Moment of inertia

1. Aug 14, 2004

mr_joshua

This problem is giving me a headache.

A disk is free to rotate about its center(frictionless) in a vertical axis. A mass m=30grams, pulls down on the disk by means of a light string that is wrapped around the disk. If the radius is 20 cm, Find "I" if the mass falls 1 meter in 2 sec.

We used rT=I(a/r)

so you get .2(.03*9.8)=I(.25/.2)

found velocity using v=s/t

and acceleration using a=v/t

Is this correct? the value for inertia I get is .045 kg m/sec^2

the book has .0222 kg m/s^2
Help

2. Aug 14, 2004

Gokul43201

Staff Emeritus
This is a little confusing ? Is the axis of the disc vertical or horizontal. If it really is vertical, I can't see how exactly the mass hangs off the edge.

If it is horizontal, then the approach you use is correct except for your calculation of the acceleration (a) of the mass.

You say, v = s/t and a = v/t . This is incorrect. The first expression gives you the average velocity while the second relies on the final velocity.

What is the correct equation for finding 'a' given s, t ? Of course, the initial velocity, u = 0. If you make this correction (using the right equation) you will find that you get I = 0.022 kg.m^2 (not kg.m/s^2, which is the unit of force)

3. Aug 15, 2004

mr_joshua

using wrong formula for accel and velocity?

I am using v=s/t to get .5 m/s

then using vf^2=Vi^2 + 2*a*s

to get .125 m/s^2....???
using this we divide by the radius of .2 to get angular accel. of .625rad/s^2

I am not seeing something.

Thanks again for all your help

4. Aug 15, 2004

mr_joshua

oops again

sorry, now I beleive that I am using the right formula,,,

s=vi*t+1/2a*t^2

to get a=.5
Ang a= 2.5

then I get I=.0235????

5. Aug 15, 2004

Gokul43201

Staff Emeritus
Yup, that's it. But you're right...it's 0.0235. Now I wonder why that little difference is there ? I'm not sure....but it looks okay, as long as the axis IS HORIZONTAL !