# Mistake in Dynamics Circular Problem's Solution

1. May 22, 2014

### alingy1

I do not understand the solutions;

Here is how I find length of string:
5m/s*1.2s/rev=6m/rev=circumference=2piR
R=0.95m.

For the force:
T=mv^2/r= 0.350 kg *5^2/0.95= 9.21N They got 9.16N?

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2. May 22, 2014

### TSny

The numbers in the problem are stated to 3 significant figures. So, make sure you keep at least 3 significant figures in your intermediate calculations. What do you get for R to 3 significant figures?

3. May 22, 2014

### alingy1

Well the answer I got is 0.95m which is almost twice as big as the radius they found.
I keep all my sig figs on my calculator.

4. May 22, 2014

### alingy1

I don't know what is going on! :O

5. May 22, 2014

### TSny

You got R = .95 m. That has only 2 significant figures. What do you get to 3 significant figures?

When you calculated T, it looks like you used r = .95 m. Again, only 2 significant figures for r. What do you get for T if you use a value for r that is accurate to 3 significant figures?

6. May 22, 2014

### TSny

OK, I did not see the answer they had for R. Their answer is wrong, yours is right. But you should give it to 3 significant figures.

7. May 22, 2014

### alingy1

R=0.955m.
T=9.16N. You were right! Ok. This was my bad.

8. May 22, 2014

### alingy1

Perfect! Everything works out great!

9. May 22, 2014

### TSny

Good!

10. May 22, 2014

### ChrisVer

their radius is mistyped- probably 4 was meant to be a 9....
as for the rest, follow TSny's post