2*Tension*String Length)/(Moment of Inertia)] .

 

[hoseA]

also watch your units... (i.e. make sure length is in m NOT cm)

 

[skinnyabbey]

thanks.it worked.

 

[Chi Meson]

D'oh!

"i've made a huge mistake."