# Homework Help: Uniform Ciruclar Motion (whirling cord)

1. Oct 29, 2007

### KatieLynn

1. The problem statement, all variables and given/known data

A 200g object is tied to the end of a cord whirled around on top of a horizontal and frictionless table top in a 2.4 diameter circle at 3 rev/s. Determine a) the velocity of the object b)t he acceleration of the object and c)t he tension in the cord.

2. Relevant equations

Vc = (2*pi*r)/T

Ac = (Vc^2)/R

Fnet = MA

Ft - Fg =MAc

3. The attempt at a solution

So for A) I got Vc = (2*pi*1.2)(1/3) which equals 22.3 m/s

Then for B) I got Ac=(22.6)^2/(1.2) which equals 425.6 m/s

Now heres where I get lost

I tried doing this but it doesn't come out right...

Fnet = MA

Ft - Fg =MAc

Ft - 196.2 = (20)(425.6)

which means Ft = 8708.2N, I know the answer should be Force of tension = 85.1N, I'm not sure what I'm doing wrong.

2. Oct 29, 2007

### Dick

200g=.2kg. Change to kg if you want the answer in newtons. Watch your units.

3. Oct 29, 2007

### PhanthomJay

you did fine in a and b. In c, however, you included the weight, but since the object is being twirled on a horizontal tabletop, the centripetal force and acceleration acts horizontally. The weight acts vertically.

4. Oct 29, 2007

### Dick

Yes, there's that too. I missed it.

5. Oct 30, 2007

### KatieLynn

Ahh that you so much. I, for some reason, was converting 200g to 20kg, now that I switched it to .2kg and took out the force normal I got the right answer. Thank youuu :)